Defining '999e999' value without using char type in C++
up vote
4
down vote
favorite
Is it possible to define 999e999
value without using the char
type?
I've tried defining it even with unsigned long long
, but the compiler keeps giving me constant too big
error.
Thanks in advance.
c++
add a comment |
up vote
4
down vote
favorite
Is it possible to define 999e999
value without using the char
type?
I've tried defining it even with unsigned long long
, but the compiler keeps giving me constant too big
error.
Thanks in advance.
c++
2
You may be interested in GMP.
– Jesper Juhl
Nov 11 at 18:48
2
Yes that's the way to go hypnokat, just like @JesperJuhl mentioned. However, I suggest you make sure first that you really need to use that big numbers!
– gsamaras
Nov 11 at 18:52
1
based on what you want to do with your number, you may be able to use opensslBN
functions too.
– Afshin
Nov 11 at 19:15
1
Out of curiosity, why would one ever need such a big number? Even if you think that every particle in the universe is itself a universe with the same number of particles, you'll have about 10^172 particles in total. Even if you wanted to count the number of particles that can fit in these universes "space", that number would still be huge! Do you need unit precision on such a... supercalifragilistic number?
– L.C.
Nov 11 at 19:28
It is just a random experiment I thought of, I'm a programming newbie. Still, thanks to that random thought, even thought I can't do that in C++ alone, I now know about the existance of GMP, so I'm still walking out with something new :)
– hypnokat
Nov 11 at 19:35
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Is it possible to define 999e999
value without using the char
type?
I've tried defining it even with unsigned long long
, but the compiler keeps giving me constant too big
error.
Thanks in advance.
c++
Is it possible to define 999e999
value without using the char
type?
I've tried defining it even with unsigned long long
, but the compiler keeps giving me constant too big
error.
Thanks in advance.
c++
c++
asked Nov 11 at 18:45
hypnokat
254
254
2
You may be interested in GMP.
– Jesper Juhl
Nov 11 at 18:48
2
Yes that's the way to go hypnokat, just like @JesperJuhl mentioned. However, I suggest you make sure first that you really need to use that big numbers!
– gsamaras
Nov 11 at 18:52
1
based on what you want to do with your number, you may be able to use opensslBN
functions too.
– Afshin
Nov 11 at 19:15
1
Out of curiosity, why would one ever need such a big number? Even if you think that every particle in the universe is itself a universe with the same number of particles, you'll have about 10^172 particles in total. Even if you wanted to count the number of particles that can fit in these universes "space", that number would still be huge! Do you need unit precision on such a... supercalifragilistic number?
– L.C.
Nov 11 at 19:28
It is just a random experiment I thought of, I'm a programming newbie. Still, thanks to that random thought, even thought I can't do that in C++ alone, I now know about the existance of GMP, so I'm still walking out with something new :)
– hypnokat
Nov 11 at 19:35
add a comment |
2
You may be interested in GMP.
– Jesper Juhl
Nov 11 at 18:48
2
Yes that's the way to go hypnokat, just like @JesperJuhl mentioned. However, I suggest you make sure first that you really need to use that big numbers!
– gsamaras
Nov 11 at 18:52
1
based on what you want to do with your number, you may be able to use opensslBN
functions too.
– Afshin
Nov 11 at 19:15
1
Out of curiosity, why would one ever need such a big number? Even if you think that every particle in the universe is itself a universe with the same number of particles, you'll have about 10^172 particles in total. Even if you wanted to count the number of particles that can fit in these universes "space", that number would still be huge! Do you need unit precision on such a... supercalifragilistic number?
– L.C.
Nov 11 at 19:28
It is just a random experiment I thought of, I'm a programming newbie. Still, thanks to that random thought, even thought I can't do that in C++ alone, I now know about the existance of GMP, so I'm still walking out with something new :)
– hypnokat
Nov 11 at 19:35
2
2
You may be interested in GMP.
– Jesper Juhl
Nov 11 at 18:48
You may be interested in GMP.
– Jesper Juhl
Nov 11 at 18:48
2
2
Yes that's the way to go hypnokat, just like @JesperJuhl mentioned. However, I suggest you make sure first that you really need to use that big numbers!
– gsamaras
Nov 11 at 18:52
Yes that's the way to go hypnokat, just like @JesperJuhl mentioned. However, I suggest you make sure first that you really need to use that big numbers!
– gsamaras
Nov 11 at 18:52
1
1
based on what you want to do with your number, you may be able to use openssl
BN
functions too.– Afshin
Nov 11 at 19:15
based on what you want to do with your number, you may be able to use openssl
BN
functions too.– Afshin
Nov 11 at 19:15
1
1
Out of curiosity, why would one ever need such a big number? Even if you think that every particle in the universe is itself a universe with the same number of particles, you'll have about 10^172 particles in total. Even if you wanted to count the number of particles that can fit in these universes "space", that number would still be huge! Do you need unit precision on such a... supercalifragilistic number?
– L.C.
Nov 11 at 19:28
Out of curiosity, why would one ever need such a big number? Even if you think that every particle in the universe is itself a universe with the same number of particles, you'll have about 10^172 particles in total. Even if you wanted to count the number of particles that can fit in these universes "space", that number would still be huge! Do you need unit precision on such a... supercalifragilistic number?
– L.C.
Nov 11 at 19:28
It is just a random experiment I thought of, I'm a programming newbie. Still, thanks to that random thought, even thought I can't do that in C++ alone, I now know about the existance of GMP, so I'm still walking out with something new :)
– hypnokat
Nov 11 at 19:35
It is just a random experiment I thought of, I'm a programming newbie. Still, thanks to that random thought, even thought I can't do that in C++ alone, I now know about the existance of GMP, so I'm still walking out with something new :)
– hypnokat
Nov 11 at 19:35
add a comment |
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
Is it possible to define
999e999
value without using the char type?
No, that's not possible using intrinsic c++ data types. That's a way to big number that could be held in either a unsigned long long
type in c++.
A long double
type would enable you to use 10 based exponents as large as you want, for modern FPU architectures.
What can be achieved with your current CPU architecture can be explored using the std::numeric_limits
facilities like this:
#include <iostream>
#include <limits>
int main() {
std::cout<< "max_exponent10: " << std::numeric_limits<long double>::max_exponent10 << std::endl;
}
Output:
max_exponent10: 4932
See the online demo
You have to use a 3rd party library (like GMP) or write your own algorithms to deal with big numbers like that.
1
Actually, wherelong double
has a bigger size than the usual 8-bytedouble
(like 80-bit on x86),long double
can store999e999
.
– geza
Nov 11 at 19:20
On x86, FPU supports 80-bit numbers. And it is usually accessible by usinglong double
. This type still can be used today on x86_64 as well.
– geza
Nov 11 at 19:24
@geza Reference please (including support from c++ standards)?
– πάντα ῥεῖ
Nov 11 at 19:26
What do you mean by support from C++ standards? C++ standard doesn't say too much about the range of floating point. It doesn't change the fact, that on x86, the FPU (i387) can use 80-bit numbers. With GCC/Clang, this is definitely accessible by usinglong double
. See: godbolt.org/z/e0HD_T (TBYTE means a 10-byte type)
– geza
Nov 11 at 19:36
@geza M'kay. I probably should refine my answer.
– πάντα ῥεῖ
Nov 11 at 19:44
|
show 6 more comments
up vote
2
down vote
In most (If not all) implementations, that constant is just too big to be represented as a unsigned long long
or long double
(Though some may just have it be floating point infinity).
You may instead be interested in std::numeric_limits<T>::infinity()
(for float
, double
or long double
) or std::numeric_limits<T>::max()
instead.
I've looked through your example, but if I read it correctly then maximum value forstd::numeric_limits<T>::max()
is1.79769e+308
, which is a lot less than999e999
, so I'm afraid I wouldn't be able to use it. Even so, could I usestd::numeric_limits<T>::infinity()
to extract a value other than infinity?
– hypnokat
Nov 11 at 19:46
1
@hypnokat It boils down to how floats are implemented in the system. The most common implementation has a largest value (Which is whatmax
returns) and aninf
value (Which is whatinfinity
returns). These, again, depend on how the float types are implemented.
– Artyer
Nov 11 at 20:03
add a comment |
up vote
2
down vote
I've tried defining it even with unsigned long long, but the compiler keeps giving me constant too big error.
Of course it does. A long long
is typically 64 bits long, which gives you log(2^64) ≅ 19
decimal digits of precision. 999e999 ≅ (10^3)^1000
, so is on the order of 3000 decimal digits long, or nearly 10,000 bits long. So 999e999
isn't just too big for a long long
, it's too big by an enormous margin.
Is it possible to define 999e999 value without using the char type?
Sure. You could define an integer-like type based on an array of some sort of integers, like long long
. You'd still need to write a set of operators to work with your new giant type, though. Also, most of the time when you're working with numbers that large, you don't need an exact representation, which is why floating point types like float
and double
are useful.
Sorry if I'm asking obvious stuff, but by defining an integer-like type do you mean I could usetypedef
? If so then how do I write my own operators from that point on?
– hypnokat
Nov 11 at 19:43
1
Sure, you could define a type liketypedef long[350] ReallyBigNumber;
and then define a bunch of functions that add, subtract, multiply, etc. values of that type. If you're working in C++, you might want to createReallyBigNumber
as a class, and then you can overloadoperator+
,operator-
, etc. so that you can use normal math operators like+
and-
. But it'd be a little silly to do all that when arbitrary precision libraries like GMP already exist.
– Caleb
Nov 11 at 20:19
@SolomonSlow You're right:999e999
is scientific notation, so should be interpreted as999*10^999
; I was thinking of 999^999, and I probably still didn't get it quite right. In a hurry just now, but will fix in a few hours.
– Caleb
Nov 11 at 21:27
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Is it possible to define
999e999
value without using the char type?
No, that's not possible using intrinsic c++ data types. That's a way to big number that could be held in either a unsigned long long
type in c++.
A long double
type would enable you to use 10 based exponents as large as you want, for modern FPU architectures.
What can be achieved with your current CPU architecture can be explored using the std::numeric_limits
facilities like this:
#include <iostream>
#include <limits>
int main() {
std::cout<< "max_exponent10: " << std::numeric_limits<long double>::max_exponent10 << std::endl;
}
Output:
max_exponent10: 4932
See the online demo
You have to use a 3rd party library (like GMP) or write your own algorithms to deal with big numbers like that.
1
Actually, wherelong double
has a bigger size than the usual 8-bytedouble
(like 80-bit on x86),long double
can store999e999
.
– geza
Nov 11 at 19:20
On x86, FPU supports 80-bit numbers. And it is usually accessible by usinglong double
. This type still can be used today on x86_64 as well.
– geza
Nov 11 at 19:24
@geza Reference please (including support from c++ standards)?
– πάντα ῥεῖ
Nov 11 at 19:26
What do you mean by support from C++ standards? C++ standard doesn't say too much about the range of floating point. It doesn't change the fact, that on x86, the FPU (i387) can use 80-bit numbers. With GCC/Clang, this is definitely accessible by usinglong double
. See: godbolt.org/z/e0HD_T (TBYTE means a 10-byte type)
– geza
Nov 11 at 19:36
@geza M'kay. I probably should refine my answer.
– πάντα ῥεῖ
Nov 11 at 19:44
|
show 6 more comments
up vote
4
down vote
accepted
Is it possible to define
999e999
value without using the char type?
No, that's not possible using intrinsic c++ data types. That's a way to big number that could be held in either a unsigned long long
type in c++.
A long double
type would enable you to use 10 based exponents as large as you want, for modern FPU architectures.
What can be achieved with your current CPU architecture can be explored using the std::numeric_limits
facilities like this:
#include <iostream>
#include <limits>
int main() {
std::cout<< "max_exponent10: " << std::numeric_limits<long double>::max_exponent10 << std::endl;
}
Output:
max_exponent10: 4932
See the online demo
You have to use a 3rd party library (like GMP) or write your own algorithms to deal with big numbers like that.
1
Actually, wherelong double
has a bigger size than the usual 8-bytedouble
(like 80-bit on x86),long double
can store999e999
.
– geza
Nov 11 at 19:20
On x86, FPU supports 80-bit numbers. And it is usually accessible by usinglong double
. This type still can be used today on x86_64 as well.
– geza
Nov 11 at 19:24
@geza Reference please (including support from c++ standards)?
– πάντα ῥεῖ
Nov 11 at 19:26
What do you mean by support from C++ standards? C++ standard doesn't say too much about the range of floating point. It doesn't change the fact, that on x86, the FPU (i387) can use 80-bit numbers. With GCC/Clang, this is definitely accessible by usinglong double
. See: godbolt.org/z/e0HD_T (TBYTE means a 10-byte type)
– geza
Nov 11 at 19:36
@geza M'kay. I probably should refine my answer.
– πάντα ῥεῖ
Nov 11 at 19:44
|
show 6 more comments
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Is it possible to define
999e999
value without using the char type?
No, that's not possible using intrinsic c++ data types. That's a way to big number that could be held in either a unsigned long long
type in c++.
A long double
type would enable you to use 10 based exponents as large as you want, for modern FPU architectures.
What can be achieved with your current CPU architecture can be explored using the std::numeric_limits
facilities like this:
#include <iostream>
#include <limits>
int main() {
std::cout<< "max_exponent10: " << std::numeric_limits<long double>::max_exponent10 << std::endl;
}
Output:
max_exponent10: 4932
See the online demo
You have to use a 3rd party library (like GMP) or write your own algorithms to deal with big numbers like that.
Is it possible to define
999e999
value without using the char type?
No, that's not possible using intrinsic c++ data types. That's a way to big number that could be held in either a unsigned long long
type in c++.
A long double
type would enable you to use 10 based exponents as large as you want, for modern FPU architectures.
What can be achieved with your current CPU architecture can be explored using the std::numeric_limits
facilities like this:
#include <iostream>
#include <limits>
int main() {
std::cout<< "max_exponent10: " << std::numeric_limits<long double>::max_exponent10 << std::endl;
}
Output:
max_exponent10: 4932
See the online demo
You have to use a 3rd party library (like GMP) or write your own algorithms to deal with big numbers like that.
edited Nov 11 at 19:50
answered Nov 11 at 18:49
πάντα ῥεῖ
71.3k972134
71.3k972134
1
Actually, wherelong double
has a bigger size than the usual 8-bytedouble
(like 80-bit on x86),long double
can store999e999
.
– geza
Nov 11 at 19:20
On x86, FPU supports 80-bit numbers. And it is usually accessible by usinglong double
. This type still can be used today on x86_64 as well.
– geza
Nov 11 at 19:24
@geza Reference please (including support from c++ standards)?
– πάντα ῥεῖ
Nov 11 at 19:26
What do you mean by support from C++ standards? C++ standard doesn't say too much about the range of floating point. It doesn't change the fact, that on x86, the FPU (i387) can use 80-bit numbers. With GCC/Clang, this is definitely accessible by usinglong double
. See: godbolt.org/z/e0HD_T (TBYTE means a 10-byte type)
– geza
Nov 11 at 19:36
@geza M'kay. I probably should refine my answer.
– πάντα ῥεῖ
Nov 11 at 19:44
|
show 6 more comments
1
Actually, wherelong double
has a bigger size than the usual 8-bytedouble
(like 80-bit on x86),long double
can store999e999
.
– geza
Nov 11 at 19:20
On x86, FPU supports 80-bit numbers. And it is usually accessible by usinglong double
. This type still can be used today on x86_64 as well.
– geza
Nov 11 at 19:24
@geza Reference please (including support from c++ standards)?
– πάντα ῥεῖ
Nov 11 at 19:26
What do you mean by support from C++ standards? C++ standard doesn't say too much about the range of floating point. It doesn't change the fact, that on x86, the FPU (i387) can use 80-bit numbers. With GCC/Clang, this is definitely accessible by usinglong double
. See: godbolt.org/z/e0HD_T (TBYTE means a 10-byte type)
– geza
Nov 11 at 19:36
@geza M'kay. I probably should refine my answer.
– πάντα ῥεῖ
Nov 11 at 19:44
1
1
Actually, where
long double
has a bigger size than the usual 8-byte double
(like 80-bit on x86), long double
can store 999e999
.– geza
Nov 11 at 19:20
Actually, where
long double
has a bigger size than the usual 8-byte double
(like 80-bit on x86), long double
can store 999e999
.– geza
Nov 11 at 19:20
On x86, FPU supports 80-bit numbers. And it is usually accessible by using
long double
. This type still can be used today on x86_64 as well.– geza
Nov 11 at 19:24
On x86, FPU supports 80-bit numbers. And it is usually accessible by using
long double
. This type still can be used today on x86_64 as well.– geza
Nov 11 at 19:24
@geza Reference please (including support from c++ standards)?
– πάντα ῥεῖ
Nov 11 at 19:26
@geza Reference please (including support from c++ standards)?
– πάντα ῥεῖ
Nov 11 at 19:26
What do you mean by support from C++ standards? C++ standard doesn't say too much about the range of floating point. It doesn't change the fact, that on x86, the FPU (i387) can use 80-bit numbers. With GCC/Clang, this is definitely accessible by using
long double
. See: godbolt.org/z/e0HD_T (TBYTE means a 10-byte type)– geza
Nov 11 at 19:36
What do you mean by support from C++ standards? C++ standard doesn't say too much about the range of floating point. It doesn't change the fact, that on x86, the FPU (i387) can use 80-bit numbers. With GCC/Clang, this is definitely accessible by using
long double
. See: godbolt.org/z/e0HD_T (TBYTE means a 10-byte type)– geza
Nov 11 at 19:36
@geza M'kay. I probably should refine my answer.
– πάντα ῥεῖ
Nov 11 at 19:44
@geza M'kay. I probably should refine my answer.
– πάντα ῥεῖ
Nov 11 at 19:44
|
show 6 more comments
up vote
2
down vote
In most (If not all) implementations, that constant is just too big to be represented as a unsigned long long
or long double
(Though some may just have it be floating point infinity).
You may instead be interested in std::numeric_limits<T>::infinity()
(for float
, double
or long double
) or std::numeric_limits<T>::max()
instead.
I've looked through your example, but if I read it correctly then maximum value forstd::numeric_limits<T>::max()
is1.79769e+308
, which is a lot less than999e999
, so I'm afraid I wouldn't be able to use it. Even so, could I usestd::numeric_limits<T>::infinity()
to extract a value other than infinity?
– hypnokat
Nov 11 at 19:46
1
@hypnokat It boils down to how floats are implemented in the system. The most common implementation has a largest value (Which is whatmax
returns) and aninf
value (Which is whatinfinity
returns). These, again, depend on how the float types are implemented.
– Artyer
Nov 11 at 20:03
add a comment |
up vote
2
down vote
In most (If not all) implementations, that constant is just too big to be represented as a unsigned long long
or long double
(Though some may just have it be floating point infinity).
You may instead be interested in std::numeric_limits<T>::infinity()
(for float
, double
or long double
) or std::numeric_limits<T>::max()
instead.
I've looked through your example, but if I read it correctly then maximum value forstd::numeric_limits<T>::max()
is1.79769e+308
, which is a lot less than999e999
, so I'm afraid I wouldn't be able to use it. Even so, could I usestd::numeric_limits<T>::infinity()
to extract a value other than infinity?
– hypnokat
Nov 11 at 19:46
1
@hypnokat It boils down to how floats are implemented in the system. The most common implementation has a largest value (Which is whatmax
returns) and aninf
value (Which is whatinfinity
returns). These, again, depend on how the float types are implemented.
– Artyer
Nov 11 at 20:03
add a comment |
up vote
2
down vote
up vote
2
down vote
In most (If not all) implementations, that constant is just too big to be represented as a unsigned long long
or long double
(Though some may just have it be floating point infinity).
You may instead be interested in std::numeric_limits<T>::infinity()
(for float
, double
or long double
) or std::numeric_limits<T>::max()
instead.
In most (If not all) implementations, that constant is just too big to be represented as a unsigned long long
or long double
(Though some may just have it be floating point infinity).
You may instead be interested in std::numeric_limits<T>::infinity()
(for float
, double
or long double
) or std::numeric_limits<T>::max()
instead.
answered Nov 11 at 19:06
Artyer
3,578625
3,578625
I've looked through your example, but if I read it correctly then maximum value forstd::numeric_limits<T>::max()
is1.79769e+308
, which is a lot less than999e999
, so I'm afraid I wouldn't be able to use it. Even so, could I usestd::numeric_limits<T>::infinity()
to extract a value other than infinity?
– hypnokat
Nov 11 at 19:46
1
@hypnokat It boils down to how floats are implemented in the system. The most common implementation has a largest value (Which is whatmax
returns) and aninf
value (Which is whatinfinity
returns). These, again, depend on how the float types are implemented.
– Artyer
Nov 11 at 20:03
add a comment |
I've looked through your example, but if I read it correctly then maximum value forstd::numeric_limits<T>::max()
is1.79769e+308
, which is a lot less than999e999
, so I'm afraid I wouldn't be able to use it. Even so, could I usestd::numeric_limits<T>::infinity()
to extract a value other than infinity?
– hypnokat
Nov 11 at 19:46
1
@hypnokat It boils down to how floats are implemented in the system. The most common implementation has a largest value (Which is whatmax
returns) and aninf
value (Which is whatinfinity
returns). These, again, depend on how the float types are implemented.
– Artyer
Nov 11 at 20:03
I've looked through your example, but if I read it correctly then maximum value for
std::numeric_limits<T>::max()
is 1.79769e+308
, which is a lot less than 999e999
, so I'm afraid I wouldn't be able to use it. Even so, could I use std::numeric_limits<T>::infinity()
to extract a value other than infinity?– hypnokat
Nov 11 at 19:46
I've looked through your example, but if I read it correctly then maximum value for
std::numeric_limits<T>::max()
is 1.79769e+308
, which is a lot less than 999e999
, so I'm afraid I wouldn't be able to use it. Even so, could I use std::numeric_limits<T>::infinity()
to extract a value other than infinity?– hypnokat
Nov 11 at 19:46
1
1
@hypnokat It boils down to how floats are implemented in the system. The most common implementation has a largest value (Which is what
max
returns) and an inf
value (Which is what infinity
returns). These, again, depend on how the float types are implemented.– Artyer
Nov 11 at 20:03
@hypnokat It boils down to how floats are implemented in the system. The most common implementation has a largest value (Which is what
max
returns) and an inf
value (Which is what infinity
returns). These, again, depend on how the float types are implemented.– Artyer
Nov 11 at 20:03
add a comment |
up vote
2
down vote
I've tried defining it even with unsigned long long, but the compiler keeps giving me constant too big error.
Of course it does. A long long
is typically 64 bits long, which gives you log(2^64) ≅ 19
decimal digits of precision. 999e999 ≅ (10^3)^1000
, so is on the order of 3000 decimal digits long, or nearly 10,000 bits long. So 999e999
isn't just too big for a long long
, it's too big by an enormous margin.
Is it possible to define 999e999 value without using the char type?
Sure. You could define an integer-like type based on an array of some sort of integers, like long long
. You'd still need to write a set of operators to work with your new giant type, though. Also, most of the time when you're working with numbers that large, you don't need an exact representation, which is why floating point types like float
and double
are useful.
Sorry if I'm asking obvious stuff, but by defining an integer-like type do you mean I could usetypedef
? If so then how do I write my own operators from that point on?
– hypnokat
Nov 11 at 19:43
1
Sure, you could define a type liketypedef long[350] ReallyBigNumber;
and then define a bunch of functions that add, subtract, multiply, etc. values of that type. If you're working in C++, you might want to createReallyBigNumber
as a class, and then you can overloadoperator+
,operator-
, etc. so that you can use normal math operators like+
and-
. But it'd be a little silly to do all that when arbitrary precision libraries like GMP already exist.
– Caleb
Nov 11 at 20:19
@SolomonSlow You're right:999e999
is scientific notation, so should be interpreted as999*10^999
; I was thinking of 999^999, and I probably still didn't get it quite right. In a hurry just now, but will fix in a few hours.
– Caleb
Nov 11 at 21:27
add a comment |
up vote
2
down vote
I've tried defining it even with unsigned long long, but the compiler keeps giving me constant too big error.
Of course it does. A long long
is typically 64 bits long, which gives you log(2^64) ≅ 19
decimal digits of precision. 999e999 ≅ (10^3)^1000
, so is on the order of 3000 decimal digits long, or nearly 10,000 bits long. So 999e999
isn't just too big for a long long
, it's too big by an enormous margin.
Is it possible to define 999e999 value without using the char type?
Sure. You could define an integer-like type based on an array of some sort of integers, like long long
. You'd still need to write a set of operators to work with your new giant type, though. Also, most of the time when you're working with numbers that large, you don't need an exact representation, which is why floating point types like float
and double
are useful.
Sorry if I'm asking obvious stuff, but by defining an integer-like type do you mean I could usetypedef
? If so then how do I write my own operators from that point on?
– hypnokat
Nov 11 at 19:43
1
Sure, you could define a type liketypedef long[350] ReallyBigNumber;
and then define a bunch of functions that add, subtract, multiply, etc. values of that type. If you're working in C++, you might want to createReallyBigNumber
as a class, and then you can overloadoperator+
,operator-
, etc. so that you can use normal math operators like+
and-
. But it'd be a little silly to do all that when arbitrary precision libraries like GMP already exist.
– Caleb
Nov 11 at 20:19
@SolomonSlow You're right:999e999
is scientific notation, so should be interpreted as999*10^999
; I was thinking of 999^999, and I probably still didn't get it quite right. In a hurry just now, but will fix in a few hours.
– Caleb
Nov 11 at 21:27
add a comment |
up vote
2
down vote
up vote
2
down vote
I've tried defining it even with unsigned long long, but the compiler keeps giving me constant too big error.
Of course it does. A long long
is typically 64 bits long, which gives you log(2^64) ≅ 19
decimal digits of precision. 999e999 ≅ (10^3)^1000
, so is on the order of 3000 decimal digits long, or nearly 10,000 bits long. So 999e999
isn't just too big for a long long
, it's too big by an enormous margin.
Is it possible to define 999e999 value without using the char type?
Sure. You could define an integer-like type based on an array of some sort of integers, like long long
. You'd still need to write a set of operators to work with your new giant type, though. Also, most of the time when you're working with numbers that large, you don't need an exact representation, which is why floating point types like float
and double
are useful.
I've tried defining it even with unsigned long long, but the compiler keeps giving me constant too big error.
Of course it does. A long long
is typically 64 bits long, which gives you log(2^64) ≅ 19
decimal digits of precision. 999e999 ≅ (10^3)^1000
, so is on the order of 3000 decimal digits long, or nearly 10,000 bits long. So 999e999
isn't just too big for a long long
, it's too big by an enormous margin.
Is it possible to define 999e999 value without using the char type?
Sure. You could define an integer-like type based on an array of some sort of integers, like long long
. You'd still need to write a set of operators to work with your new giant type, though. Also, most of the time when you're working with numbers that large, you don't need an exact representation, which is why floating point types like float
and double
are useful.
answered Nov 11 at 19:09
Caleb
108k16149238
108k16149238
Sorry if I'm asking obvious stuff, but by defining an integer-like type do you mean I could usetypedef
? If so then how do I write my own operators from that point on?
– hypnokat
Nov 11 at 19:43
1
Sure, you could define a type liketypedef long[350] ReallyBigNumber;
and then define a bunch of functions that add, subtract, multiply, etc. values of that type. If you're working in C++, you might want to createReallyBigNumber
as a class, and then you can overloadoperator+
,operator-
, etc. so that you can use normal math operators like+
and-
. But it'd be a little silly to do all that when arbitrary precision libraries like GMP already exist.
– Caleb
Nov 11 at 20:19
@SolomonSlow You're right:999e999
is scientific notation, so should be interpreted as999*10^999
; I was thinking of 999^999, and I probably still didn't get it quite right. In a hurry just now, but will fix in a few hours.
– Caleb
Nov 11 at 21:27
add a comment |
Sorry if I'm asking obvious stuff, but by defining an integer-like type do you mean I could usetypedef
? If so then how do I write my own operators from that point on?
– hypnokat
Nov 11 at 19:43
1
Sure, you could define a type liketypedef long[350] ReallyBigNumber;
and then define a bunch of functions that add, subtract, multiply, etc. values of that type. If you're working in C++, you might want to createReallyBigNumber
as a class, and then you can overloadoperator+
,operator-
, etc. so that you can use normal math operators like+
and-
. But it'd be a little silly to do all that when arbitrary precision libraries like GMP already exist.
– Caleb
Nov 11 at 20:19
@SolomonSlow You're right:999e999
is scientific notation, so should be interpreted as999*10^999
; I was thinking of 999^999, and I probably still didn't get it quite right. In a hurry just now, but will fix in a few hours.
– Caleb
Nov 11 at 21:27
Sorry if I'm asking obvious stuff, but by defining an integer-like type do you mean I could use
typedef
? If so then how do I write my own operators from that point on?– hypnokat
Nov 11 at 19:43
Sorry if I'm asking obvious stuff, but by defining an integer-like type do you mean I could use
typedef
? If so then how do I write my own operators from that point on?– hypnokat
Nov 11 at 19:43
1
1
Sure, you could define a type like
typedef long[350] ReallyBigNumber;
and then define a bunch of functions that add, subtract, multiply, etc. values of that type. If you're working in C++, you might want to create ReallyBigNumber
as a class, and then you can overload operator+
, operator-
, etc. so that you can use normal math operators like +
and -
. But it'd be a little silly to do all that when arbitrary precision libraries like GMP already exist.– Caleb
Nov 11 at 20:19
Sure, you could define a type like
typedef long[350] ReallyBigNumber;
and then define a bunch of functions that add, subtract, multiply, etc. values of that type. If you're working in C++, you might want to create ReallyBigNumber
as a class, and then you can overload operator+
, operator-
, etc. so that you can use normal math operators like +
and -
. But it'd be a little silly to do all that when arbitrary precision libraries like GMP already exist.– Caleb
Nov 11 at 20:19
@SolomonSlow You're right:
999e999
is scientific notation, so should be interpreted as 999*10^999
; I was thinking of 999^999, and I probably still didn't get it quite right. In a hurry just now, but will fix in a few hours.– Caleb
Nov 11 at 21:27
@SolomonSlow You're right:
999e999
is scientific notation, so should be interpreted as 999*10^999
; I was thinking of 999^999, and I probably still didn't get it quite right. In a hurry just now, but will fix in a few hours.– Caleb
Nov 11 at 21:27
add a comment |
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2
You may be interested in GMP.
– Jesper Juhl
Nov 11 at 18:48
2
Yes that's the way to go hypnokat, just like @JesperJuhl mentioned. However, I suggest you make sure first that you really need to use that big numbers!
– gsamaras
Nov 11 at 18:52
1
based on what you want to do with your number, you may be able to use openssl
BN
functions too.– Afshin
Nov 11 at 19:15
1
Out of curiosity, why would one ever need such a big number? Even if you think that every particle in the universe is itself a universe with the same number of particles, you'll have about 10^172 particles in total. Even if you wanted to count the number of particles that can fit in these universes "space", that number would still be huge! Do you need unit precision on such a... supercalifragilistic number?
– L.C.
Nov 11 at 19:28
It is just a random experiment I thought of, I'm a programming newbie. Still, thanks to that random thought, even thought I can't do that in C++ alone, I now know about the existance of GMP, so I'm still walking out with something new :)
– hypnokat
Nov 11 at 19:35