array of 2 dimensions, one being known and the other unknown
I need to efficiently allocate an array of two dimensions, one being known (5) and the other being unknown.
My aim is to have a continuous block of memory named block that stores my data in the following way :
block[0,0] block[0,1] block[0,2] block[0,3] block[0,4] block[1,0] ...]
The following code compiles (in C++) :
unsigned int size2=200;
auto block = new float[size2][5];
Is it correct ?
What is the type of the variable block ?
I have tried :
float[5]* block = new float[size2][5];
float[5] block = new float[size2][5];
float block[5] = new float[size2][5];
float block*[5] = new float[size2][5];
All of them leads to an error.
I am running out of imagination...
c++ arrays types
edited Nov 14 '18 at 16:32
YSC
20.8k34594
asked Nov 14 '18 at 16:15
Arnaud Mégret
1157
add a comment |
I need to efficiently allocate an array of two dimensions, one being known (5) and the other being unknown.
My aim is to have a continuous block of memory named block that stores my data in the following way :
block[0,0] block[0,1] block[0,2] block[0,3] block[0,4] block[1,0] ...]
The following code compiles (in C++) :
unsigned int size2=200;
auto block = new float[size2][5];
Is it correct ?
What is the type of the variable block ?
I have tried :
float[5]* block = new float[size2][5];
float[5] block = new float[size2][5];
float block[5] = new float[size2][5];
float block*[5] = new float[size2][5];
All of them leads to an error.
I am running out of imagination...
c++ arrays types
edited Nov 14 '18 at 16:32
YSC
20.8k34594
asked Nov 14 '18 at 16:15
Arnaud Mégret
1157
How about creating array of pointers each points to a dynamic Array?
– Ali Kanat
Nov 14 '18 at 16:23
it allocates additional memory and will fragment allocated memory
– Arnaud Mégret
Nov 14 '18 at 16:24
1
What not use vectors?
– eshirima
Nov 14 '18 at 16:24
add a comment |
I need to efficiently allocate an array of two dimensions, one being known (5) and the other being unknown.
My aim is to have a continuous block of memory named block that stores my data in the following way :
block[0,0] block[0,1] block[0,2] block[0,3] block[0,4] block[1,0] ...]
The following code compiles (in C++) :
unsigned int size2=200;
auto block = new float[size2][5];
Is it correct ?
What is the type of the variable block ?
I have tried :
float[5]* block = new float[size2][5];
float[5] block = new float[size2][5];
float block[5] = new float[size2][5];
float block*[5] = new float[size2][5];
All of them leads to an error.
I am running out of imagination...
c++ arrays types
edited Nov 14 '18 at 16:32
YSC
20.8k34594
asked Nov 14 '18 at 16:15
Arnaud Mégret
1157
I need to efficiently allocate an array of two dimensions, one being known (5) and the other being unknown.
My aim is to have a continuous block of memory named block that stores my data in the following way :
block[0,0] block[0,1] block[0,2] block[0,3] block[0,4] block[1,0] ...]
The following code compiles (in C++) :
unsigned int size2=200;
auto block = new float[size2][5];
Is it correct ?
What is the type of the variable block ?
I have tried :
float[5]* block = new float[size2][5];
float[5] block = new float[size2][5];
float block[5] = new float[size2][5];
float block*[5] = new float[size2][5];
All of them leads to an error.
I am running out of imagination...
c++ arrays types
c++ arrays types
edited Nov 14 '18 at 16:32
YSC
20.8k34594
asked Nov 14 '18 at 16:15
Arnaud Mégret
1157
edited Nov 14 '18 at 16:32
YSC
20.8k34594
asked Nov 14 '18 at 16:15
Arnaud Mégret
1157
edited Nov 14 '18 at 16:32
YSC
20.8k34594
edited Nov 14 '18 at 16:32
YSC
20.8k34594
edited Nov 14 '18 at 16:32
YSC
20.8k34594
20.8k34594
asked Nov 14 '18 at 16:15
Arnaud Mégret
1157
asked Nov 14 '18 at 16:15
Arnaud Mégret
1157
asked Nov 14 '18 at 16:15
Arnaud Mégret
1157
1157
How about creating array of pointers each points to a dynamic Array?
– Ali Kanat
Nov 14 '18 at 16:23
it allocates additional memory and will fragment allocated memory
– Arnaud Mégret
Nov 14 '18 at 16:24
1
What not use vectors?
– eshirima
Nov 14 '18 at 16:24
add a comment |
How about creating array of pointers each points to a dynamic Array?
– Ali Kanat
Nov 14 '18 at 16:23
it allocates additional memory and will fragment allocated memory
– Arnaud Mégret
Nov 14 '18 at 16:24
1
What not use vectors?
– eshirima
Nov 14 '18 at 16:24
How about creating array of pointers each points to a dynamic Array?
– Ali Kanat
Nov 14 '18 at 16:23
How about creating array of pointers each points to a dynamic Array?
– Ali Kanat
Nov 14 '18 at 16:23
it allocates additional memory and will fragment allocated memory
– Arnaud Mégret
Nov 14 '18 at 16:24
it allocates additional memory and will fragment allocated memory
– Arnaud Mégret
Nov 14 '18 at 16:24
1
1
What not use vectors?
– eshirima
Nov 14 '18 at 16:24
What not use vectors?
– eshirima
Nov 14 '18 at 16:24
add a comment |
3 Answers
3
active
oldest
votes
unsigned int size2=200;
auto block = new float[size2][5];
Is it correct?
Yes it is correct.
What is the type of the variable block?
Using auto here is the best you can do. But for learning purpose, the type of block is float(*)[5]. To define a variable with that type, you must use the following syntax:
unsigned int size2=200;
float(*block)[5] = new float[size2][5];
Demo: https://godbolt.org/z/4iJMA5
"How can I read such a thing as float(*block)[5]?" you might ask. It's simple, use the C++ Guru Snail Rule.
answered Nov 14 '18 at 16:29
YSC
20.8k34594
You can do something similar defining atypedefallowing a somewhat more natural pointer to array allocation, e.g.typedef float row_t[5]; ... row_t *block = new row_t[200];(whether that is more clear or not is arguable)
– David C. Rankin
Nov 14 '18 at 17:54
add a comment |
unsigned int size2=200;
auto block = new float[size2][5];
Is it correct ?
Yes, although a bare owning pointer is a bad idea. It's better to use std::vector for dynamic arrays.
What is the type of the variable block ?
It is float (*)[5] i.e. pointer to an array of 5 float.
As I mentioned, I recommend using std::vector instead to avoid pitfalls with memory management:
std::vector<std::array<float, 5>> block(size2);
answered Nov 14 '18 at 16:37
eerorika
76.8k556117
add a comment |
I'm a bit old-school, but the simplest is really
new float[size2 * 5];
and access it with
my_array[y * 5 + x]
You can even add an inline helper function
int at(int x, int y) { return y * 5 + x;}
...
my_array[at(x,y)]
And ideally combine with a std::vector instead of a dynamic array.
answered Nov 14 '18 at 16:28
Jeffrey
1,458520
I don't see how this is better than the questioner's own version tbh. Using this is more complicated.
– Galik
Nov 14 '18 at 17:21
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
unsigned int size2=200;
auto block = new float[size2][5];
Is it correct?
Yes it is correct.
What is the type of the variable block?
Using auto here is the best you can do. But for learning purpose, the type of block is float(*)[5]. To define a variable with that type, you must use the following syntax:
unsigned int size2=200;
float(*block)[5] = new float[size2][5];
Demo: https://godbolt.org/z/4iJMA5
"How can I read such a thing as float(*block)[5]?" you might ask. It's simple, use the C++ Guru Snail Rule.
answered Nov 14 '18 at 16:29
YSC
20.8k34594
You can do something similar defining atypedefallowing a somewhat more natural pointer to array allocation, e.g.typedef float row_t[5]; ... row_t *block = new row_t[200];(whether that is more clear or not is arguable)
– David C. Rankin
Nov 14 '18 at 17:54
add a comment |
unsigned int size2=200;
auto block = new float[size2][5];
Is it correct?
Yes it is correct.
What is the type of the variable block?
Using auto here is the best you can do. But for learning purpose, the type of block is float(*)[5]. To define a variable with that type, you must use the following syntax:
unsigned int size2=200;
float(*block)[5] = new float[size2][5];
Demo: https://godbolt.org/z/4iJMA5
"How can I read such a thing as float(*block)[5]?" you might ask. It's simple, use the C++ Guru Snail Rule.
answered Nov 14 '18 at 16:29
YSC
20.8k34594
You can do something similar defining atypedefallowing a somewhat more natural pointer to array allocation, e.g.typedef float row_t[5]; ... row_t *block = new row_t[200];(whether that is more clear or not is arguable)
– David C. Rankin
Nov 14 '18 at 17:54
add a comment |
unsigned int size2=200;
auto block = new float[size2][5];
Is it correct?
Yes it is correct.
What is the type of the variable block?
Using auto here is the best you can do. But for learning purpose, the type of block is float(*)[5]. To define a variable with that type, you must use the following syntax:
unsigned int size2=200;
float(*block)[5] = new float[size2][5];
Demo: https://godbolt.org/z/4iJMA5
"How can I read such a thing as float(*block)[5]?" you might ask. It's simple, use the C++ Guru Snail Rule.
answered Nov 14 '18 at 16:29
YSC
20.8k34594
unsigned int size2=200;
auto block = new float[size2][5];
Is it correct?
Yes it is correct.
What is the type of the variable block?
Using auto here is the best you can do. But for learning purpose, the type of block is float(*)[5]. To define a variable with that type, you must use the following syntax:
unsigned int size2=200;
float(*block)[5] = new float[size2][5];
Demo: https://godbolt.org/z/4iJMA5
"How can I read such a thing as float(*block)[5]?" you might ask. It's simple, use the C++ Guru Snail Rule.
answered Nov 14 '18 at 16:29
YSC
20.8k34594
edited Nov 14 '18 at 16:37
answered Nov 14 '18 at 16:29
YSC
20.8k34594
answered Nov 14 '18 at 16:29
YSC
20.8k34594
answered Nov 14 '18 at 16:29
YSC
20.8k34594
20.8k34594
You can do something similar defining atypedefallowing a somewhat more natural pointer to array allocation, e.g.typedef float row_t[5]; ... row_t *block = new row_t[200];(whether that is more clear or not is arguable)
– David C. Rankin
Nov 14 '18 at 17:54
add a comment |
You can do something similar defining atypedefallowing a somewhat more natural pointer to array allocation, e.g.typedef float row_t[5]; ... row_t *block = new row_t[200];(whether that is more clear or not is arguable)
– David C. Rankin
Nov 14 '18 at 17:54
You can do something similar defining a
typedef allowing a somewhat more natural pointer to array allocation, e.g. typedef float row_t[5]; ... row_t *block = new row_t[200]; (whether that is more clear or not is arguable)– David C. Rankin
Nov 14 '18 at 17:54
You can do something similar defining a
typedef allowing a somewhat more natural pointer to array allocation, e.g. typedef float row_t[5]; ... row_t *block = new row_t[200]; (whether that is more clear or not is arguable)– David C. Rankin
Nov 14 '18 at 17:54
add a comment |
unsigned int size2=200;
auto block = new float[size2][5];
Is it correct ?
Yes, although a bare owning pointer is a bad idea. It's better to use std::vector for dynamic arrays.
What is the type of the variable block ?
It is float (*)[5] i.e. pointer to an array of 5 float.
As I mentioned, I recommend using std::vector instead to avoid pitfalls with memory management:
std::vector<std::array<float, 5>> block(size2);
answered Nov 14 '18 at 16:37
eerorika
76.8k556117
add a comment |
unsigned int size2=200;
auto block = new float[size2][5];
Is it correct ?
Yes, although a bare owning pointer is a bad idea. It's better to use std::vector for dynamic arrays.
What is the type of the variable block ?
It is float (*)[5] i.e. pointer to an array of 5 float.
As I mentioned, I recommend using std::vector instead to avoid pitfalls with memory management:
std::vector<std::array<float, 5>> block(size2);
answered Nov 14 '18 at 16:37
eerorika
76.8k556117
add a comment |
unsigned int size2=200;
auto block = new float[size2][5];
Is it correct ?
Yes, although a bare owning pointer is a bad idea. It's better to use std::vector for dynamic arrays.
What is the type of the variable block ?
It is float (*)[5] i.e. pointer to an array of 5 float.
As I mentioned, I recommend using std::vector instead to avoid pitfalls with memory management:
std::vector<std::array<float, 5>> block(size2);
answered Nov 14 '18 at 16:37
eerorika
76.8k556117
unsigned int size2=200;
auto block = new float[size2][5];
Is it correct ?
Yes, although a bare owning pointer is a bad idea. It's better to use std::vector for dynamic arrays.
What is the type of the variable block ?
It is float (*)[5] i.e. pointer to an array of 5 float.
As I mentioned, I recommend using std::vector instead to avoid pitfalls with memory management:
std::vector<std::array<float, 5>> block(size2);
answered Nov 14 '18 at 16:37
eerorika
76.8k556117
answered Nov 14 '18 at 16:37
eerorika
76.8k556117
answered Nov 14 '18 at 16:37
eerorika
76.8k556117
answered Nov 14 '18 at 16:37
eerorika
76.8k556117
76.8k556117
add a comment |
add a comment |
I'm a bit old-school, but the simplest is really
new float[size2 * 5];
and access it with
my_array[y * 5 + x]
You can even add an inline helper function
int at(int x, int y) { return y * 5 + x;}
...
my_array[at(x,y)]
And ideally combine with a std::vector instead of a dynamic array.
answered Nov 14 '18 at 16:28
Jeffrey
1,458520
I don't see how this is better than the questioner's own version tbh. Using this is more complicated.
– Galik
Nov 14 '18 at 17:21
add a comment |
I'm a bit old-school, but the simplest is really
new float[size2 * 5];
and access it with
my_array[y * 5 + x]
You can even add an inline helper function
int at(int x, int y) { return y * 5 + x;}
...
my_array[at(x,y)]
And ideally combine with a std::vector instead of a dynamic array.
answered Nov 14 '18 at 16:28
Jeffrey
1,458520
I don't see how this is better than the questioner's own version tbh. Using this is more complicated.
– Galik
Nov 14 '18 at 17:21
add a comment |
I'm a bit old-school, but the simplest is really
new float[size2 * 5];
and access it with
my_array[y * 5 + x]
You can even add an inline helper function
int at(int x, int y) { return y * 5 + x;}
...
my_array[at(x,y)]
And ideally combine with a std::vector instead of a dynamic array.
answered Nov 14 '18 at 16:28
Jeffrey
1,458520
I'm a bit old-school, but the simplest is really
new float[size2 * 5];
and access it with
my_array[y * 5 + x]
You can even add an inline helper function
int at(int x, int y) { return y * 5 + x;}
...
my_array[at(x,y)]
And ideally combine with a std::vector instead of a dynamic array.
answered Nov 14 '18 at 16:28
Jeffrey
1,458520
answered Nov 14 '18 at 16:28
Jeffrey
1,458520
answered Nov 14 '18 at 16:28
Jeffrey
1,458520
answered Nov 14 '18 at 16:28
Jeffrey
1,458520
1,458520
I don't see how this is better than the questioner's own version tbh. Using this is more complicated.
– Galik
Nov 14 '18 at 17:21
add a comment |
I don't see how this is better than the questioner's own version tbh. Using this is more complicated.
– Galik
Nov 14 '18 at 17:21
I don't see how this is better than the questioner's own version tbh. Using this is more complicated.
– Galik
Nov 14 '18 at 17:21
I don't see how this is better than the questioner's own version tbh. Using this is more complicated.
– Galik
Nov 14 '18 at 17:21
add a comment |
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How about creating array of pointers each points to a dynamic Array?
– Ali Kanat
Nov 14 '18 at 16:23
it allocates additional memory and will fragment allocated memory
– Arnaud Mégret
Nov 14 '18 at 16:24
1
What not use vectors?
– eshirima
Nov 14 '18 at 16:24