Use type of a variable as a template argument
I did the following:
long long int x = 0;
int digits_of_x = std::numeric_limits<long long int>::digits;
And it works fine. However this can easily introduce an error if someone changes the type of x. So I would prefer to do it like this:
long long int x = 0;
int digits_of_x = std::numeric_limits<typeof(x)>::digits;
I found the GCC extension typeof() but I want to use some standard function (up to C++17). Is there such a function?
c++ c++11 types c++14 c++17
asked Nov 19 '18 at 23:07
SilicomancerSilicomancer
4,09933475
add a comment |
I did the following:
long long int x = 0;
int digits_of_x = std::numeric_limits<long long int>::digits;
And it works fine. However this can easily introduce an error if someone changes the type of x. So I would prefer to do it like this:
long long int x = 0;
int digits_of_x = std::numeric_limits<typeof(x)>::digits;
I found the GCC extension typeof() but I want to use some standard function (up to C++17). Is there such a function?
c++ c++11 types c++14 c++17
asked Nov 19 '18 at 23:07
SilicomancerSilicomancer
4,09933475
add a comment |
I did the following:
long long int x = 0;
int digits_of_x = std::numeric_limits<long long int>::digits;
And it works fine. However this can easily introduce an error if someone changes the type of x. So I would prefer to do it like this:
long long int x = 0;
int digits_of_x = std::numeric_limits<typeof(x)>::digits;
I found the GCC extension typeof() but I want to use some standard function (up to C++17). Is there such a function?
c++ c++11 types c++14 c++17
asked Nov 19 '18 at 23:07
SilicomancerSilicomancer
4,09933475
I did the following:
long long int x = 0;
int digits_of_x = std::numeric_limits<long long int>::digits;
And it works fine. However this can easily introduce an error if someone changes the type of x. So I would prefer to do it like this:
long long int x = 0;
int digits_of_x = std::numeric_limits<typeof(x)>::digits;
I found the GCC extension typeof() but I want to use some standard function (up to C++17). Is there such a function?
c++ c++11 types c++14 c++17
c++ c++11 types c++14 c++17
asked Nov 19 '18 at 23:07
SilicomancerSilicomancer
4,09933475
asked Nov 19 '18 at 23:07
SilicomancerSilicomancer
4,09933475
edited Nov 19 '18 at 23:12
Silicomancer
asked Nov 19 '18 at 23:07
SilicomancerSilicomancer
4,09933475
asked Nov 19 '18 at 23:07
SilicomancerSilicomancer
4,09933475
asked Nov 19 '18 at 23:07
SilicomancerSilicomancer
4,09933475
4,09933475
add a comment |
add a comment |
1 Answer
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You want to use decltype, not typeof. Available since C++11:
long long int x = 0;
int digits_of_x = std::numeric_limits<decltype(x)>::digits;
answered Nov 19 '18 at 23:21
acmacm
6,99832456
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You want to use decltype, not typeof. Available since C++11:
long long int x = 0;
int digits_of_x = std::numeric_limits<decltype(x)>::digits;
answered Nov 19 '18 at 23:21
acmacm
6,99832456
add a comment |
You want to use decltype, not typeof. Available since C++11:
long long int x = 0;
int digits_of_x = std::numeric_limits<decltype(x)>::digits;
answered Nov 19 '18 at 23:21
acmacm
6,99832456
add a comment |
You want to use decltype, not typeof. Available since C++11:
long long int x = 0;
int digits_of_x = std::numeric_limits<decltype(x)>::digits;
answered Nov 19 '18 at 23:21
acmacm
6,99832456
You want to use decltype, not typeof. Available since C++11:
long long int x = 0;
int digits_of_x = std::numeric_limits<decltype(x)>::digits;
answered Nov 19 '18 at 23:21
acmacm
6,99832456
answered Nov 19 '18 at 23:21
acmacm
6,99832456
answered Nov 19 '18 at 23:21
acmacm
6,99832456
answered Nov 19 '18 at 23:21
acmacm
6,99832456
6,99832456
add a comment |
add a comment |
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