Type depends on condition in template
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Can I make a template function with arguments, which types are depends on template argument? (Code below is just to explain, what I want)
#include <complex>
template <bool two>
void foo( (if (two) ? double* : std::complex<double>* >) input, size_t n)
{
for (size_t i = 0; i < n; ++n)
input[i] *= two ? 2.0 : 1.0;
}
void foo_double(double *input, size_t n){
foo<true>(input,n);
}
void foo_complex (std::complex<double> *input, size_t n){
foo<false>(input,n);
}
I thought std::conditional would help, but I guess, I don't know how to use it correctly here (code below can't be compiled)
#include <type_traits>
#include <complex>
template <bool two>
void foo(std::conditional<two, double*, std::complex<double>* > input, size_t n)
{
for (size_t i = 0; i < n; ++n)
input[i] *= two ? 2.0 : 1.0;
}
void foo_double(double *input, size_t n){
foo<true>(input,n);
}
void foo_complex (std::complex<double> *input, size_t n){
foo<false>(input,n);
}
It would be very nice, if someone find solution without higher than c++11, so I will be able to compile it both in vs2012 and with gcc-6+. But some examples with c++14 or higher will be good experience too.
Thanks =)
c++ c++11 templates
edited Nov 11 at 22:15
max66
33.7k63762
asked May 5 '17 at 16:21
Sklert
177411
|
show 4 more comments
up vote
0
down vote
favorite
Can I make a template function with arguments, which types are depends on template argument? (Code below is just to explain, what I want)
#include <complex>
template <bool two>
void foo( (if (two) ? double* : std::complex<double>* >) input, size_t n)
{
for (size_t i = 0; i < n; ++n)
input[i] *= two ? 2.0 : 1.0;
}
void foo_double(double *input, size_t n){
foo<true>(input,n);
}
void foo_complex (std::complex<double> *input, size_t n){
foo<false>(input,n);
}
I thought std::conditional would help, but I guess, I don't know how to use it correctly here (code below can't be compiled)
#include <type_traits>
#include <complex>
template <bool two>
void foo(std::conditional<two, double*, std::complex<double>* > input, size_t n)
{
for (size_t i = 0; i < n; ++n)
input[i] *= two ? 2.0 : 1.0;
}
void foo_double(double *input, size_t n){
foo<true>(input,n);
}
void foo_complex (std::complex<double> *input, size_t n){
foo<false>(input,n);
}
It would be very nice, if someone find solution without higher than c++11, so I will be able to compile it both in vs2012 and with gcc-6+. But some examples with c++14 or higher will be good experience too.
Thanks =)
c++ c++11 templates
edited Nov 11 at 22:15
max66
33.7k63762
asked May 5 '17 at 16:21
Sklert
177411
4
Why do you need to even do this? Why nottemplate<typename T> void foo(T input, size_t n) { ...}?
– NathanOliver
May 5 '17 at 16:23
std::conditionalis the right choice. Did you look in the docs to see how to use it? If you don't have access to C++11, you should be able to implement it yourself (it's documented if you need help).
– Rakete1111
May 5 '17 at 16:24
usestd::conditionallike this:std::conditional<some_bool, type_true, type_false>::type
– qxz
May 5 '17 at 16:25
@NathanOliver,twois used in function body. Type of input depends on bool, so I want to reduce template arguments
– Sklert
May 5 '17 at 16:29
3
std::conditional_t</*..*/>in C++14,typename std::conditional</*..*/>::typefor C++11.
– Jarod42
May 5 '17 at 16:30
|
show 4 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Can I make a template function with arguments, which types are depends on template argument? (Code below is just to explain, what I want)
#include <complex>
template <bool two>
void foo( (if (two) ? double* : std::complex<double>* >) input, size_t n)
{
for (size_t i = 0; i < n; ++n)
input[i] *= two ? 2.0 : 1.0;
}
void foo_double(double *input, size_t n){
foo<true>(input,n);
}
void foo_complex (std::complex<double> *input, size_t n){
foo<false>(input,n);
}
I thought std::conditional would help, but I guess, I don't know how to use it correctly here (code below can't be compiled)
#include <type_traits>
#include <complex>
template <bool two>
void foo(std::conditional<two, double*, std::complex<double>* > input, size_t n)
{
for (size_t i = 0; i < n; ++n)
input[i] *= two ? 2.0 : 1.0;
}
void foo_double(double *input, size_t n){
foo<true>(input,n);
}
void foo_complex (std::complex<double> *input, size_t n){
foo<false>(input,n);
}
It would be very nice, if someone find solution without higher than c++11, so I will be able to compile it both in vs2012 and with gcc-6+. But some examples with c++14 or higher will be good experience too.
Thanks =)
c++ c++11 templates
edited Nov 11 at 22:15
max66
33.7k63762
asked May 5 '17 at 16:21
Sklert
177411
Can I make a template function with arguments, which types are depends on template argument? (Code below is just to explain, what I want)
#include <complex>
template <bool two>
void foo( (if (two) ? double* : std::complex<double>* >) input, size_t n)
{
for (size_t i = 0; i < n; ++n)
input[i] *= two ? 2.0 : 1.0;
}
void foo_double(double *input, size_t n){
foo<true>(input,n);
}
void foo_complex (std::complex<double> *input, size_t n){
foo<false>(input,n);
}
I thought std::conditional would help, but I guess, I don't know how to use it correctly here (code below can't be compiled)
#include <type_traits>
#include <complex>
template <bool two>
void foo(std::conditional<two, double*, std::complex<double>* > input, size_t n)
{
for (size_t i = 0; i < n; ++n)
input[i] *= two ? 2.0 : 1.0;
}
void foo_double(double *input, size_t n){
foo<true>(input,n);
}
void foo_complex (std::complex<double> *input, size_t n){
foo<false>(input,n);
}
It would be very nice, if someone find solution without higher than c++11, so I will be able to compile it both in vs2012 and with gcc-6+. But some examples with c++14 or higher will be good experience too.
Thanks =)
c++ c++11 templates
c++ c++11 templates
edited Nov 11 at 22:15
max66
33.7k63762
asked May 5 '17 at 16:21
Sklert
177411
edited Nov 11 at 22:15
max66
33.7k63762
asked May 5 '17 at 16:21
Sklert
177411
edited Nov 11 at 22:15
max66
33.7k63762
edited Nov 11 at 22:15
max66
33.7k63762
edited Nov 11 at 22:15
max66
33.7k63762
33.7k63762
asked May 5 '17 at 16:21
Sklert
177411
asked May 5 '17 at 16:21
Sklert
177411
asked May 5 '17 at 16:21
Sklert
177411
177411
4
Why do you need to even do this? Why nottemplate<typename T> void foo(T input, size_t n) { ...}?
– NathanOliver
May 5 '17 at 16:23
std::conditionalis the right choice. Did you look in the docs to see how to use it? If you don't have access to C++11, you should be able to implement it yourself (it's documented if you need help).
– Rakete1111
May 5 '17 at 16:24
usestd::conditionallike this:std::conditional<some_bool, type_true, type_false>::type
– qxz
May 5 '17 at 16:25
@NathanOliver,twois used in function body. Type of input depends on bool, so I want to reduce template arguments
– Sklert
May 5 '17 at 16:29
3
std::conditional_t</*..*/>in C++14,typename std::conditional</*..*/>::typefor C++11.
– Jarod42
May 5 '17 at 16:30
|
show 4 more comments
4
Why do you need to even do this? Why nottemplate<typename T> void foo(T input, size_t n) { ...}?
– NathanOliver
May 5 '17 at 16:23
std::conditionalis the right choice. Did you look in the docs to see how to use it? If you don't have access to C++11, you should be able to implement it yourself (it's documented if you need help).
– Rakete1111
May 5 '17 at 16:24
usestd::conditionallike this:std::conditional<some_bool, type_true, type_false>::type
– qxz
May 5 '17 at 16:25
@NathanOliver,twois used in function body. Type of input depends on bool, so I want to reduce template arguments
– Sklert
May 5 '17 at 16:29
3
std::conditional_t</*..*/>in C++14,typename std::conditional</*..*/>::typefor C++11.
– Jarod42
May 5 '17 at 16:30
4
4
Why do you need to even do this? Why not
template<typename T> void foo(T input, size_t n) { ...} ?– NathanOliver
May 5 '17 at 16:23
Why do you need to even do this? Why not
template<typename T> void foo(T input, size_t n) { ...} ?– NathanOliver
May 5 '17 at 16:23
std::conditional is the right choice. Did you look in the docs to see how to use it? If you don't have access to C++11, you should be able to implement it yourself (it's documented if you need help).– Rakete1111
May 5 '17 at 16:24
std::conditional is the right choice. Did you look in the docs to see how to use it? If you don't have access to C++11, you should be able to implement it yourself (it's documented if you need help).– Rakete1111
May 5 '17 at 16:24
use
std::conditional like this: std::conditional<some_bool, type_true, type_false>::type– qxz
May 5 '17 at 16:25
use
std::conditional like this: std::conditional<some_bool, type_true, type_false>::type– qxz
May 5 '17 at 16:25
@NathanOliver,
two is used in function body. Type of input depends on bool, so I want to reduce template arguments– Sklert
May 5 '17 at 16:29
@NathanOliver,
two is used in function body. Type of input depends on bool, so I want to reduce template arguments– Sklert
May 5 '17 at 16:29
3
3
std::conditional_t</*..*/> in C++14, typename std::conditional</*..*/>::type for C++11.– Jarod42
May 5 '17 at 16:30
std::conditional_t</*..*/> in C++14, typename std::conditional</*..*/>::type for C++11.– Jarod42
May 5 '17 at 16:30
|
show 4 more comments
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
You only have to add a ::type and a typename
template <bool two>
// ......*typename*.....................................................*::type*
void foo (typename std::conditional<two, double*, std::complex<double>* >::type input, size_t n)
{
for (size_t i = 0; i < n; ++n)
input[i] *= two ? 2.0 : 1.0;
}
answered May 5 '17 at 16:27
max66
33.7k63762
Thank you! It works! I will mark this answer after time limit =)
– Sklert
May 5 '17 at 16:32
@Sklert in C++14, you can instead just change your code tostd::conditional_tand not add thetypenameand::type.
– Yakk - Adam Nevraumont
May 5 '17 at 17:18
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You only have to add a ::type and a typename
template <bool two>
// ......*typename*.....................................................*::type*
void foo (typename std::conditional<two, double*, std::complex<double>* >::type input, size_t n)
{
for (size_t i = 0; i < n; ++n)
input[i] *= two ? 2.0 : 1.0;
}
answered May 5 '17 at 16:27
max66
33.7k63762
Thank you! It works! I will mark this answer after time limit =)
– Sklert
May 5 '17 at 16:32
@Sklert in C++14, you can instead just change your code tostd::conditional_tand not add thetypenameand::type.
– Yakk - Adam Nevraumont
May 5 '17 at 17:18
add a comment |
up vote
2
down vote
accepted
You only have to add a ::type and a typename
template <bool two>
// ......*typename*.....................................................*::type*
void foo (typename std::conditional<two, double*, std::complex<double>* >::type input, size_t n)
{
for (size_t i = 0; i < n; ++n)
input[i] *= two ? 2.0 : 1.0;
}
answered May 5 '17 at 16:27
max66
33.7k63762
Thank you! It works! I will mark this answer after time limit =)
– Sklert
May 5 '17 at 16:32
@Sklert in C++14, you can instead just change your code tostd::conditional_tand not add thetypenameand::type.
– Yakk - Adam Nevraumont
May 5 '17 at 17:18
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You only have to add a ::type and a typename
template <bool two>
// ......*typename*.....................................................*::type*
void foo (typename std::conditional<two, double*, std::complex<double>* >::type input, size_t n)
{
for (size_t i = 0; i < n; ++n)
input[i] *= two ? 2.0 : 1.0;
}
answered May 5 '17 at 16:27
max66
33.7k63762
You only have to add a ::type and a typename
template <bool two>
// ......*typename*.....................................................*::type*
void foo (typename std::conditional<two, double*, std::complex<double>* >::type input, size_t n)
{
for (size_t i = 0; i < n; ++n)
input[i] *= two ? 2.0 : 1.0;
}
answered May 5 '17 at 16:27
max66
33.7k63762
answered May 5 '17 at 16:27
max66
33.7k63762
answered May 5 '17 at 16:27
max66
33.7k63762
answered May 5 '17 at 16:27
max66
33.7k63762
33.7k63762
Thank you! It works! I will mark this answer after time limit =)
– Sklert
May 5 '17 at 16:32
@Sklert in C++14, you can instead just change your code tostd::conditional_tand not add thetypenameand::type.
– Yakk - Adam Nevraumont
May 5 '17 at 17:18
add a comment |
Thank you! It works! I will mark this answer after time limit =)
– Sklert
May 5 '17 at 16:32
@Sklert in C++14, you can instead just change your code tostd::conditional_tand not add thetypenameand::type.
– Yakk - Adam Nevraumont
May 5 '17 at 17:18
Thank you! It works! I will mark this answer after time limit =)
– Sklert
May 5 '17 at 16:32
Thank you! It works! I will mark this answer after time limit =)
– Sklert
May 5 '17 at 16:32
@Sklert in C++14, you can instead just change your code to
std::conditional_t and not add the typename and ::type.– Yakk - Adam Nevraumont
May 5 '17 at 17:18
@Sklert in C++14, you can instead just change your code to
std::conditional_t and not add the typename and ::type.– Yakk - Adam Nevraumont
May 5 '17 at 17:18
add a comment |
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4
Why do you need to even do this? Why not
template<typename T> void foo(T input, size_t n) { ...}?– NathanOliver
May 5 '17 at 16:23
std::conditionalis the right choice. Did you look in the docs to see how to use it? If you don't have access to C++11, you should be able to implement it yourself (it's documented if you need help).– Rakete1111
May 5 '17 at 16:24
use
std::conditionallike this:std::conditional<some_bool, type_true, type_false>::type– qxz
May 5 '17 at 16:25
@NathanOliver,
twois used in function body. Type of input depends on bool, so I want to reduce template arguments– Sklert
May 5 '17 at 16:29
3
std::conditional_t</*..*/>in C++14,typename std::conditional</*..*/>::typefor C++11.– Jarod42
May 5 '17 at 16:30