Do I need to put constexpr after else-if?

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Inspired by this answer, I tried to copy and paste (and add testing in main()) this code:

template<typename T>
std::tuple<int, double> foo(T a) 
 if constexpr (std::is_same_v<int, T>)
 return a, 0.0;

 else if (std::is_same_v<double, T>)
 return 0, a;

 else
 return 0, 0.0;


int main()

 auto [x, y] = foo("");

 std::cout << x << " " << y;

This is very straightforward - if T is deduced as int, we want to return a tuple of [a, 0.0]. If T is deduced as double, we want to return a tuple of [0, a]. Otherwise, we want to return [0, 0.0].

As you can see, in the main() function, I am calling foo with const char* argument, which should result in x and y being 0. That is not the case.

While trying to compile it, I encountered a strange error:

error: could not convert '0, a' from '<brace-enclosed initializer list>' to 'std::tuple<int, double>'

And I was like what?. Why on earth would I want that... I specifically used std::is_same to enable return 0, a only when the type of a is deduced as double.

So I quickly ran to cppreference on if-constexpr. At the bottom of the page, above Notes, we can see this snippet of code:

extern int x; // no definition of x required
int f() 
if constexpr (true)
 return 0;
else if (x)
 return x;
else
 return -x;

I thought to myself oookay..? I can't really see what's wrong with the original code. They use the same syntax and semantics....

But I was curious. I was curious if maybe something odd (at that time) might fix that issue, so I changed the original code to:

template<typename T>
std::tuple<int, double> foo(T a) 
 if constexpr (std::is_same_v<int, T>)
 return a, 0.0;

 else if constexpr (std::is_same_v<double, T>) // notice the additional constexpr here
 return 0, a;

 else
 return 0, 0.0;


int main()

 auto [x, y] = foo("");

 std::cout << x << " " << y;

And voilÃƒÂ ! The code compiled and executed as expected. So, my question is - Do we need to put constexpr after every if statement in if-else block in these kind of situations? Or is it just my compiler? I am using GCC 7.3.

asked 1 hour ago

Fureeish

1,9951719

add a commentÂ |Â

up vote
6
down vote

favorite

Inspired by this answer, I tried to copy and paste (and add testing in main()) this code:

template<typename T>
std::tuple<int, double> foo(T a) 
 if constexpr (std::is_same_v<int, T>)
 return a, 0.0;

 else if (std::is_same_v<double, T>)
 return 0, a;

 else
 return 0, 0.0;


int main()

 auto [x, y] = foo("");

 std::cout << x << " " << y;

As you can see, in the main() function, I am calling foo with const char* argument, which should result in x and y being 0. That is not the case.

While trying to compile it, I encountered a strange error:

error: could not convert '0, a' from '<brace-enclosed initializer list>' to 'std::tuple<int, double>'

And I was like what?. Why on earth would I want that... I specifically used std::is_same to enable return 0, a only when the type of a is deduced as double.

So I quickly ran to cppreference on if-constexpr. At the bottom of the page, above Notes, we can see this snippet of code:

extern int x; // no definition of x required
int f() 
if constexpr (true)
 return 0;
else if (x)
 return x;
else
 return -x;

I thought to myself oookay..? I can't really see what's wrong with the original code. They use the same syntax and semantics....

But I was curious. I was curious if maybe something odd (at that time) might fix that issue, so I changed the original code to:

template<typename T>
std::tuple<int, double> foo(T a) 
 if constexpr (std::is_same_v<int, T>)
 return a, 0.0;

 else if constexpr (std::is_same_v<double, T>) // notice the additional constexpr here
 return 0, a;

 else
 return 0, 0.0;


int main()

 auto [x, y] = foo("");

 std::cout << x << " " << y;

asked 1 hour ago

Fureeish

1,9951719

add a commentÂ |Â

up vote
6
down vote

favorite

Inspired by this answer, I tried to copy and paste (and add testing in main()) this code:

template<typename T>
std::tuple<int, double> foo(T a) 
 if constexpr (std::is_same_v<int, T>)
 return a, 0.0;

 else if (std::is_same_v<double, T>)
 return 0, a;

 else
 return 0, 0.0;


int main()

 auto [x, y] = foo("");

 std::cout << x << " " << y;

As you can see, in the main() function, I am calling foo with const char* argument, which should result in x and y being 0. That is not the case.

While trying to compile it, I encountered a strange error:

error: could not convert '0, a' from '<brace-enclosed initializer list>' to 'std::tuple<int, double>'

And I was like what?. Why on earth would I want that... I specifically used std::is_same to enable return 0, a only when the type of a is deduced as double.

So I quickly ran to cppreference on if-constexpr. At the bottom of the page, above Notes, we can see this snippet of code:

extern int x; // no definition of x required
int f() 
if constexpr (true)
 return 0;
else if (x)
 return x;
else
 return -x;

I thought to myself oookay..? I can't really see what's wrong with the original code. They use the same syntax and semantics....

But I was curious. I was curious if maybe something odd (at that time) might fix that issue, so I changed the original code to:

template<typename T>
std::tuple<int, double> foo(T a) 
 if constexpr (std::is_same_v<int, T>)
 return a, 0.0;

 else if constexpr (std::is_same_v<double, T>) // notice the additional constexpr here
 return 0, a;

 else
 return 0, 0.0;


int main()

 auto [x, y] = foo("");

 std::cout << x << " " << y;

asked 1 hour ago

Fureeish

1,9951719

Inspired by this answer, I tried to copy and paste (and add testing in main()) this code:

template<typename T>
std::tuple<int, double> foo(T a) 
 if constexpr (std::is_same_v<int, T>)
 return a, 0.0;

 else if (std::is_same_v<double, T>)
 return 0, a;

 else
 return 0, 0.0;


int main()

 auto [x, y] = foo("");

 std::cout << x << " " << y;

As you can see, in the main() function, I am calling foo with const char* argument, which should result in x and y being 0. That is not the case.

While trying to compile it, I encountered a strange error:

error: could not convert '0, a' from '<brace-enclosed initializer list>' to 'std::tuple<int, double>'

And I was like what?. Why on earth would I want that... I specifically used std::is_same to enable return 0, a only when the type of a is deduced as double.

So I quickly ran to cppreference on if-constexpr. At the bottom of the page, above Notes, we can see this snippet of code:

extern int x; // no definition of x required
int f() 
if constexpr (true)
 return 0;
else if (x)
 return x;
else
 return -x;

I thought to myself oookay..? I can't really see what's wrong with the original code. They use the same syntax and semantics....

But I was curious. I was curious if maybe something odd (at that time) might fix that issue, so I changed the original code to:

template<typename T>
std::tuple<int, double> foo(T a) 
 if constexpr (std::is_same_v<int, T>)
 return a, 0.0;

 else if constexpr (std::is_same_v<double, T>) // notice the additional constexpr here
 return 0, a;

 else
 return 0, 0.0;


int main()

 auto [x, y] = foo("");

 std::cout << x << " " << y;

c++ if-statement c++17 if-constexpr

asked 1 hour ago

Fureeish

1,9951719

asked 1 hour ago

Fureeish

1,9951719

asked 1 hour ago

Fureeish

1,9951719

asked 1 hour ago

Fureeish

1,9951719

asked 1 hour ago

Fureeish

1,9951719

add a commentÂ |Â

1 Answer
1

active

oldest

votes

up vote
9
down vote

accepted

Do we need to put constexpr after every if statement in if-else block in these kind of situations?

Yes. The else-if block is a lie :), there are only if blocks and else blocks. This is how your code is seen by the compiler:

if constexpr (std::is_same_v<int, T>)
 return a, 0.0;
else // 
 if (std::is_same_v<double, T>)
 return 0, a;
 else
 return 0, 0.0;
//

else if (/*...*/) is just a formatting convention that everyone uses. As such, you can clearly see that the second constexpr is needed.

answered 1 hour ago

Rakete1111

32.2k973108

1

Thank you! So now it's time to correct the original piece of code that made me come across this. Gonna accept the answer asap
â€“Â Fureeish
1 hour ago

3

@Fureeish Also note the cppref's example is correct, the missing constexpr is there by design to show that you don't have to have a definition of x because the else block (see answer) is discarded.
â€“Â Rakete1111
1 hour ago

Question: Defining function foo as constexpr will make everything inside a constexpr or still all ifs should be decorated with constexpr?
â€“Â Marek R
1 hour ago

@MarekR The inner if's still need to be decorated. Just because foo is constexpr doesn't mean that everything inside the function is constexpr (as you can still call the function at runtime).
â€“Â Rakete1111
1 hour ago

Defining a function foo as constexpr will simply make it possible to be evaluated at compile time. You still need to use constexpr inside it to achieve the same goals
â€“Â Fureeish
1 hour ago

add a commentÂ |Â

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1 Answer
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active

oldest

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up vote
9
down vote

accepted

Do we need to put constexpr after every if statement in if-else block in these kind of situations?

Yes. The else-if block is a lie :), there are only if blocks and else blocks. This is how your code is seen by the compiler:

if constexpr (std::is_same_v<int, T>)
 return a, 0.0;
else // 
 if (std::is_same_v<double, T>)
 return 0, a;
 else
 return 0, 0.0;
//

else if (/*...*/) is just a formatting convention that everyone uses. As such, you can clearly see that the second constexpr is needed.

answered 1 hour ago

Rakete1111

32.2k973108

1

Thank you! So now it's time to correct the original piece of code that made me come across this. Gonna accept the answer asap
â€“Â Fureeish
1 hour ago

3

@Fureeish Also note the cppref's example is correct, the missing constexpr is there by design to show that you don't have to have a definition of x because the else block (see answer) is discarded.
â€“Â Rakete1111
1 hour ago

Question: Defining function foo as constexpr will make everything inside a constexpr or still all ifs should be decorated with constexpr?
â€“Â Marek R
1 hour ago

@MarekR The inner if's still need to be decorated. Just because foo is constexpr doesn't mean that everything inside the function is constexpr (as you can still call the function at runtime).
â€“Â Rakete1111
1 hour ago

Defining a function foo as constexpr will simply make it possible to be evaluated at compile time. You still need to use constexpr inside it to achieve the same goals
â€“Â Fureeish
1 hour ago

add a commentÂ |Â

up vote
9
down vote

accepted

Do we need to put constexpr after every if statement in if-else block in these kind of situations?

Yes. The else-if block is a lie :), there are only if blocks and else blocks. This is how your code is seen by the compiler:

if constexpr (std::is_same_v<int, T>)
 return a, 0.0;
else // 
 if (std::is_same_v<double, T>)
 return 0, a;
 else
 return 0, 0.0;
//

else if (/*...*/) is just a formatting convention that everyone uses. As such, you can clearly see that the second constexpr is needed.

answered 1 hour ago

Rakete1111

32.2k973108

1

Thank you! So now it's time to correct the original piece of code that made me come across this. Gonna accept the answer asap
â€“Â Fureeish
1 hour ago

3

@Fureeish Also note the cppref's example is correct, the missing constexpr is there by design to show that you don't have to have a definition of x because the else block (see answer) is discarded.
â€“Â Rakete1111
1 hour ago

Question: Defining function foo as constexpr will make everything inside a constexpr or still all ifs should be decorated with constexpr?
â€“Â Marek R
1 hour ago

@MarekR The inner if's still need to be decorated. Just because foo is constexpr doesn't mean that everything inside the function is constexpr (as you can still call the function at runtime).
â€“Â Rakete1111
1 hour ago

Defining a function foo as constexpr will simply make it possible to be evaluated at compile time. You still need to use constexpr inside it to achieve the same goals
â€“Â Fureeish
1 hour ago

add a commentÂ |Â

up vote
9
down vote

accepted

Do we need to put constexpr after every if statement in if-else block in these kind of situations?

Yes. The else-if block is a lie :), there are only if blocks and else blocks. This is how your code is seen by the compiler:

if constexpr (std::is_same_v<int, T>)
 return a, 0.0;
else // 
 if (std::is_same_v<double, T>)
 return 0, a;
 else
 return 0, 0.0;
//

else if (/*...*/) is just a formatting convention that everyone uses. As such, you can clearly see that the second constexpr is needed.

answered 1 hour ago

Rakete1111

32.2k973108

Do we need to put constexpr after every if statement in if-else block in these kind of situations?

Yes. The else-if block is a lie :), there are only if blocks and else blocks. This is how your code is seen by the compiler:

if constexpr (std::is_same_v<int, T>)
 return a, 0.0;
else // 
 if (std::is_same_v<double, T>)
 return 0, a;
 else
 return 0, 0.0;
//

else if (/*...*/) is just a formatting convention that everyone uses. As such, you can clearly see that the second constexpr is needed.

answered 1 hour ago

Rakete1111

32.2k973108

answered 1 hour ago

Rakete1111

32.2k973108

answered 1 hour ago

Rakete1111

32.2k973108

answered 1 hour ago

Rakete1111

32.2k973108

1

Thank you! So now it's time to correct the original piece of code that made me come across this. Gonna accept the answer asap
â€“Â Fureeish
1 hour ago

3

@Fureeish Also note the cppref's example is correct, the missing constexpr is there by design to show that you don't have to have a definition of x because the else block (see answer) is discarded.
â€“Â Rakete1111
1 hour ago

Question: Defining function foo as constexpr will make everything inside a constexpr or still all ifs should be decorated with constexpr?
â€“Â Marek R
1 hour ago

@MarekR The inner if's still need to be decorated. Just because foo is constexpr doesn't mean that everything inside the function is constexpr (as you can still call the function at runtime).
â€“Â Rakete1111
1 hour ago

Defining a function foo as constexpr will simply make it possible to be evaluated at compile time. You still need to use constexpr inside it to achieve the same goals
â€“Â Fureeish
1 hour ago

add a commentÂ |Â

1

Thank you! So now it's time to correct the original piece of code that made me come across this. Gonna accept the answer asap
â€“Â Fureeish
1 hour ago

3

@Fureeish Also note the cppref's example is correct, the missing constexpr is there by design to show that you don't have to have a definition of x because the else block (see answer) is discarded.
â€“Â Rakete1111
1 hour ago

Question: Defining function foo as constexpr will make everything inside a constexpr or still all ifs should be decorated with constexpr?
â€“Â Marek R
1 hour ago

@MarekR The inner if's still need to be decorated. Just because foo is constexpr doesn't mean that everything inside the function is constexpr (as you can still call the function at runtime).
â€“Â Rakete1111
1 hour ago

Defining a function foo as constexpr will simply make it possible to be evaluated at compile time. You still need to use constexpr inside it to achieve the same goals
â€“Â Fureeish
1 hour ago

Thank you! So now it's time to correct the original piece of code that made me come across this. Gonna accept the answer asap
â€“Â Fureeish
1 hour ago

@Fureeish Also note the cppref's example is correct, the missing constexpr is there by design to show that you don't have to have a definition of x because the else block (see answer) is discarded.
â€“Â Rakete1111
1 hour ago

Question: Defining function foo as constexpr will make everything inside a constexpr or still all ifs should be decorated with constexpr?
â€“Â Marek R
1 hour ago

@MarekR The inner if's still need to be decorated. Just because foo is constexpr doesn't mean that everything inside the function is constexpr (as you can still call the function at runtime).
â€“Â Rakete1111
1 hour ago

Defining a function foo as constexpr will simply make it possible to be evaluated at compile time. You still need to use constexpr inside it to achieve the same goals
â€“Â Fureeish
1 hour ago

add a commentÂ |Â

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