CODEWITHSUNDEEP
c++functiongccc++14auto
msc
msc

Reputation: 34608

Use of 'auto func(int)' before deduction of 'auto' in C++14

I have compiled following program in GCC using C++14. 

#include <iostream>
using namespace std;

auto func(int i);

int main() 
{
    auto ret = func(5);
    return 0;
}

auto func(int i) 
{
  if (i == 1)
    return i;              
  else
    return func(i-1) + i;  
}

But, I get the following error.

In function 'int main()': 8:16: error: use of 'auto func(int)' before
deduction of 'auto'
 auto ret = func(5);

So, what am I missing here?

Upvotes: 40

Views: 32422

Answers (4)

Alexis Wilke
Alexis Wilke

Reputation: 20731

In your example, there is really no reason that you could not just move the implementation of the function before main():

#include <iostream>
using namespace std;  // you should avoid this, too

auto func(int i)
{
  if (i == 1)
    return i;
  else
    return func(i-1) + i;
}

int main()
{
    auto ret = func(5);
    return 0;
}

Otherwise you just can't use the auto keyword. In particular, you can't use auto in a recursive function which does not return anything. You have to use void. And that applies to lambda functions. For example:

int main()
{
    auto func = [](int i)
    {
        // ... do something with `i` ...
        if(i > 0)
        {
            func(i - 1);   // <-- same error here
        }
    }

    auto ret(func(5));
    return 0;
}

The call func(i - 1) has a problem. To fix it you have to replace the auto by the actual type:

int main()
{
    typedef std::function<void(int)> func_t;
    func_t func = [](int i)
    {
    ...

If you really want a function which supports varying return types you want to use a template anyway, not auto. This is really only to help you with less typing, not so much as a way to allow "any type".

Upvotes: 1

Griwes
Griwes

Reputation: 9029

This is [dcl.spec.auto/11]:

If the type of an entity with an undeduced placeholder type is needed to determine the type of an expression, the program is ill-formed. Once a non-discarded return statement has been seen in a function, however, the return type deduced from that statement can be used in the rest of the function, including in other return statements. [ Example:

auto n = n;                     // error, n's type is unknown
auto f();
void g() { &f; }                // error, f's return type is unknown
auto sum(int i) {
  if (i == 1)
    return i;                   // sum's return type is int
  else
    return sum(i-1)+i;          // OK, sum's return type has been deduced
}

 — end example ]

To translate this into English: the compiler needs to know the return type before you can use the function. In case of auto used like this, this is typically achieved by moving the definition before the point of use. If you don't actually need to use return type deduction, you can keep the definition after the use if you provide a signature, including the return type, in the declaration.

Upvotes: 48

songyuanyao
songyuanyao

Reputation: 172934

When auto is used as the return type in a function declaration that does not use the trailing return type syntax, the keyword auto indicates that the return type will be deduced from the operand of its return statement. That means the deduction can't be performed until the definition of the function func(), but before that it has been used in main().

You could move the definition before main(), or use trailing return type syntax to specify the return type on the declaration.

Upvotes: 6

Bartek Banachewicz
Bartek Banachewicz

Reputation: 39380

Clang has a much better error message for that one:

main.cpp:8:16: error: function 'func' with deduced return type cannot be used before it is defined
    auto ret = func(5);
               ^

I guess that's self-explanatory.

Upvotes: 35

Related Questions