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

Question

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?

Answer 1

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.

Answer 2

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.