Confusion about auto keyword in C++ [duplicate]

Question

I'm confused by the following piece of code:

#include <iostream>
using namespace std;

int *foo()
{
   //Operation
}

int main ()
{
        auto int ret = foo();
}

I compiled the above code under GCC, but I got the following error:

error: two or more data types in declaration of 'ret'
         auto int ret = foo();

But, if I remove int type, like so:

auto ret = foo();

then it runs successfully.

auto is a storage class and int is a data type, then why do I get the error "two or more data types" in the first case?

Answer 1

auto is not a storage class. It used to be, before C++11. But it was completely useless, so the keyword was re-purposed to allow automatic type inference. So when you say:

auto int ret = foo();

You are basically declaring the object to have 2 types (or possibly the same type twice), and that is an error. And when you say:

auto ret = foo();

The type of ret is determined by whatever the function foo returns, which is int* in this case.

Answer 2

auto is a storage class

That used to be true before C++11 but not any more.

Starting from C++11, the meaning of the word has been changed. It is now used to automatically deduce types. See http://www.stroustrup.com/C++11FAQ.html#auto.

Why in first case I got an error "two or more data types"?

By using

auto int ret = foo();

you are trying to specify two types for ret -- one deduced and the other explicitly specified.

If you want to use an explicitly specified type, you can use:

int ret = *foo(); // Since foo() returns a pointer.

or

int* ret = foo();

or you can let the compiler deduce the type by using:

auto ret = foo();  

Answer 3

auto is NOT a storage class (not anymore since C++11) C++11 brings the keyword allowing the compiler to infer what is the type required for the variable you are declaring.

so basically doing auto int myVar is as invalid as string double myVar2 or bool long myVar3... the variable can have only one data type defining it, and in your case the keyword auto does that...

how to get rid off the error:

Remove the int type and just use auto, doing so will let the compiler to *AUTO***matically infer that type of the variable **ret is exactly that from what foo() returns :) just genial!

auto ret = foo();

from the doc:

For variables, specifies that the type of the variable that is being declared will be automatically deduced from its initializer. For functions, specifies that the return type is a trailing return type or will be deduced from its return statements (since C++14). for non-type template parameters, specifies that the type will be deduced from the argument

Answer 4

You wrote that:

auto is a storage class

but this is no more true in C++11 (or later). The auto keyword has been reused for something completely different (some limited kind of type inference). The former C++03 or C99 auto storage class is now (in C++11) always implicit and should not use the auto keyword.

^{(if you like type inference, C++11 is not doing it very well, but C++14 or C++17 have progressed on it; Ocaml has a much more powerful and interesting Hindley-Milner type inference which is more "global"; that is why I wrote "some limited kind")}