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What is the meaning of auto when using C++ trailing return type?

Tags:

c++

c++11

Instead of usual

void foo (void ) {
    cout << "Meaning of life: " << 42 << endl;
}

C++11 allows is an alternative, using the Trailing Return

auto bar (void) -> void {
    cout << "More meaning: " << 43 << endl;
}

In the latter - what is auto designed to represent?

Another example, consider function

auto func (int i) -> int (*)[10] {

}

Same question, what is the meaning of auto in this example?

like image 323
James Raitsev Avatar asked Sep 01 '12 02:09

James Raitsev


3 Answers

In general, the new keyword auto in C++11 indicates that the type of the expression (in this case the return type of a function) should be inferred from the result of the expression, in this case, what occurs after the ->.

Without it, the function would have no type (thus not being a function), and the compiler would end up confused.

like image 191
Richard J. Ross III Avatar answered Nov 17 '22 13:11

Richard J. Ross III


Consider the code:

template<typename T1, typename T2>
Tx Add(T1 t1, T2 t2)
{
    return t1+t2;
}

Here the return type depends on expression t1+t2, which in turn depends on how Add is called. If you call it as:

Add(1, 1.4);

T1 would be int, and T2 would be double. The resulting type is now double (int+double). And hence Tx should (must) be specified using auto and ->

 template<typename T1, typename T2>
    auto Add(T1 t1, T2 t2) -> decltype(t1+t2)
    {
        return t1+t2;
    }

You can read about it in my article.

like image 30
Ajay Avatar answered Nov 17 '22 12:11

Ajay


I think the answer is relatively straightforward, and not covered in other answers here.

Basically, without the auto, there are ambiguities, so The Committee decided "you need auto here to avoid those ambiguities".

class A {
    T B;
};

class B;

A* f();

f()->B;

Now, what does f()->B mean? Is it a function with no parameters that returns a B (with trailing return type syntax), or is it a function call to A* f()?

If we didn't have the auto required at the start of the trailing return type syntax, we wouldn't be able to tell. With the auto, it's clear that this is a function call.

So, to avoid the unclearness here, we simply require auto at the start of any function declaration that uses a trailing return type.

like image 2
CoffeeTableEspresso Avatar answered Nov 17 '22 12:11

CoffeeTableEspresso