Array decay to pointers in templates

Question

Please consider this code:

#include 

template
void f(T x) {
    std::cout << sizeof(T) << \'\\n\';
}

int main         


        

Answer 1

The behaviour of this code is explained by C++14 [temp.deduct.call]:

Deducing template arguments from a function call

Template argument deduction is done by comparing each function template parameter type (call it P) with the type of the corresponding argument of the call (call it A) as described below

and then below:

If P is not a reference type:

• If A is an array type, the pointer type produced by the array-to-pointer standard conversion (4.2) is used in place of A for type deduction;

For the call f(array);, we have A = int[27]. A is an array type. So the deduced type T is int *, according to this last bullet point.

We can see from the qualifier "If P is not a reference type" that this behaviour could perhaps be avoided by making P a reference type. For the code:

template
void f(T (&x)[N])

the symbol P means T(&)[N], which is a reference type; and it turns out that there are no conversions applied here. T is deduced to int, with the type of x being int(&)[N].

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Note that this only applies to function templates where the type is deduced from the argument. The behaviour is covered by separate parts of the specification for explicitly-provided function template parameters, and class templates.