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c++compiler-errorslanguage-lawyerfunction-declarationdefault-arguments

Function default argument value depending on argument name in C++

If one defines a new variable in C++, then the name of the variable can be used in the initialization expression, for example:

int x = sizeof(x);

And what about default value of a function argument? Is it allowed there to reference the argument by its name? For example:

void f(int y = sizeof(y)) {}

This function is accepted in Clang, but rejected in GCC with the error:

'y' was not declared in this scope

Demo: https://gcc.godbolt.org/z/YsvYnhjTb

Which compiler is right here?

Solution

  • According to the C++17 standard (11.3.6 Default arguments)

    9 A default argument is evaluated each time the function is called with no argument for the corresponding parameter. A parameter shall not appear as a potentially-evaluated expression in a default argument. Parameters of a function declared before a default argument are in scope and can hide namespace and class member name

    It provides the following example:

    int h(int a, int b = sizeof(a)); // OK, unevaluated operand

    So, this function declaration

    void f(int y = sizeof(y)) {}

    is correct because, in this expression sizeof(y), y is not an evaluated operand, based on C++17 8.3.3 Sizeof:

    1 The sizeof operator yields the number of bytes in the object representation of its operand. The operand is either an expression, which is an unevaluated operand (Clause 8), or a parenthesized type-id.

    and C++17 6.3.2 Point of declaration:

    1 The point of declaration for a name is immediately after its complete declarator (Clause 11) and before its initializer (if any), except as noted below.