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Subject: Does [expr.call]/1 correctly authorize the snippet below to compile?
From: J.A. Belloc (jabelloc_at_[hidden])
Date: 2019-10-10 08:08:21

[expr.call]/1 <http://eel.is/c++draft/expr.call#1> (emphasis is mine):

A function call is a postfix expression followed by parentheses containing
> a possibly empty, comma-separated list of *initializer-clause
> <http://eel.is/c++draft/dcl.init#nt:initializer-clause>*s which
> constitute the arguments to the function.
> <http://eel.is/c++draft/expr.call#1.sentence-1>

> The postfix expression shall have function type or function pointer type.
> <http://eel.is/c++draft/expr.call#1.sentence-2>
> For a call to a non-member function or to a static member function, the
> postfix expression shall either be an lvalue that refers to a function (in
> which case the function-to-pointer standard conversion ([conv.func]
> <http://eel.is/c++draft/conv.func>) is suppressed on the postfix
> expression), or *have function pointer type*.
> <http://eel.is/c++draft/expr.call#1.sentence-3>

My impression reading this paragraph is that the sentence "have a function
pointer type" is referring to an lvalue having function to pointer type,
not to a prvalue obtained from a function call returning a pointer to

Thus, I don't think [expr.call]/1 correctlly allows the following snippet
to compile (demo <http://coliru.stacked-crooked.com/a/c4a74e9721c05b48>):


void f(int a) {
    std::cout << "f(" << a << ")\n";

// g is a function of () returning a pointer to a function of (int)
returning void

void(*g())(int) {
    return f; // f is function of (int) returning void.

int main()
   g()(1); // The expression g() has type pointer to function, but is a


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