Can a void-returning function g return f(); when f returns void?

别说谁变了你拦得住时间么 提交于 2019-11-30 08:00:35
haccks

Anything after return is an expression.

6.8.6:1 Jump statements

Syntax  

   ...
   return expressionopt; 

And standard says that:

A return statement with an expression shall not appear in a function whose return type is void. ....

f() is also an expression here. The compiler should raise a warning

[Warning] ISO C forbids 'return' with expression, in function returning void [-pedantic]

This clearly is a constraint violation, in particular in view of

6.3.2.2 void: The (nonexistent) value of a void expression (an expression that has type void) shall not be used in any way,

That means that the incomplete type void is a dead end that cannot be reused for any purpose whatsoever.

It clearly states A return statement without an expression shall only appear in a function whose return type is void, try and execute this:

void g()
{
    return; // does not return any expression at all
}
void f()
{
    return g();
}


int main(void) {
    f();
    return 0;
}
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