The ambiguous star operator on function name

Question

Just a minimum working example:

#include <stdio.h>

void foo(char* str)
{
    printf("%s\n", str);
}

main()
{
   foo("foo");
   (* foo)("* foo");
}

which outputs

foo
* foo

I thought the function name should be the address of the routine in the code segment, so the star operator should return an executable instruction. But how does this operator work here actually? Thanks for shedding some light.

May as well toss `(&foo)("& foo")` in there while you're at it. — WhozCraig, Sep 16 '14 at 08:00

unwind · Accepted Answer · 2014-09-16T08:28:27.930

0

In C, functions are not values, you cannot "dereference a pointer to a function" and get something (a value) that makes sense.

Therefore, trying to dereference and call such a pointer has no additional effect, it's the same as calling through the pointer in the first place.

edited Sep 16 '14 at 08:28

answered Sep 16 '14 at 08:00

unwind

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@Angew Thanks, I clarified that I was mainly talking about C. – unwind Sep 16 '14 at 08:28
C++, on the other hand, *does* differentiate between a pointer to function and a reference to function. You get the latter by dereferencing the former. It's just that operator `()` is defined for pointers to functions, and an implicit conversion from (reference to) function to pointer to function exists. – Angew is no longer proud of SO Sep 16 '14 at 08:32
Reworded the comment accordingly :-) – Angew is no longer proud of SO Sep 16 '14 at 08:33

The ambiguous star operator on function name

1 Answers1