Function Pointers (ambigous syntax) [duplicate]

Question

In the following code:

/* mylog.c */
#include <stdio.h>
#include <stdlib.h> /* for atoi(3) */

int mylog10(int n)
{
    int log = 0;
    while (n > 0)
    {
        log++;
        n /= 10;
    }
    return log;
}

int mylog2(int n)
{
    int log = 0;
    while (n > 0)
    {
        log++;
        n >>= 1;
    }
    return log;
}

int main(int argc, const char* argv[])
{
    int (*logfunc)(int); /* function pointer */
    int n = 0, log;

    if (argc > 1)
    {
        n = atoi(argv[1]);
    }

    logfunc = &mylog10; /* is unary '&' operator needed? */

    log = logfunc(n);
    printf("%d\n", log);
    return 0;
}

in the line

logfunc = &mylog10;

I've noticed that the unary & (address of) operator is optional, and the program compiles and runs the same way either with or without it (in Linux with GCC 4.2.4). Why? Is this a compiler-specific issue, or perhaps two different language standards being accepted by the compiler? Thanks.

Answer 1

You are correct that the & is optional. Functions, like arrays, can be automatically converted into pointers. It's neither compiler-specific nor the result of different language standards. From the standard, Section 6.3.2.1, paragraph 4:

A function designator is an expression that has function type. Except when it is the operand of the sizeof operator or the unary & operator, a function designator with type "function returning type" is converted to an expression that has type "pointer to function returning type".

Answer 2

Operator & is indeed optional when taking the address of a function in your context (assigning it to something). It is not compiler-specific, it follows from the formal definition of the language.

Symmetrically, operator * is optional when invoking the function through a pointer. In your example, you could invoke the function as either (*logfunc)(n) or logfunc(n). You used the latter, but the former would work as well.