ANSI C function pointers aren't like other pointers

Is this C program valid? What does it do?

#include 

int main( int argc, char *argv[] ) {
    (*puts)( "foo" );
    (&puts)( "foo" );
    (***puts)( "foo" );
    (*&**&puts)( "foo" );
}

Surprisingly, this is perfectly valid C. The authors of the ANSI C specification decided there wasn't anything reasonable you could do with a function pointer other than dereference it, so there is no difference between a function and a function address. All of the above lines have exactly the same meaning, which we can see using the Compiler Explorer: https://godbolt.org/z/UmaKcM

main:                                   # @main
        push    rbp
        mov     rbp, rsp
        sub     rsp, 32
        movabs  rax, offset .L.str
        mov     dword ptr [rbp - 4], edi
        mov     qword ptr [rbp - 16], rsi
        mov     rdi, rax
        call    puts
        movabs  rdi, offset .L.str
        mov     dword ptr [rbp - 20], eax # 4-byte Spill
        call    puts
        movabs  rdi, offset .L.str
        mov     dword ptr [rbp - 24], eax # 4-byte Spill
        call    puts
        movabs  rdi, offset .L.str
        mov     dword ptr [rbp - 28], eax # 4-byte Spill
        call    puts
        xor     ecx, ecx
        mov     dword ptr [rbp - 32], eax # 4-byte Spill
        mov     eax, ecx
        add     rsp, 32
        pop     rbp
        ret
.L.str:
        .asciz  "foo"

The & and * operations have no effect at all on a function pointer. You can apply as many as you want.

Don't try this, though:

    (*&&*puts)( "foo" );

C's lexing rules mean that && is parsed as the logical AND operator, not two "address of" operators.

Source: https://www.quora.com/In-C-is-it-possible-to-pass-a-function-to-another-function/answer/Petr-Skocik