How could my code tell a compile-time constant versus a variable?

Question

Here\'s my problem. I have a BINARY_FLAG macro:

#define BINARY_FLAG( n ) ( static_cast( 1 << ( n ) ) )

Whic

Answer 1

This is simpler than you think :)

Let's have a look:

#include <cassert>

static inline int FLAG(int n) {
    assert(n>=0 && n<32);
    return 1<<n;
}

int test1(int n) {
    return FLAG(n);
}
int test2() {
    return FLAG(5);
}

I don't use MSVC, but I compiled with Mingw GCC 4.5:

g++ -c -S -O3 08042.cpp

The resulting code for first method looks like:

__Z5test1i:
    pushl   %ebp
    movl    %esp, %ebp
    subl    $24, %esp
    movl    8(%ebp), %ecx
    cmpl    $31, %ecx
    ja  L4
    movl    $1, %eax
    sall    %cl, %eax
    leave
    ret
L4:
    movl    $4, 8(%esp)
    movl    $LC0, 4(%esp)
    movl    $LC1, (%esp)
    call    __assert
    .p2align 2,,3

And the second:

__Z5test2v:
    pushl   %ebp
    movl    %esp, %ebp
    movl    $32, %eax
    leave
    ret

See? The compiler is smart enough to do it for you. No need for macros, no need for metaprogramming, no need for C++0x. As simple as that.

Check if MSVC does the same... But look - it's really easy for the compiler to evaluate a constant expression and drop the unused conditional branch. Check it if you want to be sure... But generally - trust your tools.

Answer 2

I suggest you use two macros. BINARY_FLAG CONST_BINARY_FLAG That will make your code easier to grasp for others. You do know, at the time of writing, if it is a const or not. And I would in no case worry about runtime overhead. Your optimizer, at least in VS, will sort that out for you.

Answer 3

It's not possible to pass an argument to a macro or function and determine if it's compile time constant or a variable.

The best way is that you #define BINARY_FLAG(n) with compile time code and place that macro everywhere and then compile it. You will receive compiler-errors at the places where n is going to be runtime. Now, you can replace those macros with your runtime macro BINARY_FLAG_RUNTIME(n). This is the only feasible way.