Determining if an integer is a power of 2

Method from Bit Twiddling Hacks: (x != 0) && (x & (x-1) == 0). GCC compiles this to following code:

; input/ouput: eax
; destroys: ebx

        test    %eax,  %eax     ; x == 0?
        jz      1f

        leal -1(%eax), %ebx     ; ebx := x-1
        test    %eax,  %ebx     ; ZF  := (eax & ebx == 0)

        setz     %al
        movzx    %al, %eax       ; eax := ZF
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We can use also BSF and BSR instructions, that determines position of first and last bit=1. If number is power of 2, then just one bit is set, and thus these positions are equal. BSx sets also ZF flag if input value is zero.

; input/output: eax
; destroys: ebx, edx

        bsf     %eax, %ebx      ; ebx := LSB's position if eax != 0, ZF = 1 if eax = 0
        jz      1f
        bsr     %eax, %edx      ; edx := MSB's position

        cmp     %ebx, %edx      ; ZF  := (ebx = edx)

        setz    %al
        movzx   %al, %eax       ; eax := ZF
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