Is x >= 0 more efficient than x > -1?

Question

Doing a comparison in C++ with an int is x >= 0 more efficient than x > -1?

Answer 1

short answer: no.

longer answer to provide some educational insight: it depends entirely on your compiler, allthough i bet that every sane compiler creates identical code for the 2 expressions.

example code:

int func_ge0(int a) {
    return a >= 0;
}   

int func_gtm1(int a) {
    return a > -1; 
}

and then compile and compare the resulting assembler code:

   % gcc -S -O2 -fomit-frame-pointer foo.cc

yields this:

_Z8func_ge0i:
.LFB0:
    .cfi_startproc
    .cfi_personality 0x0,__gxx_personality_v0
    movl    4(%esp), %eax
    notl    %eax
    shrl    $31, %eax
    ret
    .cfi_endproc

vs.

_Z9func_gtm1i:
.LFB1:
    .cfi_startproc
    .cfi_personality 0x0,__gxx_personality_v0
    movl    4(%esp), %eax
    notl    %eax
    shrl    $31, %eax
    ret
    .cfi_endproc

(compiler: g++-4.4)

conclusion: don't try to outsmart the compiler, concentrate on algorithms and data structures, benchmark and profile real bottlenecks, if in doubt: check the output of the compiler.