Multiply by 0 optimization

Question

Suppose i have:

double f(const double *r) {     return 0*(r[0]*r[1]); } 

should compiler be able to optimize out the segment, or does it still have to perform operation, in case the values might be inf or nan?

gcc -O3 -S test.c:          .file   "test.c"         .text         .p2align 4,,15 .globl f         .type   f, @function f: .LFB0:         .cfi_startproc         movsd   (%rdi), %xmm0         mulsd   8(%rdi), %xmm0         mulsd   .LC0(%rip), %xmm0         ret         .cfi_endproc .LFE0:         .size   f, .-f         .section        .rodata.cst8,"aM",@progbits,8         .align 8 .LC0:         .long   0         .long   0         .ident  "GCC: (Ubuntu 4.4.3-4ubuntu5) 4.4.3"         .section        .note.GNU-stack,"",@progbits 

seems no elimination?

aha:

gcc -O3  -ffast-math  -S test.c          .file   "test.c"         .text         .p2align 4,,15 .globl f         .type   f, @function f: .LFB0:         .cfi_startproc         xorpd   %xmm0, %xmm0         ret         .cfi_endproc .LFE0:         .size   f, .-f         .ident  "GCC: (Ubuntu 4.4.3-4ubuntu5) 4.4.3"         .section        .note.GNU-stack,"",@progbits 

Answer 1

It isn't only inf and NaN that prevent the optimization there, it's also the sign - 0.0 * something negative is -0.0, otherwise it's 0.0, so you actually have to compute the sign of r[0]*r[1].