Why would gcc movl to the same register?

Question

For this piece of C code:

uint64_t roundUp(uint64_t value, uint32_t blockSize) 
{
    return (value + blockSize - 1) & ~(blockSize - 1);
}

gcc 4.6 -O3 generated the following assembly:

roundUp(unsigned long, unsigned int):
.LFB0:
    .cfi_startproc
    movl    %esi, %edx
    movl    %esi, %esi
    leaq    -1(%rdi,%rsi), %rax
    negl    %edx
    andl    %edx, %eax
    ret
    .cfi_endproc

Could anyone tell me why would it want to do this?

movl    %esi, %esi

Answer 1

That clears the upper 32 bits. When you write to a 32 bit register in x86-64, the upper 32 bits are cleared automatically. Since esi contains a 32 bit parameter, the upper 32 bits could contain any value, so they need to be cleared before rsi can be used.