search for: negq

...the base being the last item and subtracting from the last item, via a negative induction variable. I imagine the result would look something like the following: movq %rsi, %rbx ; rbx becomes the value n leaq (%rdi, %rbx, 4), %r15 ; <base> = p + (n * 4) negq %rbx ; set rbx to be -n, so we count from -n to 0 LBB0_1: movl (%r15,%rbx,4), %edi callq g addq $1, %rbx jne .LBB0_1 I realize this is a micro-op saving a single cycle. But this reduces the instruction count, one less instr to decode in...

...gt; item and subtracting from the last item, via a negative induction variable. I > imagine the result would look something like the following: > movq %rsi, %rbx ; rbx becomes the value n > leaq (%rdi, %rbx, 4), %r15 ; <base> = p + (n * 4) > negq %rbx ; set rbx to be -n, so we count from -n to 0 > LBB0_1: > movl (%r15,%rbx,4), %edi > callq g > addq $1, %rbx > jne .LBB0_1 > > I realize this is a micro-op saving a single cycle. But this reduces the instruction...

...# =>This Inner Loop Header: Depth=1 movq -8(%rbp), %rcx movq %rsp, %rax addq $-16, %rax movq %rax, %rsp testq %rcx, %rcx jns .LBB8_3 # BB#2: # %abs_then # in Loop: Header=BB8_1 Depth=1 negq %rcx .LBB8_3: # %abs_else # in Loop: Header=BB8_1 Depth=1 movq %rcx, (%rax) movq (%rax), %rax movq %rax, -16(%rbp) movq -8(%rbp), %rax incq %rax movq %rax, -8(%rbp) cmpq $600001, %rax # imm = 0x927C1 jl .LBB8_1 # BB#...