Open HowardHinnant opened 8 years ago
Free code:
// src/gsfxx/util.cc double TimePointToSeconds(system_clock::time_point time_point) { return duration<double>{time_point.time_since_epoch()}.count(); }
What you have is correct, but this has <chrono> do the conversion, and is actually smaller and a little more efficient:
<chrono>
.cfi_def_cfa_register %rbp cvtsi2sdq %rdi, %xmm0 divsd LCPI0_0(%rip), %xmm0 popq %rbp req
vs:
.cfi_def_cfa_register %rbp imulq $1000, %rdi, %rcx ## imm = 0x3E8 movabsq $4835703278458516699, %rdx ## imm = 0x431BDE82D7B634DB movq %rdi, %rax imulq %rdx movq %rdx, %rax shrq $63, %rax sarq $18, %rdx addq %rax, %rdx cvtsi2sdq %rdx, %xmm1 cvtsi2sdq %rcx, %xmm2 movsd LCPI0_0(%rip), %xmm0 ## xmm0 = mem[0],zero mulsd %xmm1, %xmm0 addsd %xmm2, %xmm0 mulsd LCPI0_1(%rip), %xmm0 addsd %xmm1, %xmm0 popq %rbp req
Explanation: duration<double> is a double-based duration representing seconds. The proposed code gets the duration out of time_point and converts it to double-based seconds, and then retrieves the double with .count().
duration<double>
time_point
.count()
Free code:
What you have is correct, but this has
<chrono>do the conversion, and is actually smaller and a little more efficient:vs:
Explanation:
duration<double>is a double-based duration representing seconds. The proposed code gets the duration out oftime_pointand converts it to double-based seconds, and then retrieves the double with.count().