我写了一个简单的程序，在循环中执行一堆 NOP 指令，令我惊讶的是它每秒执行大约 10600000000 条指令，或大约 10Ghz，而我的 CPU 只有 2.2GHz。
这怎么可能？ CPU 是将它们视为单个 mega-NOP，还是我只是发现了“指令级并行性”的含义？
什么是每秒指令数的更好衡量标准？执行添加指令仅达到 414900000/s，我的 CPU 报告的 bogomips 的十分之一:4390.03
C 代码:
#include <stdio.h>
#include <stdint.h>
#include <time.h>
#define ten(a) a a a a a a a a a a
#define hundred(a) ten(a) ten(a) ten(a) ten(a) ten(a) ten(a) ten(a) \
ten(a) ten(a) ten(a)
#define ITER 10000000
int main(void) {
uint64_t i=0;
uint64_t t=time(NULL);
while(1) {
for(int j=0; j<ITER;j++) {
hundred(asm volatile ("nop");)
}
i+=ITER*100;
printf("%lu/%lu\n", i, time(NULL)-t);
}
return 0;
}
编译后的程序集:
.file "gbloopinc.c"
.section .rodata
.LC0:
.string "%lu/%lu\n"
.text
.globl main
.type main, @function
main:
.LFB0:
.cfi_startproc
pushq %rbp
.cfi_def_cfa_offset 16
.cfi_offset 6, -16
movq %rsp, %rbp
.cfi_def_cfa_register 6
subq $32, %rsp
movq $0, -16(%rbp)
movl $0, %edi
call time
movq %rax, -8(%rbp)
.L4:
movl $0, -20(%rbp)
jmp .L2
.L3:
#APP
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
# 15 "gbloopinc.c" 1
nop
# 0 "" 2
#NO_APP
addl $1, -20(%rbp)
.L2:
cmpl $9999999, -20(%rbp)
jle .L3
addq $1000000000, -16(%rbp)
movl $0, %edi
call time
subq -8(%rbp), %rax
movq %rax, %rdx
movq -16(%rbp), %rax
movq %rax, %rsi
movl $.LC0, %edi
movl $0, %eax
call printf
jmp .L4
.cfi_endproc
.LFE0:
.size main, .-main
.ident "GCC: (Ubuntu 5.4.0-6ubuntu1~16.04.2) 5.4.0 20160609"
.section .note.GNU-stack,"",@progbits
最佳答案
这与多核无关。核心不是“端口”。
每个时钟 4 个 NOP 是您的超标量/乱序 CPU 的问题/报废流水线宽度。 NOP 甚至不需要执行单元/执行端口(ALU 或加载或存储)，因此您甚至不受整数执行单元数量的限制。甚至 Core2(英特尔的第一个 4 宽 x86 CPU)也可以每个时钟运行 4 个 NOP。
如您所料，这是指令级并行的一个示例。 NOP 当然没有输入依赖性。
在您的 Sandybridge CPU(每个内核有 3 个 ALU 执行单元)上，您可以每个时钟运行 3 个 ADD 和一个加载或存储指令，因为它的流水线宽度是 4 微指令。参见 Agner Fog's microarch pdf和 x86 中的其他链接tag wiki .在独立的 ADD 指令流上，例如
add eax, eax
add ebx, ebx
add ecx, ecx
add edx, edx
...
您会在 SnB 上看到每个时钟吞吐量大约 3 个，在整数 ALU 执行端口上出现瓶颈。 Haswell 可以以每个时钟 4 个 ADD 的速度运行它，因为它有一个可以处理非 vector 整数操作(和分支)的第 4 个 ALU 执行端口。
无序 CPU 通常具有比执行单元数量更宽的前端和发布/退出宽度。一旦有空闲的执行单元，就会有更多的指令被解码并准备好执行，这会增加它们的利用率。否则，如果执行由于串行依赖性而停滞或减慢，则乱序机器只能提前看到当前正在执行的内容。 (例如 add eax,eax/add eax,eax 需要第一个 add 的输出作为第二个 add 的输入，因此每个时钟只能运行一个 insn。 )
关于我的 CPU 可以在每个 CPU 周期执行多个 NOP 吗？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/39643882/