Jas*_*n R 15 c++ x86 assembly compiler-optimization
我有一些模板密集的C++代码,我想确保编译器尽可能地优化,因为它在编译时有大量的信息.为了评估它的性能,我决定看一下它生成的目标文件的反汇编.以下是我从中得到的片段objdump -dC
:
0000000000000000 <bar<foo, 0u>::get(bool)>:
0: 41 57 push %r15
2: 49 89 f7 mov %rsi,%r15
5: 41 56 push %r14
7: 41 55 push %r13
9: 41 54 push %r12
b: 55 push %rbp
c: 53 push %rbx
d: 48 81 ec 68 02 00 00 sub $0x268,%rsp
14: 48 89 7c 24 10 mov %rdi,0x10(%rsp)
19: 48 89 f7 mov %rsi,%rdi
1c: 89 54 24 1c mov %edx,0x1c(%rsp)
20: e8 00 00 00 00 callq 25 <bar<foo, 0u>::get(bool)+0x25>
25: 84 c0 test %al,%al
27: 0f 85 eb 00 00 00 jne 118 <bar<foo, 0u>::get(bool)+0x118>
2d: 48 c7 44 24 08 00 00 movq $0x0,0x8(%rsp)
34: 00 00
36: 4c 89 ff mov %r15,%rdi
39: 4d 8d b7 30 01 00 00 lea 0x130(%r15),%r14
40: e8 00 00 00 00 callq 45 <bar<foo, 0u>::get(bool)+0x45>
45: 84 c0 test %al,%al
47: 88 44 24 1b mov %al,0x1b(%rsp)
4b: 0f 85 ef 00 00 00 jne 140 <bar<foo, 0u>::get(bool)+0x140>
51: 80 7c 24 1c 00 cmpb $0x0,0x1c(%rsp)
56: 0f 85 24 03 00 00 jne 380 <bar<foo, 0u>::get(bool)+0x380>
5c: 48 8b 44 24 10 mov 0x10(%rsp),%rax
61: c6 00 00 movb $0x0,(%rax)
64: 80 7c 24 1b 00 cmpb $0x0,0x1b(%rsp)
69: 75 25 jne 90 <bar<foo, 0u>::get(bool)+0x90>
6b: 48 8b 74 24 10 mov 0x10(%rsp),%rsi
70: 4c 89 ff mov %r15,%rdi
73: e8 00 00 00 00 callq 78 <bar<foo, 0u>::get(bool)+0x78>
78: 48 8b 44 24 10 mov 0x10(%rsp),%rax
7d: 48 81 c4 68 02 00 00 add $0x268,%rsp
84: 5b pop %rbx
85: 5d pop %rbp
86: 41 5c pop %r12
88: 41 5d pop %r13
8a: 41 5e pop %r14
8c: 41 5f pop %r15
8e: c3 retq
8f: 90 nop
90: 4c 89 f7 mov %r14,%rdi
93: e8 00 00 00 00 callq 98 <bar<foo, 0u>::get(bool)+0x98>
98: 83 f8 04 cmp $0x4,%eax
9b: 74 f3 je 90 <bar<foo, 0u>::get(bool)+0x90>
9d: 85 c0 test %eax,%eax
9f: 0f 85 e4 08 00 00 jne 989 <bar<foo, 0u>::get(bool)+0x989>
a5: 49 83 87 b0 01 00 00 addq $0x1,0x1b0(%r15)
ac: 01
ad: 49 8d 9f 58 01 00 00 lea 0x158(%r15),%rbx
b4: 48 89 df mov %rbx,%rdi
b7: e8 00 00 00 00 callq bc <bar<foo, 0u>::get(bool)+0xbc>
bc: 49 8d bf 80 01 00 00 lea 0x180(%r15),%rdi
c3: e8 00 00 00 00 callq c8 <bar<foo, 0u>::get(bool)+0xc8>
c8: 48 89 df mov %rbx,%rdi
cb: e8 00 00 00 00 callq d0 <bar<foo, 0u>::get(bool)+0xd0>
d0: 4c 89 f7 mov %r14,%rdi
d3: e8 00 00 00 00 callq d8 <bar<foo, 0u>::get(bool)+0xd8>
d8: 83 f8 04 cmp $0x4,%eax
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这个特定函数的反汇编继续进行,但我注意到的一件事就是call
像这样的指令数量相对较多:
20: e8 00 00 00 00 callq 25 <bar<foo, 0u>::get(bool)+0x25>
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这些指令始终与操作码一起在e8 00 00 00 00
整个生成的代码中频繁出现,而且据我所知,只不过是无操作; 他们似乎都只是接受了下一个指令.这就引出了一个问题,那么,为什么生成所有这些指令有充分的理由呢?
我担心生成的代码的指令缓存占用量,因此在整个函数中多次浪费5个字节似乎适得其反.它看起来有点重量级nop
,除非编译器试图保留某种内存对齐或其他东西.如果是这样的话,我不会感到惊讶.
我使用g ++ 4.8.5编译了我的代码-O3 -fomit-frame-pointer
.为了它的价值,我看到了使用clang 3.7的类似代码生成.
Pas*_*uoq 23
的00 00 00 00
在(相对)目标地址e8 00 00 00 00
是为了由连接器来填充.这并不意味着电话会通过.它只是意味着您正在拆卸尚未链接的目标文件.
另外,调用下一条指令,如果这是链接阶段之后的最终结果,则不会是无操作,因为它会更改堆栈(某些暗示这不是您的情况).