为了回答这个问题,我们将检查X86和X64 jit为每种情况生成的汇编代码。
X86,如果/那么
32: foreach (int i in array)
0000007c 33 D2 xor edx,edx
0000007e 83 7E 04 00 cmp dword ptr [esi+4],0
00000082 7E 1C jle 000000A0
00000084 8B 44 96 08 mov eax,dword ptr [esi+edx*4+8]
33: {
34: if (i > 0)
00000088 85 C0 test eax,eax
0000008a 7E 08 jle 00000094
35: {
36: value += 2;
0000008c 83 C3 02 add ebx,2
0000008f 83 D7 00 adc edi,0
00000092 EB 06 jmp 0000009A
37: }
38: else
39: {
40: value += 3;
00000094 83 C3 03 add ebx,3
00000097 83 D7 00 adc edi,0
0000009a 42 inc edx
32: foreach (int i in array)
0000009b 39 56 04 cmp dword ptr [esi+4],edx
0000009e 7F E4 jg 00000084
30: for (int x = 0; x < iterations; x++)
000000a0 41 inc ecx
000000a1 3B 4D F0 cmp ecx,dword ptr [ebp-10h]
000000a4 7C D6 jl 0000007C
X86,三元
59: foreach (int i in array)
00000075 33 F6 xor esi,esi
00000077 83 7F 04 00 cmp dword ptr [edi+4],0
0000007b 7E 2D jle 000000AA
0000007d 8B 44 B7 08 mov eax,dword ptr [edi+esi*4+8]
60: {
61: value += i > 0 ? 2 : 3;
00000081 85 C0 test eax,eax
00000083 7F 07 jg 0000008C
00000085 BA 03 00 00 00 mov edx,3
0000008a EB 05 jmp 00000091
0000008c BA 02 00 00 00 mov edx,2
00000091 8B C3 mov eax,ebx
00000093 8B 4D EC mov ecx,dword ptr [ebp-14h]
00000096 8B DA mov ebx,edx
00000098 C1 FB 1F sar ebx,1Fh
0000009b 03 C2 add eax,edx
0000009d 13 CB adc ecx,ebx
0000009f 89 4D EC mov dword ptr [ebp-14h],ecx
000000a2 8B D8 mov ebx,eax
000000a4 46 inc esi
59: foreach (int i in array)
000000a5 39 77 04 cmp dword ptr [edi+4],esi
000000a8 7F D3 jg 0000007D
57: for (int x = 0; x < iterations; x++)
000000aa FF 45 E4 inc dword ptr [ebp-1Ch]
000000ad 8B 45 E4 mov eax,dword ptr [ebp-1Ch]
000000b0 3B 45 F0 cmp eax,dword ptr [ebp-10h]
000000b3 7C C0 jl 00000075
X64,如果/那么
32: foreach (int i in array)
00000059 4C 8B 4F 08 mov r9,qword ptr [rdi+8]
0000005d 0F 1F 00 nop dword ptr [rax]
00000060 45 85 C9 test r9d,r9d
00000063 7E 2B jle 0000000000000090
00000065 33 D2 xor edx,edx
00000067 45 33 C0 xor r8d,r8d
0000006a 4C 8B 57 08 mov r10,qword ptr [rdi+8]
0000006e 66 90 xchg ax,ax
00000070 42 8B 44 07 10 mov eax,dword ptr [rdi+r8+10h]
33: {
34: if (i > 0)
00000075 85 C0 test eax,eax
00000077 7E 07 jle 0000000000000080
35: {
36: value += 2;
00000079 48 83 C5 02 add rbp,2
0000007d EB 05 jmp 0000000000000084
0000007f 90 nop
37: }
38: else
39: {
40: value += 3;
00000080 48 83 C5 03 add rbp,3
00000084 FF C2 inc edx
00000086 49 83 C0 04 add r8,4
32: foreach (int i in array)
0000008a 41 3B D2 cmp edx,r10d
0000008d 7C E1 jl 0000000000000070
0000008f 90 nop
30: for (int x = 0; x < iterations; x++)
00000090 FF C1 inc ecx
00000092 41 3B CC cmp ecx,r12d
00000095 7C C9 jl 0000000000000060
X64,三元
59: foreach (int i in array)
00000044 4C 8B 4F 08 mov r9,qword ptr [rdi+8]
00000048 45 85 C9 test r9d,r9d
0000004b 7E 2F jle 000000000000007C
0000004d 45 33 C0 xor r8d,r8d
00000050 33 D2 xor edx,edx
00000052 4C 8B 57 08 mov r10,qword ptr [rdi+8]
00000056 8B 44 17 10 mov eax,dword ptr [rdi+rdx+10h]
60: {
61: value += i > 0 ? 2 : 3;
0000005a 85 C0 test eax,eax
0000005c 7F 07 jg 0000000000000065
0000005e B8 03 00 00 00 mov eax,3
00000063 EB 05 jmp 000000000000006A
00000065 B8 02 00 00 00 mov eax,2
0000006a 48 63 C0 movsxd rax,eax
0000006d 4C 03 E0 add r12,rax
00000070 41 FF C0 inc r8d
00000073 48 83 C2 04 add rdx,4
59: foreach (int i in array)
00000077 45 3B C2 cmp r8d,r10d
0000007a 7C DA jl 0000000000000056
57: for (int x = 0; x < iterations; x++)
0000007c FF C1 inc ecx
0000007e 3B CD cmp ecx,ebp
00000080 7C C6 jl 0000000000000048
首先:为什么X86代码比X64慢这么多?
这是由于代码的以下特征:
X64有几个额外的寄存器可用,每个寄存器都是64位的。这允许X64 JIT完全使用寄存器来执行内部循环,而不是从数组中加载i,而X86 JIT在循环中放置了几个堆栈操作(内存访问)。
value是一个64位整数,在X86上需要2条机器指令(add和adc),但在X64上只需要1条机器指令(add)。
第二:为什么三元运算符在X86和X64上都比较慢?
This is due to a subtle difference in the order of operations impacting the JIT's optimizer. To JIT the ternary operator, rather than directly coding 2 and 3 in the add machine instructions themselves, the JIT creating an intermediate variable (in a register) to hold the result. This register is then sign-extended from 32-bits to 64-bits before adding it to value. Since all of this is performed in registers for X64, despite the significant increase in complexity for the ternary operator the net impact is somewhat minimized.
另一方面,X86 JIT受到更大程度的影响,因为在内部循环中添加一个新的中间值会导致它“溢出”另一个值,导致在内部循环中至少有2个额外的内存访问(参见X86三进制代码中对[ebp-14h]的访问)。