貌似是前者
return (a+b)*(a+b).method public hidebysig instance int32 Calc1(int32 a, int32 b) cil managed
{
.maxstack 8
L_0000: ldarg.1
L_0001: ldarg.2
L_0002: add
L_0003: ldarg.1
L_0004: ldarg.2
L_0005: add
L_0006: mul
L_0007: ret
}c=a+b; return c*c;
.method public hidebysig instance int32 Calc2(int32 a, int32 b) cil managed
{
.maxstack 2
.locals init (
[0] int32 c)
L_0000: ldarg.1
L_0001: ldarg.2
L_0002: add
L_0003: stloc.0
L_0004: ldloc.0
L_0005: ldloc.0
L_0006: mul
L_0007: ret
}IL指令都是8条 不过 JIT的反汇编差别就很大了
20: return (a + b) * (a + b);
00000000 55pushebp
00000001 8B ECmovebp,esp
00000003 83 EC 08 subesp,8
00000006 89 4D FC movdword ptr [ebp-4],ecx
00000009 89 55 F8 movdword ptr [ebp-8],edx
0000000c 83 3D 34 14 AD 03 00 cmpdword ptr ds:[03AD1434h],0
00000013 74 05je0000001A
00000015 E8 79 4E 35 76call76354E93
0000001a 8B 45 F8 moveax,dword ptr [ebp-8]
0000001d 03 45 08 addeax,dword ptr [ebp+8]
00000020 8B 55 F8 movedx,dword ptr [ebp-8]
00000023 03 55 08 addedx,dword ptr [ebp+8]
00000026 0F AF C2 imuleax,edx
00000029 8B E5movesp,ebp
0000002b 5Dpopebp
0000002c C2 04 00 ret4它很主动的用edx去保存临时结果...orz
25: int c = a + b;
00000000 55pushebp
00000001 8B ECmovebp,esp
00000003 83 EC 0C subesp,0Ch
00000006 89 4D FC movdword ptr [ebp-4],ecx
00000009 89 55 F8 movdword ptr [ebp-8],edx
0000000c 83 3D 34 14 AD 03 00 cmpdword ptr ds:[03AD1434h],0
00000013 74 05je0000001A
00000015 E8 39 4E 35 76call76354E53
0000001a 33 D2xoredx,edx
0000001c 89 55 F4 movdword ptr [ebp-0Ch],edx
0000001f 8B 45 F8 moveax,dword ptr [ebp-8]
00000022 03 45 08 addeax,dword ptr [ebp+8]
00000025 89 45 F4 movdword ptr [ebp-0Ch],eax
26: return c * c;
00000028 8B 45 F4 moveax,dword ptr [ebp-0Ch]
0000002b 0F AF 45 F4imuleax,dword ptr [ebp-0Ch]
0000002f 8B E5movesp,ebp
00000031 5Dpopebp
00000032 C2 04 00 ret4计算出a+b后又扔回了内存 就显得重复了..能直接mov到edx应该就和上面一样了...中间两句编译的也很奇怪...先对edx清0然后再mov回栈区 其实执行的是c=0初始化..
.net的安全性要求使得效率略微有些降低
以上调试环境 vs10 .net4 consoleApp release CPU是桌面版i3