为什么不看看您实际上得到了什么?
这是C#中的简单代码块:
static void Main(string[] args)
{
int i = 0;
try
{
i = 1;
Console.WriteLine(i);
return;
}
finally
{
Console.WriteLine("finally.");
}
}
这是调试版本中生成的IL:
.method private hidebysig static void Main(string[] args) cil managed
{
.entrypoint
.maxstack 1
.locals init ([0] int32 i)
L_0000: nop
L_0001: ldc.i4.0
L_0002: stloc.0
L_0003: nop
L_0004: ldc.i4.1
L_0005: stloc.0
L_0006: ldloc.0
L_0007: call void [mscorlib]System.Console::WriteLine(int32)
L_000c: nop
L_000d: leave.s L_001d
L_000f: nop
L_0010: ldstr "finally."
L_0015: call void [mscorlib]System.Console::WriteLine(string)
L_001a: nop
L_001b: nop
L_001c: endfinally
L_001d: nop
L_001e: ret
.try L_0003 to L_000f finally handler L_000f to L_001d
}
这是在调试中运行时由JIT生成的程序集:
00000000 push ebp
00000001 mov ebp,esp
00000003 push edi
00000004 push esi
00000005 push ebx
00000006 sub esp,34h
00000009 mov esi,ecx
0000000b lea edi,[ebp-38h]
0000000e mov ecx,0Bh
00000013 xor eax,eax
00000015 rep stos dword ptr es:[edi]
00000017 mov ecx,esi
00000019 xor eax,eax
0000001b mov dword ptr [ebp-1Ch],eax
0000001e mov dword ptr [ebp-3Ch],ecx
00000021 cmp dword ptr ds:[00288D34h],0
00000028 je 0000002F
0000002a call 59439E21
0000002f xor edx,edx
00000031 mov dword ptr [ebp-40h],edx
00000034 nop
int i = 0;
00000035 xor edx,edx
00000037 mov dword ptr [ebp-40h],edx
try
{
0000003a nop
i = 1;
0000003b mov dword ptr [ebp-40h],1
Console.WriteLine(i);
00000042 mov ecx,dword ptr [ebp-40h]
00000045 call 58DB2EA0
0000004a nop
return;
0000004b nop
0000004c mov dword ptr [ebp-20h],0
00000053 mov dword ptr [ebp-1Ch],0FCh
0000005a push 4E1584h
0000005f jmp 00000061
}
finally
{
00000061 nop
Console.WriteLine("finally.");
00000062 mov ecx,dword ptr ds:[036E2088h]
00000068 call 58DB2DB4
0000006d nop
}
0000006e nop
0000006f pop eax
00000070 jmp eax
00000072 nop
}
00000073 nop
00000074 lea esp,[ebp-0Ch]
00000077 pop ebx
00000078 pop esi
00000079 pop edi
0000007a pop ebp
0000007b ret
0000007c mov dword ptr [ebp-1Ch],0
00000083 jmp 00000072
现在,如果我注释掉try,finally和return,我将从JIT获得几乎相同的程序集。您将看到的差异是跳转到finally块中,以及一些代码来确定在执行finally之后要去哪里。所以您在谈论TINY的差异。在发行版中,进入finally的跳转将得到优化-大括号是nop指令,因此这将成为到下一条指令(也是nop)的跳转-这是一个简单的窥孔优化。流行eax然后是jmp eax同样便宜。
{
00000000 push ebp
00000001 mov ebp,esp
00000003 push edi
00000004 push esi
00000005 push ebx
00000006 sub esp,34h
00000009 mov esi,ecx
0000000b lea edi,[ebp-38h]
0000000e mov ecx,0Bh
00000013 xor eax,eax
00000015 rep stos dword ptr es:[edi]
00000017 mov ecx,esi
00000019 xor eax,eax
0000001b mov dword ptr [ebp-1Ch],eax
0000001e mov dword ptr [ebp-3Ch],ecx
00000021 cmp dword ptr ds:[00198D34h],0
00000028 je 0000002F
0000002a call 59549E21
0000002f xor edx,edx
00000031 mov dword ptr [ebp-40h],edx
00000034 nop
int i = 0;
00000035 xor edx,edx
00000037 mov dword ptr [ebp-40h],edx
i = 1;
0000003a mov dword ptr [ebp-40h],1
Console.WriteLine(i);
00000041 mov ecx,dword ptr [ebp-40h]
00000044 call 58EC2EA0
00000049 nop
Console.WriteLine("finally.");
0000004a mov ecx,dword ptr ds:[034C2088h]
00000050 call 58EC2DB4
00000055 nop
}
00000056 nop
00000057 lea esp,[ebp-0Ch]
0000005a pop ebx
0000005b pop esi
0000005c pop edi
0000005d pop ebp
0000005e ret
因此,您说的是非常小的最终尝试成本。在这个问题上很少有问题域。如果您正在执行类似memcpy的操作,并在要复制的每个字节上进行尝试/最后操作,然后继续复制数百MB的数据,我可以认为这是一个问题,但是在大多数情况下?微不足道。