我在IL中犯了错误吗？我没看到？

Question

我正在使用编译器System.Reflection.Emit,我得到JIT限制错误,我无法弄清楚.问题出现在我的函数句柄实现中.即生成代码

function foo() { }
f = foo;
f();

由于我无法控制的规范,语言是动态类型的,所以我不知道f在编译时会有多少参数.为了解决这个问题,我生成了一个新方法,而不是发出一个Ldftnfor foo,?foo它接受一个在调用表达式中给出的参数数组,然后将它们推送到eval堆栈中foo.CLR允许这样做吗？

我现在得到的是"JIT遇到内部限制"异常(或者"CLR检测到无效程序",如果我保存程序集并运行它而不是从内存中调用它),堆栈跟踪显示它发生在?foo.这就是我正在生成的IL.

.method private instance class [MylibInterop]MylibInterop.MylibValue 
        '?foo'(class [MylibInterop]MylibInterop.MylibValue[] A_1) cil managed
{
  // Code size       90 (0x5a)
  .maxstack  10
  .locals init (int32 V_0,
           int32 V_1)
  IL_0000:  ldarg.1
  IL_0001:  call       instance int32 [mscorlib]System.Array::get_Length()
  IL_0006:  stloc.0
  IL_0007:  ldloc.0
  IL_0008:  ldc.i4     0x0
  IL_000d:  ble        IL_001d
  IL_0012:  ldstr      "Too many arguments to lambda call"
  IL_0017:  newobj     instance void [mscorlib]System.Exception::.ctor(string)
  IL_001c:  throw
  IL_001d:  ldarg.0
  IL_001e:  ldc.i4.0
  IL_001f:  stloc.1
  IL_0020:  ldloc.0
  IL_0021:  newobj     instance void [MylibInterop]MylibInterop.MylibValue::.ctor(int32)
  IL_0026:  ldloc.1
  IL_0027:  ldloc.0
  IL_0028:  bge        IL_003d
  IL_002d:  ldarg.1
  IL_002e:  ldloc.1
  IL_002f:  ldelem     [MylibInterop]MylibInterop.MylibValue
  IL_0034:  ldloc.1
  IL_0035:  ldc.i4.1
  IL_0036:  add
  IL_0037:  stloc.1
  IL_0038:  br         IL_0026
  IL_003d:  ldloc.0
  IL_003e:  stloc.1
  IL_003f:  ldloc.1
  IL_0040:  ldc.i4     0x0
  IL_0045:  bge        IL_0054
  IL_004a:  ldnull
  IL_004b:  ldloc.1
  IL_004c:  ldc.i4.1
  IL_004d:  add
  IL_004e:  stloc.1
  IL_004f:  br         IL_003f
  IL_0054:  call       instance class [MylibInterop]MylibInterop.MylibValue debug.Program::foo(class [MylibInterop]MylibInterop.MylibValue)
  IL_0059:  ret
} // end of method Program::'?foo'

Answer 1

@leppie在评论中得到了它:堆栈需要是确定性的; 它不在我生成的代码中(即使我知道它正在推动正确数量的args).我能够解决这个问题,因为编译器有足够的信息来展开循环(因此生成的IL中的常量),从而产生一个确定性的堆栈.