Why is IL.Emit method adding additional nop instructions?

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Category:Languages

I have this code that emits some IL instructions that calls string.IndexOf on a null object:

MethodBuilder methodBuilder = typeBuilder.DefineMethod(                                              "Foo",                                              MethodAttributes.Public,                                              typeof(void), Array.Empty<Type>()); var methodInfo = typeof(string).GetMethod("IndexOf", new[] {typeof(char)}); ILGenerator ilGenerator = methodBuilder.GetILGenerator();  ilGenerator.Emit(OpCodes.Ldnull); ilGenerator.Emit(OpCodes.Ldc_I4_S, 120); ilGenerator.Emit(OpCodes.Call, methodInfo); ilGenerator.Emit(OpCodes.Ret); 

This is the generated IL code:

.method public instance int32  Foo() cil managed {   // Code size       12 (0xc)   .maxstack  2   IL_0000:  ldnull   IL_0001:  ldc.i4.s   120   IL_0003:  nop   IL_0004:  nop   IL_0005:  nop   IL_0006:  call       instance int32 [mscorlib]System.String::IndexOf(char)   IL_000b:  ret } // end of method MyDynamicType::Foo 

As you can see there are three nop instructions before the call instruction.

First I thought about Debug/Release build but this is not compiler generated code, I am emitting raw IL code and expect to see it as is.

So my question is why are there three nop instruction when I hadn't emitted any?

 


ILGenerator is not very advanced, if you use the Emit(OpCode, Int32) overload it will put the entire int32 in the instruction stream, no matter if the opcode is Ldc_I4 (which actually takes 4 bytes of immediate) or Ldc_I4_S (which doesn't).

So make sure to use the right overload:

ilGenerator.Emit(OpCodes.Ldc_I4_S, (byte)120); 

The lemmas for the opcodes in the documentation specify which overload of Emit is the right one to use.


In the reference source, Emit with an int argument does this:

public virtual void Emit(OpCode opcode, int arg)  {     // Puts opcode onto the stream of instructions followed by arg     EnsureCapacity(7);     InternalEmit(opcode);     PutInteger4(arg); } 

Where PutInteger4 writes four bytes to the byte array in which the IL is built up.

The documentation of Emit says that the extra bytes will be Nop instructions, but that's only if they are actually zero. If the value being passed is "more wrong" (with the high bytes different from zero) then the effects can be worse, from invalid opcodes to operations that subtly corrupt results.

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