问题
Does the compiler optimize out any multiplications by 1? That is, consider:
int a = 1;
int b = 5 * a;
Will the expression 5 * a be optimized into just 5? If not, will it if a is defined as:
const int a = 1;
回答1:
It will pre-calculate any constant expressions when it compiles, including string concatenation. Without the const it will be left alone.
Your first example compiles to this IL:
.maxstack 2
.locals init ([0] int32, [1] int32)
ldc.i4.1 //load 1
stloc.0 //store in 1st local variable
ldc.i4.5 //load 5
ldloc.0 //load 1st variable
mul // 1 * 5
stloc.1 // store in 2nd local variable
The second example compiles to this:
.maxstack 1
.locals init ( [0] int32 )
ldc.i4.5 //load 5
stloc.0 //store in local variable
回答2:
Constant propagation is one of the most common and easiest optimisations.
回答3:
Looking at the code generated by the mono compiler, the version with the non-const a performs the multiplication at run time. That is, the multiplication is not optimized out. If you make a const, then the multiplication is optimized out.
The Microsoft compiler might have a more aggressive compiler, the best solution is to look at the code generated by the compiler to see what it is doing.
回答4:
What the compiler would optimise here is not multiplication by 1 per-se, but rather arithmetic with values known at compile-time. So yeah, a compiler would optimise out all the maths in your example, with or without the const.
Edit: A competent compiler, I should say.
来源：https://stackoverflow.com/questions/160848/net-multiplication-optimization