该编译器优化任何乘法1?也就是考虑:
INT A = 1;
INT B = 5 *一个;
请问EX pression 5 *一个被优化成只有5?如果不是,将它,如果一个被定义为:
const int的一个= 1;
解决方案
这将pre-计算任何常量EX pressions时汇编,其中包括字符串连接。如果没有在 常量
将被单独留在家中。
您第一个例子编译这个IL:
.maxstack 2
.locals的init([0] INT32,[1] INT32)
ldc.i4.1 //负载1
stloc.0 //店1日局部变量
ldc.i4.5 //负荷5
ldloc.0 //装载第一个变量
MUL // 1 * 5
stloc.1 //存储在第二局部变量
第二个例子编译到这一点:
.maxstack 1
.locals的init([0] INT32)
ldc.i4.5 //负荷5
stloc.0 //存储到本地变量
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;
解决方案
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