罗斯林(Roslyn)没有优化多个增量的原因吗? [英] Is there a reason why Roslyn does not optimize multiple increments?

问题描述

我试图查看 Roslyn 如何优化以下代码段:

I was trying to see how Roslyn optimizes the following snippet:

代码

public int F(int n) {
    ++n;       
    ++n;       
    ++n;       
    ++n;       

    return n;
}              

asm

C.F(Int32)
    L0000: inc edx
    L0002: inc edx
    L0004: inc edx
    L0006: inc edx
    L0008: mov eax, edx
    L000a: ret

问题

为什么Roslyn不能像MSVC这样的提前C编译器那样对其进行优化? 4 x INC 较慢(即使假设消除了移动，延迟为4个周期，而1个为4个周期，并且吞吐量需要多于4个oups； C 等效"；其中:

int
f(void *dummy_this, int n) {
        ++n;        
        ++n;        
        ++n;        
        ++n;        

        return n;
}

来自MSVC的

asm 或具有 __ attribute __((ms_abi))的GCC，以使用与C#asm相同的Windows x64调用约定: https://godbolt.org/z/sK6h7KKcn

asm from MSVC, or GCC with __attribute__((ms_abi)) to use the same Windows x64 calling convention as the C# asm: https://godbolt.org/z/sK6h7KKcn

f:
        lea     eax, [rdx+4]
        ret

推荐答案

编译器确实进行了优化.但是 n 是一个参数，因此无法修改.JIT编译器必须修改参数值的副本.

The compiler does optimize. n is a parameter though, so it can't be modified. The JIT compiler must modify a copy of the parameter's value.

如果在递增之前将值分配给变量，则Roslyn编译器将消除增量.来自

If the value is assigned to a variable before incrementing, the Roslyn compiler will eliminate the increments. From this Sharplab.io snippet, this C# code :

public int F(int i) {
    var n=i;
    ++n;       
    ++n;       
    ++n;       
    ++n;       

    return n;
} 

将被翻译为

public int F(int i)
{
    return i + 1 + 1 + 1 + 1;
}

并最终编译为以下汇编代码:

And eventually compiled to this assembly code:

C.F(Int32)
    L0000: lea eax, [edx+4]
    L0003: ret

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