为什么设置字段比获取字段慢很多倍？

Question

我已经知道设置字段比设置局部变量要慢得多,但是看起来设置带局部变量的字段比用字段设置局部变量慢得多.为什么是这样？在任何一种情况下都使用该字段的地址.

public class Test
{
    public int A = 0;
    public int B = 4;

    public void Method1() // Set local with field
    {
        int a = A;

        for (int i = 0; i < 100; i++)
        {
            a += B;
        }

        A = a;
    }

    public void Method2() // Set field with local
    {
        int b = B;

        for (int i = 0; i < 100; i++)
        {
            A += b;
        }
    }
}

10e + 6次迭代的基准测试结果如下:

Method1: 28.1321 ms
Method2: 162.4528 ms

Answer 1

在我的机器上运行它,我得到了类似的时间差异,但是看看10M迭代的JITted代码,很明显为什么会出现这种情况:

方法A:

mov     r8,rcx
; "A" is loaded into eax
mov     eax,dword ptr [r8+8]
xor     edx,edx
; "B" is loaded into ecx
mov     ecx,dword ptr [r8+0Ch]
nop     dword ptr [rax]
loop_start:
; Partially unrolled loop, all additions done in registers
add     eax,ecx
add     eax,ecx
add     eax,ecx
add     eax,ecx
add     edx,4
cmp     edx,989680h
jl      loop_start
; Store the sum in eax back to "A"
mov     dword ptr [r8+8],eax
ret

和方法B:

; "B" is loaded into edx
mov     edx,dword ptr [rcx+0Ch]
xor     r8d,r8d
nop word ptr [rax+rax]
loop_start:
; Partially unrolled loop, but each iteration requires reading "A" from memory
; adding "B" to it, and then writing the new "A" back to memory.
mov     eax,dword ptr [rcx+8]
add     eax,edx
mov     dword ptr [rcx+8],eax
mov     eax,dword ptr [rcx+8]
add     eax,edx
mov     dword ptr [rcx+8],eax
mov     eax,dword ptr [rcx+8]
add     eax,edx
mov     dword ptr [rcx+8],eax
mov     eax,dword ptr [rcx+8]
add     eax,edx
mov     dword ptr [rcx+8],eax
add     r8d,4
cmp     r8d,989680h
jl      loop_start
rep ret

从程序集中可以看出,方法A的速度要快得多,因为A和B的值都放在寄存器中,并且所有的添加都在那里发生而没有对内存的中间写入.另一方面,方法B在每次迭代中产生加载并存储到存储器中的"A" .