为什么如此复杂的code发出由两个功率分配有符号整数？ [英] Why is such complex code emitted for dividing a signed integer by a power of two?

问题描述

当我编译这个code。与VC ++ 10：

When I compile this code with VC++10:

DWORD ran = rand();
return ran / 4096;

我得到这个拆卸：

I get this disassembly:

299: {
300:    DWORD ran = rand();
  00403940  call        dword ptr [__imp__rand (4050C0h)]  
301:    return ran / 4096;
  00403946  shr         eax,0Ch  
302: }
  00403949  ret

它是干净简洁和由二的幂与一个逻辑右移取代表决。

which is clean and concise and replaced a division by a power of two with a logical right shift.

然而，当我编译这个code：

Yet when I compile this code:

int ran = rand();
return ran / 4096;

我得到这个拆卸：

I get this disassembly:

299: {
300:    int ran = rand();
  00403940  call        dword ptr [__imp__rand (4050C0h)]  
301:    return ran / 4096;
  00403946  cdq  
  00403947  and         edx,0FFFh  
  0040394D  add         eax,edx  
  0040394F  sar         eax,0Ch  
302: }
  00403952  ret

这是做算术右移之前执行某些操作。

that performs some manipulations before doing a right arithmetic shift.

什么是需要这些额外的操作？为什么算术移位不够？

What's the need for those extra manipulations? Why is an arithmetic shift not enough?

推荐答案

其原因是无符号除以2 ^ n可以非常简单地实现，而符号的除法稍微复杂一些。

The reason is that unsigned division by 2^n can be implemented very simply, whereas signed division is somewhat more complex.

unsigned int u;
int v;

U / 4096 等同于 U＆GT;＆GT; 12 为 U 所有可能的值。

u / 4096 is equivalent to u >> 12 for all possible values of u.

V / 4096 是不会等同于 V＆GT;＆GT; 12 - 它发生故障时 V＆LT; 0 ，由于舍入方向是相对于移动时负数参与分工不同。

v / 4096 is NOT equivalent to v >> 12 - it breaks down when v < 0, as the rounding direction is different for shifting versus division when negative numbers are involved.

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