什么是该C成语是什么意思？ [英] What does this C idiom mean?

问题描述

可能重复：结果
  <一href=\"http://stackoverflow.com/questions/1349542/john-carmacks-unusual-fast-inverse-square-root-quake-iii\">John卡马克大局;不寻常的快速平方根倒数（雷神之锤III）

我穿过这片code一个博客的最近来了 - 这是从Quake3的引擎。它的目的是快速计算逆平方根使用牛顿 - Rhapson方法

I came across this piece of code a blog recently - it is from the Quake3 Engine. It is meant to calculate the inverse square root fast using the Newton-Rhapson method.

float InvSqrt (float x){
    float xhalf = 0.5f*x;
    int i = *(int*)&x;
    i = 0x5f3759df - (i>>1);
    x = *(float*)&i;
    x = x*(1.5f - xhalf*x*x);
    return x;
}

什么是做的原因INT I = *为（int *）及X; ？做 INT I =（int）的X; ，而不是给出一个完全不同的结果。

What is the reason for doing int i = *(int*)&x;? Doing int i = (int) x; instead gives a completely different result.

推荐答案

INT I = *（INT *）及X; 不转换 X 为int - 它的作用是让浮动的实际的位的 X ，这是通常重新psented如超出你的想象完全是另外一个4字节的值$ p $。

int i = *(int*)&x; doesn't convert x to an int -- what it does is get the actual bits of the float x, which is usually represented as a whole other 4-byte value than you'd expect.

有关参考，这样做，这是一个非常糟糕的主意，除非你知道浮点值究竟是如何重新在内存psented $ P $。

For reference, doing this is a really bad idea unless you know exactly how float values are represented in memory.

这篇关于什么是该C成语是什么意思？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！