为什么我的计数领先零“程序故障？ [英] Why is my 'count leading zero' program malfunctioning?

问题描述

这里是返回Hacker's Delight书中前导零数的代码：

  #include< iostream> 
 using namespace std; 
 
 int nlz（unsigned x）{
 int n; 
 if（x == 0）return（32）; 
 n = 1; 
 if（（x>> 16）== 0）{n = n + 16; x = x <16;} 
 
 if（（x>> 24）== 0）{n = n + 8; x = x< 8;} 
 if（（x>>> 28）== 0）{n = n + 4; x = x< 4;} 
 if（（x>> 30）== 0）{n = n + 2; x = x< 2;} 
 n = n  - （x> 31）; 
 return n; 
} 
 
 int main（）{
 int x; 
 cin>> x; 
 cout<<< nlz（x）<< endl; 
 
 return 0; 
} 
  

，当我输入8时，它返回8， return 3 yes？

8 // 1000

解决方案

在无符号整数中为零的前导比特的数量，并且其假定为32比特的整数。

<8>是0000 0000 0000 0000 0000 0000 0000 1000二进制，它应该返回28，因为在前1位之前有28个前导位零。如果你在一个整数不是32位的地方运行它，它将无法工作。

Here is code which returns number of leading zeros from Hacker's Delight book:

#include <iostream>
using namespace std;

int nlz(unsigned x) {
int n;
if (x == 0) return(32);
n = 1;
if ((x >> 16) == 0) {n = n +16; x = x <<16;}

if ((x >> 24) == 0) {n = n + 8; x = x << 8;}
if ((x >> 28) == 0) {n = n + 4; x = x << 4;}
if ((x >> 30) == 0) {n = n + 2; x = x << 2;}
n = n - (x >> 31);
return n;
}

int main(){
    int x;
    cin>>x;
    cout<<nlz(x)<<endl;

     return 0;
}

and when I enter number 8 it return 8 and is it correct maybe it should return 3 yes?

8//1000

解决方案

It returns the number of leading bits that is zero in an unsigned integer , and it assumes an integer is 32 bits.

8 is 0000 0000 0000 0000 0000 0000 0000 1000 binary, and it should return 28 for that, as there's 28 leading bits zero before the first 1 bit. If you're running this on something where an integer is not 32 bits, it won't work.

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