Question

我想显示从输入接收的整数的四个字节中的每个字节的二进制值。我尝试用位操作符来做这个但是它运行得不好。谁能告诉我怎么能这样做？

代码如下所示：

void main()
{
int x;
int n;

printf("Please introduce an integer number");
scanf("%d",&x);
for (n=0; n<=3; n++) {
    printf("byte %d of 0x%X is 0x%X\n",n,x,getByte(x,n));   
}
}


int getByte(int x, int n)
{
return (x >> (n << 3)) & 1;
}

但它返回：字节0为1 字节1为0 字节2为0 字节3为0，而不是当我插入8时。

Answer 1

不使用位运算符的解决方案......

#include <iostream>
#include <string> 
#include <math.h> 
using namespace std;

string digits[2] = {"0","1"};

// return an integer in binary format
string num2bin(int n){ return n<2 ? digits[n] : num2bin(n/2) + digits[n%2]; }

// get the b-th byte of an integer
int getByte(long n, int b){ return (long)(n/pow(256,b)) % 256; }

int main() {
    long n = 987654321; // or whatever

    // loop through the bytes in order
    for (int b=0;b<4;b++) {
        cout << "byte " << b << " = " << num2bin(getByte(n,b)) << endl;
    }
 }

输出：

byte 0 = 10110001
byte 1 = 1101000
byte 2 = 11011110
byte 3 = 111010

Answer 2

#include <bitset>
#include <iostream>

int i;
std::cin>>i;
std::cout<<std::bitset<8*sizeof(int)>(i);

Answer 3

int x = ...;
int b0 = x&255;
int b1 = (x >> 8)&255;
int b2 = (x >> 16)&255;
int b3 = (x >> 24)&255;

从整数c ++中提取每个字节的二进制值

3 个答案: