c-如何在4字节int中打印出每个字节的内容

我设计了一个C代码来检查机器的字节序.
它运作良好.但是,它不能在4字节int中打印出每个字节的内容.

#include<iostream>
using namespace std;
bool f()
{
    int a = 1;
    char *p = (char*)&a;
    for (int i = 0 ; i < 4 ; ++i)
            cout << "p[" << i << "] is " << hex << *p++ << "   ";
    cout << endl ;
    p -= 4;
    if (*p == 1) return true ; // it is little endian
    else return false;  // it is big endian
}

int main()
{
    cout << "it is little endian ? " << f() << endl ;
    return 0 ;
}

输出：

 p[0] is    p[1] is    p[2] is    p[3] is
 it is little endian ? 1

为什么输出为空？

谢谢

解决方法:

问题在于* p的类型为char,因此流尝试将其值打印为ASCII字符(这可能不是可见字符的值).如果将其强制转换为int,则将获得您期望的结果：

cout << "p[" << i << "] is " << hex << static_cast<int>(*p++) << "   ";

标签：byte,endianness,linux,c-4
来源： https://codeday.me/bug/20191201/2079215.html