在C ++中将int转换为字节数组

Question

I am trying to use implement the LSB lookup method suggested by Andrew Grant in an answer to this question: Position of least significant bit that is set

However, it's resulting in a segmentation fault. Here is a small program demonstrating the problem:

#include <iostream>

typedef unsigned char Byte;

int main()  
{  
    int value = 300;  
    Byte* byteArray = (Byte*)value;  
    if (byteArray[0] > 0)  
    {  
        std::cout<< "This line is never reached. Trying to access the array index results in a seg-fault." << std::endl;  
    }  
    return 0;  
}  

What am I doing wrong?
I've read that it's not good practice to use 'C-Style' casts in C++. Should I use reinterpret_cast<Byte*>(value) instead? This still results in a segmentation fault, though.

Answer 1

Use this:

(Byte*) &value;

You don't want a pointer to address 300, you want a pointer to where 300 is stored. So, you use the address-of operator & to get the address of value .

Answer 2

虽然Erik回答了你的整体问题，但作为一个后续内容我会强调说 - 是的，应该使用reinterpret_cast而不是C风格的演员。

Byte* byteArray = reinterpret_cast<Byte*>(&value);

Answer 3

The line should be: Byte* byteArray = (Byte*)&value;

You should not have to put the (void *) in front of it.

-Chert

Answer 4

char *array=(char*)(void*)&value;

基本上，您将指针指向字符串的开头并将其重新转换为指向字节的指针。

Answer 5

@Erik already fixed your primary problem, but there is a subtle one that you still have. If you are only looking for the least significant bit , there is no need to bother with the cast at all.

int main()
{
    int value = 300;       
    if (value & 0x00000001)       
    {           
        std::cout<< "LSB is set" << std::endl;
    }
    return 0;
}