如何在C++中获取char的整数值？

Question

我想取存储在32位无符号int中的值,将其放入四个字符中,然后将每个字符的整数值存储在一个字符串中.

我认为第一部分是这样的:

char a = orig << 8;
char b = orig << 8;
char c = orig << 8;
char d = orig << 8;

Answer 1

假设"orig"是包含您的值的32位变量.

我想你想做这样的事情:

unsigned char byte1=orig&0xff;
unsigned char byte2=(orig>>8)&0xff;
unsigned char byte3=(orig>>16)&0xff;
unsigned char byte4=(orig>>24)&0xff;

char myString[256];
sprintf(myString,"%x %x %x %x",byte1,byte2,byte3,byte4);

顺便说一句,我不确定这是否总是正确的.(编辑:确实,它是字节序正确的,因为bitshift操作不应受字节顺序的影响)

希望这可以帮助.

Answer 2

如果你真的想先提取单个字节:

unsigned char a = orig & 0xff;
unsigned char b = (orig >> 8) & 0xff;
unsigned char c = (orig >> 16) & 0xff;
unsigned char d = (orig >> 24) & 0xff;

要么:

unsigned char *chars = (unsigned char *)(&orig);
unsigned char a = chars[0];
unsigned char b = chars[1];
unsigned char c = chars[2];
unsigned char d = chars[3];

或者使用unsigned long和four chars的并集:

union charSplitter {
    struct {
        unsigned char a, b, c, d;
    } charValues;

    unsigned int intValue;
};

charSplitter splitter;
splitter.intValue = orig;
// splitter.charValues.a will give you first byte etc.

更新:正如Friol指出的那样,解决方案2和3不是字节序无关的; 哪些字节a,b,c和d代表依赖于CPU架构.