Question

我想了解打印出一个char和一个未签名的char的结果。

#include <iostream>
#include <string>

using namespace std;

int main()
{
    int8_t a = 0xA1;
    uint8_t b = 0xA1;
    printf("0x%x,", a);
    printf("0x%x,", b);
    std::cout << std::hex << a << ",";
    std::cout << std::hex << b << std::endl;
}

结果

0xffffffa1,0xa1,�,�

我不明白为什么带符号的char变成uint或int，为什么std :: hex失败了。

Answer 1

int8_t a = 0xA1;导致签名溢出，因此行为未定义。如果打开了正确的编译器标志，您将get类似于：

error: overflow in conversion from 'int' to 'int8_t' {aka 'signed char'} changes value from '161' to '-95' [-Werror=overflow]

    8 |     int8_t a = 0xA1;

      |                ^~~~

此外，%x期望unsigned int。这也会导致未定义的行为。您打算执行以下操作：

#include <iostream>

int main()
{
    int8_t a = 42; // Doesnt overflow
    uint8_t b = 42;
    std::printf("%#x,", static_cast<unsigned int>(a));
    std::printf("%#x,", static_cast<unsigned int>(b));
    std::cout << std::hex << static_cast<unsigned int>(a) << ",";
    std::cout << std::hex << static_cast<unsigned int>(b) << std::endl;
}

输出：0x2a,0x2a,2a,2a

打印十六进制字符

1 个答案: