c - C 中的 Printf 打印 fffffffe1 而不是 e1

Question

我很困惑，我有一个小程序，我在不同的地址位置打印值。

int main ()
{
 // unsigned int x = 0x15711056;
  unsigned int x = 0x15b11056;
  char *c = (char*) &x;

  printf ("*c is: 0x%x\n", *c);
  printf("size of %d\n", sizeof(x));
  printf("Value at first address %x\n", *(c+0));
  printf("Value at second address %x\n", *(c+1));
  printf("Value at third address %x\n", *(c+2));
  printf("Value at fourth address %x\n", *(c+3));

对于注释的 unsigned int x，printf 值符合预期，即

  printf("Value at first address %x\n", *(c+0)) = 56
  printf("Value at second address %x\n", *(c+1))= 10
  printf("Value at third address %x\n", *(c+2))= 71
  printf("Value at fourth address %x\n", *(c+3))= 15

但是对于未注释的 int x 为什么我得到的结果低于 *(c+2) 它应该是 b1 而不是 ffffffb1。请帮助我理解这一点，我正在在线 IDE 上运行它 https://www.onlinegdb.com/online_c_compiler 。我的电脑是 i7 intel。

  printf("Value at first address %x\n", *(c+0)) = 56
  printf("Value at second address %x\n", *(c+1))= 10
  printf("Value at third address %x\n", *(c+2))= ffffffb1
  printf("Value at fourth address %x\n", *(c+3))= 15

Answer 1

该值的符号为0xB1，二进制为10110001，您需要使用unsigned char指针:

unsigned char *c = (unsigned char*) &x;

您的代码适用于 0x7F 之前的任何字节。