无需转换即可从有符号整数转换为无符号整数

Question

int isNegative(int x) {
  return ((unsigned) x)>> 31;
}

我正在编写一个采用 32 位的函数，如果 x<0 则返回 1，否则返回 0。如何在不强制转换的情况下将有符号整数转换为无符号整数。

Answer 1

OP 有不同的隐含问题

a function that takes 32 bits and returns 1 if x<0, and 0 otherwise.

int isNegative(int x) {
  return x < 0;
}
// or maybe return bool/_Bool
bool isNegative(int x) {
  return x < 0;
}
// or pedantically
int isNegative(int_least32_t x) {
  return x < 0;
}
// or pedantically and nearly universally portable,
int isNegative(int32_t x) {
  return x < 0;
}

converting from signed int to unsigned int without casting (and)
How do I convert a signed int to an unsigned int without casting.

简单赋值

 int i;
 unsigned u = i;

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尝试使用 >>> 将这两种风险实现定义的行为结合起来，应该避免，除非给出令人信服的理由。

EXAMPLE An example of implementation-defined behavior is the propagation of the high-order bit when a signed integer is shifted right. C11§3.4.2 2