是否签名为无符号转换，然后返回定义的整数行为？ [英] Is signed to unsigned conversion, and back, defined behaviour for integers?

问题描述

#include <cstdint>
#include <iostream>

int main() {
  uint32_t i = -64;
  int32_t j = i;

  std::cout << j;
  return 0;
}

我尝试过的大多数编译器都会创建输出 -64 ，但这是定义的行为吗？

Most compilers I've tried will create programs that output -64, but is this defined behaviour?

• 是否将有符号整数赋值给无符号整数 uint32_t i = -64; 已定义的行为？


• 有符号整数赋值 int32_t j = i; ，当 i 等于 4294967232 ，定义的行为？

• Is the assignment of a signed integer to and unsigned integer uint32_t i = -64; defined behaviour?
• Is the signed integer assignment int32_t j = i;, when i equals 4294967232, defined behaviour?

推荐答案

对于无符号整数超范围转换，定义结果;对于有符号整数，它是实现定义的。

For unsigned integer out-of-range conversion, the result is defined; for signed integers, it's implementation-defined.

C ++ 11（ISO / IEC 14882：2011）§4.7完整转换[转换。 integer / 2]

如果目标类型是无符号的，则结果值是与源整数一致的最小无符号整数（模2 ^ n，其中n是数字用于表示无符号类型的位）。 [注意：在二进制补码表示中，此转换是概念性的，并且位模式没有变化（如果没有截断）。 -end note]

C++11(ISO/IEC 14882:2011) §4.7 Integral conversions [conv.integral/2]

If the destination type is unsigned, the resulting value is the least unsigned integer congruent to the source integer (modulo 2^n where n is the number of bits used to represent the unsigned type). [ Note: In a two’s complement representation, this conversion is conceptual and there is no change in the bit pattern (if there is no truncation). —end note ]

如果目标类型是有符号的，如果它可以用目标类型（和位字段宽度）表示，则该值不变;否则，该值是实现定义的。

If the destination type is signed, the value is unchanged if it can be represented in the destination type (and bit-field width); otherwise, the value is implementation-defined.

对于C ++ 14，此文本保持不变。

This text remains the same for C++14.

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