c++ - 有符号到无符号转换，并返回，整数的定义行为？

Question

#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 的程序，但这是定义的行为吗？

• 将有符号整数分配给无符号整数uint32_t i = -64;是否定义了行为？
• 有符号整数赋值int32_t j = i;，当i等于4294967232时，定义了行为吗？

Answer 1

对于无符号整数越界转换，定义结果；对于有符号整数，它是实现定义的。

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 ]

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.

此文本在 C++14 中保持不变。