有符号整数和无符号整数之间的转换 [英] Conversion between signed integer and unsigned integer

问题描述

如果我把一个无符号整数转换为有符号整数，然后回传，我保证得到原始值？例如，对于根据C ++标准在任何平台上的任何 x ，此函数是否始终返回 true ？

  bool f（unsigned int x）
 {
 return x == static_cast< unsigned int>（static_cast< int> ;（X））; 
} 
  

这个怎么办？

  bool g（int x）
 {
 return x == static_cast< int>（static_cast< unsigned int>（x））; 
} 
  

解决方案

这不能保证 f 和 g 。

这是标准说的：

4.7积分转换

2. 如果目标类型是无符号，则结果值是与源
   整数一致的最小无符号整数（模2 ⁿ 是用于表示无符号类型的位数。 [注意：在二元的
   补码表示中，这种转换是概念性的，并且位模式没有变化（如果没有截断）。 - end note]


3. 如果目标类型是signed，则该值不会改变，如果它可以在目标类型中表示;

第2节讲述函数 g 。 x 到 unsigned 的转换可能导致不使用二的补码表示的系统上的表示更改，因此将

第3节介绍函数 f 。当数字在 signed int 的范围之外时，结果是实现定义的，因此不允许有人对将其转换回unsigned的结果提出任何声明。

If I cast an unsigned integer to a signed integer then cast back, am I guaranteed to get the original value? For example, does this function always return true for any x on any platform according to the C++ standard?

bool f(unsigned int x)
{
    return x == static_cast<unsigned int>(static_cast<int>(x));
}

What about this one?

bool g(int x)
{
    return x == static_cast<int>(static_cast<unsigned int>(x));
}

解决方案

The answer is "no, this is not guaranteed" for both f and g.

Here is what the standard says about it:

4.7 Integral conversions

2. If the destination type is unsigned, the resulting value is the least unsigned integer congruent to the source integer (modulo 2ⁿ 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 ]
3. If the destination type is signed, the value is unchanged if it can be represented in the destination type; otherwise, the value is implementation-defined.

Section 2 talks about function g. The cast of x to unsigned can cause change of representation on systems that do not use two's complement representation, so converting the number back may get you a different value.

Section 3 talks about function f. When the number is outside signed int's range, the result is implementation-defined, so one is not allowed to make any claims about the result of converting it back to unsigned.

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