解释签名为unsigned [英] interpret signed as unsigned

问题描述

我有这样的值：

int64_t s_val = SOME_SIGNED_VALUE;

如何获得

uint64_t u_val

$ c> s_val ，但是被视为无符号？

that has exactly the same bit pattern as s_val, but is treated as unsigned?

这可能很简单，但是在Stackoverflow和其他地方

This may be really simple, but after looking on Stackoverflow and elsewhere I haven't turned up the answer.

推荐答案

int64_t s_val = SOME_SIGNED_VALUE;
uint64_t u_val = static_cast<uint64_t>(s_val);

C ++ Standard 4.7 / 2声明：

C++ Standard 4.7/2 states that:

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是用于表示无符号类型）。 [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）。 ]

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). ]

另一方面，Standard说由 reinterpret_cast 是实现定义的。[注：它可能或可能不会产生与原始值不同的表示。]（5.2.10 / 3）。所以，我建议使用 static_cast 。

From the other hand, Standard says that "The mapping performed by reinterpret_cast is implementation-defined. [Note: it might, or might not, produce a representation different from the original value. ]" (5.2.10/3). So, I'd recommend to use static_cast.

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