将uint16_t解释为int16_t [英] Interpreting a uint16_t as a int16_t

问题描述

有一个可移植的和安全的方法来解释由 boost :: uint16_t 作为 boost :: int16_t ？我有一个 uint16_t ，我知道代表一个有符号的16位整数编码为小端。我需要对这个值进行一些有符号的算术，那么有没有说服编译器它已经是一个有符号的值？

如果我没有错误，

解决方案

code> static_cast 会转换该值，

如果你正在寻找一个不同于cast的东西，那么将它的内存表示复制到 boost :: int16_t ，因为它代表了它的开头。 / p>

编辑：如果您必须在大端机上工作，只需将字节向后复制。使用 std :: copy 和 std :: reverse 。

Is there a portable and safe way to interpret the bit-pattern made by a boost::uint16_t as a boost::int16_t? I have a uint16_t, which I know represents a signed 16-bit integer encoded as little-endian. I need to do some signed arithmetic on this value, so is there anyway to convince the compiler that it already is a signed value?

If I a not mistaken, a static_cast<int16_t> would convert the value, perhaps changing its bit-pattern.

解决方案

If you are looking for something different than a cast, then copy its memory representation to that of a boost::int16_t since its what it represents to begin with.

Edit: If you have to make it work on a big endian machine, simply copy the bytes backwards. Use std::copy and std::reverse.

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