交换“字节顺序”浮动和双打

问题描述

 <$ c $ 
我想切换float和double值的字节顺序 c> float const v {1.f}; 
 
 swap（reinterpret_cast< :: std :: uint32_t const&>（v））; 
  

是否存在更好的方法来进行交换，而无需投射？
$ b

编辑： swap（）是gcc内置函数的C ++包装器，我没有在这里包含它。 b
$ b

  uint16_t __builtin_bswap16（uint16_t x）
 uint32_t __builtin_bswap32（uint32_t x）
 uint64_t __builtin_bswap64（uint64_t x）
  

一些数据格式需要交换字节数，如CBOR。

解决方案

虽然尝试和避免强制转换是一个很好的做法，但这样的用法也是强制转换的原因。一个endian swap是一个原始数据操作，所以为了做到这一点，你必须删除输入信息，我会说，如果它不是endian正确的开始，那么它不是一个浮动，不应该以这种方式键入而在错误的endianess状态。

I'd like to switch the "endianness" of float and double values, it works OK, by doing something like:

float const v{1.f};

swap(reinterpret_cast<::std::uint32_t const&>(v));

Does there exist a better way to do the swap, without a cast?

EDIT: swap() is a C++ wrapper for gcc's built-in functions, I did not include it here.

uint16_t __builtin_bswap16 (uint16_t x)
uint32_t __builtin_bswap32 (uint32_t x)
uint64_t __builtin_bswap64 (uint64_t x)

Swapping of endianess is needed for a some data formats, like CBOR.

解决方案

While it is good practice to try and avoid casts, it's uses like this that are the reason casts exist. An endian swap is a raw data operation so in order to do it you must strip away the typing information, I would say that if it is not endian correct to begin with, then it is not a float and should never have been typed that way while in the wrong endianess state.

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