reinterpret_cast in C++

Question

 uint32_t r,g,b;
 r = (uint32_t)145;
 g = (uint32_t)131;
 b = (uint32_t)139;
 uint32_t rgb = ((uint32_t)r << 16 | (uint32_t)g << 8 | (uint32_t)b);
 float rgbf = *reinterpret_cast<float*>(&rgb);
 uint32_t rgbnew = *(reinterpret_cast<uint32_t *>(&rgbf));
 uint8_t rnew = (rgbnew >> 16) & 0x0000ff;
 uint8_t gnew = (rgbnew >> 8) & 0x0000ff;
 uint8_t bnew = (rgbnew) & 0x0000ff;

当我尝试运行这段代码时，在

行有分段错误

RGB = ((uint32_t)r <<16 | (uint32_t)g <<8 | (uint32_t)b);

事实上，在一个地方它运行正常。

Answer 1

请尝试编译带有所有警告和调试信息的代码(例如Linux上的g++ -Wall -g)并改进它直到没有警告。学习使用调试器(即Linux上的gdb)

我猜故障可能在

 float rgbf = *reinterpret_cast<float*>(&rgb);

，因为如果rgb(即uint32_t)和float没有相同的对齐或大小约束，这可能会触发故障。某些系统(处理器、abi、编译器)可能有不同的&不兼容的约束

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顺便说一句，你的代码可以很好地与GCC 4.7在Debian/GNU/Linux/x86-64调用

   g++-4.7 -std=c++11 -Wall -g ramji.cc -o ramji

当我有

 #include <cstdint>
 #include <iostream>
 int main(int argc, char**argv)
 {
   uint32_t r,g,b;
   r = (uint32_t)145;
   g = (uint32_t)131;
   b = (uint32_t)139;
   std::cout << "r=" << r << " g=" << g << " b=" << b << std::endl;
   uint32_t rgb = ((uint32_t)r << 16 | (uint32_t)g << 8 | (uint32_t)b);
   std::cout << "rgb=" << rgb << std::endl;
   float rgbf = *reinterpret_cast<float*>(&rgb);
   std::cout << "rgbf=" << rgbf << std::endl;
   uint32_t rgbnew = *(reinterpret_cast<uint32_t *>(&rgbf));
   std::cout << "rgbnew=" << rgb << std::endl;
   uint8_t rnew = (rgbnew >> 16) & 0x0000ff;
   uint8_t gnew = (rgbnew >> 8) & 0x0000ff;
   uint8_t bnew = (rgbnew) & 0x0000ff;
   std::cout << "rnew=" << rnew << " gnew=" << gnew 
        << " bnew=" << bnew << std::endl;
   return 0;
 }  

没有警告，执行时没有崩溃。

注意<cstdint>需要c++ 11兼容的编译器