combine 4 bytes into a 32 bit number

Comments

• Tsag

  January 2012

  Should work . . .
  
  I used your code and wrote it in the following way both gave correct outputs for me.
  
  [code]
  #include
  
  int
  main(
  void)
  {
  // 8-bit values
  char a = 0x56;
  char b = 0x78;
  char c = 0x9A;
  char d = 0xBC;
  
  // combine into a single 32-bit value
  unsigned int value = d | (c<<8) | (b<<16) | (a<<24);
  std::cout << std::hex << value << std::endl;
  }
  [/code]
  
  output is:
  
  0x5678ABC
  
  Compiled in both x64 and x86, result is the same.
  --
  What output are you getting?
  
• anoopakes

  January 2013

  Hi.
  
  
  
  this may not work because you are using signed char.
  
  change it to unsigned char..