c++ - Integer |= Char; operation ignoring high order byte in Integer -

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Just a quick and specific question, it stumped me for about half an hour. 

  four * bytes = {0x01, 0xD8}; Int value = 0; Value = bytes [0]; // result1 (0x0001) is the value   
 The last operation is interested in me, how is it changing the integer signed in -40 in 256?  
  Edit : For example, a large part of the example code has changed   
 
 
 In your case is equivalent to  char   signed four, which means that when you value  0xD8  to  Char , then this is a negative number <.  
  |  Common arithmetic conversions  that occur during the operation are cost-protection, so the negative number is preserved.  
 To solve the problem, you can either create all your data types as  unsigned  when you have binary arithmetic or you can call  value | = ((Unsigned char) buffer [0]) you can type  or  value | = Buffer [0] & amp; 0xFF .

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