Why does it make a difference if left and right shift are used together in one expression or not?

Question

I have the following code:

unsigned char x = 255;
printf(\"%x\\n\", x); // ff

unsigned char tmp = x << 7;
unsign         


        

Answer 1

The shift operator is not defined for the char types. The value of any char operand is converted to int and the result of the expression is converted the char type. So, when you put the left and right shift operators in the same expression the calculation will be performed as type int (without loosing any bit), and the result will be converted to char.