[SOLVED] Why is the expression true after bit shifting the value and condition with && in C

Why is the expression true after bit shifting the value and condition with && in C

Have a look at the example:

unsigned char c = 64; /* 0100 0000 */
c = (c << 2 && 512); /* 512: 10 0000 0000  */

If I shift c two times to the left, I should get from 0100 0000 (64) to this 0001 0000 0000 (256). So my 8-bit char c has only zeroes. At the end, I am wondering why the expression (c << 2 && 512) is true (1) and not false (0), because my c has only zeroes.

Solution

From the C Standard (6.5.7 Bitwise shift operators)

3 The integer promotions are performed on each of the operands. The type of the result is that of the promoted left operand.

So in this expression

c << 2

the object c is promoted to the type int and the result also has the type int.

As the both operands of the logical AND operator are not equal to 0 then the whole expression evaluates to 1.

From the C Standard (6.5.13 Logical AND operator)

3 The && operator shall yield 1 if both of its operands compare unequal to 0; otherwise, it yields 0. The result has type int.

It seems you are confusing the logical operator AND && with the bitwise operator AND &.

If you will write

c = (c << 2 & 512);

then the variable c will have the value 0.