CODEWITHSUNDEEP
cbit-shift
ipkiss
ipkiss

Reputation: 13671

Right shift operator in C

I got the following code:

int main(int argc, char *argv[])
{
    char c = 128;

    c = c >> 1;

    printf("c = %d\n", c);

    return 0;
}

Running the above code on Windows XP 32 bit, I got the result: -64. Why -64?

Upvotes: 3

Views: 12062

Answers (4)

dldnh
dldnh

Reputation: 8961

Variable c is signed. Changing the declaration to unsigned char c... will yield a result of 64.

Upvotes: 0

Guffa
Guffa

Reputation: 700910

Because the char type is a signed 8-bit integer (in the implementation of C that you are using). If you try to store the value 128 in it, it will actually be -128.

The bits for that would be:

10000000

Shifting a negative number will keep the sign bit set (as your implementation uses an arithmetic shift):

11000000

The result is -64.

Upvotes: 9

JMacArthur
JMacArthur

Reputation: 101

The C standard doesn't specify whether char is signed or unsigned. In this case it looks like you're getting a signed char, with a range from -128 to +127. Assigning 128 to it rolls round and leaves you with -128, so c>>1 is -64.

If you specify c as "unsigned char", c>>1 will be 64.

As the comment says, right-shifting a negative value is undefined by the standard so it's just luck that it comes out as -64.

Upvotes: 3

doron
doron

Reputation: 28932

You are using type char which by default is signed. Signed chars have a range of -128 to 127. which means char c = 128 really sets c to -128. (This is because most processors use two's complement to represent negative numbers) Thus when you shift right you get -64.

Bottom line is that when doing bit manipulations, use unsigned types to get the results you expect.

Upvotes: 0

Related Questions