right Shift Operator infinite loop

Question

Why this code is going into infinite loop?

While it will work on other machine. Also the machine is `little endian. It will go on printing -1 values;

void printfbit(int n)
{
    while (n) {
        if (n & 1)
            printf("1");
        else
            printf("0");
        n = n >> 1;
        printf("\t %d\n",n);
    }
    //printf("\n");
}

Closely related to [Shift operator in C](http://stackoverflow.com/q/7622/1708801) — Shafik Yaghmour, Jul 20 '15 at 11:55

Michael · Answer 1 · 2015-07-20T12:20:04.490

7

From the C standard (see section 6.5.7 Bitwise shift operators):

The result of E1 >> E2 is E1 right-shifted E2 bit positions. If E1 has an unsigned type or if E1 has a signed type and a nonnegative value, the value of the result is the integral part of the quotient of E1 / 2^E2. If E1 has a signed type and a negative value, the resulting value is implementation-defined.

The behavior you're seeing with an infinite loop is due to the right-shift semantics of that particular implementation being: right-shifting a signed integer preserves the sign bit.

Hence, for any negative input you'll eventually end up with 0xffffffff... (== -1), and the condition for continuing your loop will always be fulfilled.

An example:

Original input:      0x80000000
After one shift:     0xC0000000
After two shifts:   0xE0000000
After three shifts: 0xF0000000
After four shifts:   0xF8000000
etc

edited Jul 20 '15 at 12:20

answered Jul 20 '15 at 11:42

Michael

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could you please tell us how it works for other machines? – Dayal rai Jul 20 '15 at 11:44
On _what_ other machine? And for which compiler? – Michael Jul 20 '15 at 11:45
i was just extending the question from OP. – Dayal rai Jul 20 '15 at 11:46
This works the same for all machines that use the two's complent method for representing negative numbes (that is, the left-most bit indicates the sign of number). – Paul Ogilvie Jul 20 '15 at 11:47
1

This is implementation defined behavior see the link in my comment to the question. Preserving the sign bit is one possible behavior but it is not required behavior. – Shafik Yaghmour Jul 20 '15 at 11:56

score 0 · Answer 2 · edited Jul 20 '15 at 11:46

0

Right-shifting signed integer will preserve sign bit. That's the rule applied here, leading to the above mentioned behaviour

edited Jul 20 '15 at 11:46

nhahtdh

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answered Jul 20 '15 at 11:44

Sohil Omer

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right Shift Operator infinite loop

2 Answers2