bitwise shifiting question

if i have int temp=(1<<31)>>31. How come the开发者_开发问答 temp becomes -1? how do i get around this problem? thanks

Ints are signed by default, which usually means that the high bit is reserved to indicate whether the integer is negative or not. Look up Two's complement for an explanation of how this works.

Here's the upshot:

[steven@sexy:~]% cat test.c
#include <stdint.h>
#include <stdio.h>

int main(int argc, char **argv[]) {
    uint32_t uint;
    int32_t sint;
    int64_t slong;
    uint = (((uint32_t)1)<<31) >> 31;
    sint = (1<<31) >> 31;
    slong = (1L << 31) >> 31;
    printf("signed 32 = %d, unsigned 32 = %u, signed 64 = %ld\n", sint, uint, slong);
}

[steven@sexy:~]% ./test
signed 32 = -1, unsigned 32 = 1, signed 64 = 1

Notice how you can avoid this problem either by using an "unsigned" int (allowing the use of all 32 bits), or by going to a larger type which you don't overflow.

In your case, the 1 in your expression is a signed type - so when you upshift it by 31, its sign changes. Then downshifting causes the sign bit to be duplicated, and you end up with a bit pattern of 0xffffffff.

You can fix it like this:

int temp = (1UL << 31) >> 31;

GCC warns about this kind of error if you have -Wall turned on.

int is signed.

what 'problem' - what are you trying to do ?

int i = (1<<31); // i = -2147483648
i>>31;  // i = -1

unsigned int i = (1<<31); // i = 2147483648
i>>31;  // i = 1

ps ch is a nice command line 'c' intepreter for windows that lets you try this sort of stuff without compiling, it also gives you a unix command shell. See http://www.drdobbs.com/184402054

When you do (1<<31), the MSB which is the sign-bit is set(becomes 1). Then when you do the right shift, it is sign extended. Hence, you get -1. Solution: (1UL << 31) >> 31.

bit to indicate the sign is set when you do "such" left shift on a integer variable. Hence the result.