Does passing an unsigned int to a function make it lose bits?

Question

My question is quite similar to the one here except that I’m working with C.

I wrote some code to rotate an unsigned int; that is, the function bitRotate() (code below).

The function works very well when, instead of the printfs and scanfs, I directly put the literals I want to use, e.g. bitRotate(0xabcdef00,8);in the main function.

However, when I pass x as an argument as in the following code, abcdef00 that was got from the user, the x gets corrupted to ab000000. I checked and double checked and debugged my code multiple times and I am pretty sure the error is in this part but I don’t understand why.

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#define WIDTH sizeof(unsigned int)*CHAR_BIT

unsigned int bitRotate(unsigned int, char );

int main()
{

    unsigned int x;
    char n;
    while(1)
    {
        printf("Enter x: ");
        scanf("%x", &x);
        printf("Enter n: ");
        scanf("%d", &n);
        printf("%x\n", bitRotate(x,n));
    }

    return 0;
}

unsigned int bitRotate(unsigned int value, char n)
{

    char un = abs(n);
    unsigned int fallen = ~(0u);

    if(un == WIDTH)
        return value;
    else if (un < WIDTH)
    {

        if (n < 0)
        {
            fallen >>= (WIDTH - n);
            fallen = value & fallen;
            fallen <<= (WIDTH - n);
            value >>= n;
        }
        else
        {
            fallen <<= (WIDTH - n);
            fallen = value & fallen;
            fallen >>= (WIDTH - n);
            value <<= n;
        }
        value |= fallen;
        return value;

    }
    else
        return 0;

}

Answer 1

The bitRotate function is fine and so is the way you call it.

The actual culprit is here:

char n;
scanf("%d", &n);  // <<<<<

You provide the wrong format specifier to scanf which results in undefined behaviour. Your compiler most likely warned you about this.

The %d format specifier needs an pointer to an int (which usually takes 32 bits), but you provide the pointer to a char which takes 8 bits. Therefore scanf most likely clobbers the memory adjacent to the address of n.

You want this:

int n;
scanf("%d", &n);

Answer 2

Your code has undefined behaviour.

In this call

scanf("%d", &n);

you are using a wrong conversion specifier with an object of the type char. Moreover the type char can behave either as a signed char or as unsigned char depending on a compiler option.

Within the function there is used a wrong expression when n is negative

fallen >>= (WIDTH - n);
                 ^^^
fallen <<= (WIDTH - n);
                 ^^^

I think you mean

fallen >>= (WIDTH + n);
                 ^^^
fallen <<= (WIDTH + n);
                 ^^^

In any case the function can be written simpler. Here is a demonstrative program.

#include <stdio.h>
#include <limits.h>

unsigned int bitRotate(unsigned int value, int n )
{
    const int Width = sizeof( unsigned int ) * CHAR_BIT;

    n %= Width;

    if ( n < 0 )
    {
        value = ( value >> -n ) | ( value << ( Width + n ) );
    }
    else if ( n > 0 )
    {
        value = ( value << n ) | ( value >> ( Width - n ) );

    }

    return value;
}

int main(void) 
{
    while ( 1 )
    {
        unsigned int x;

        printf( "Enter a hexadecimal value of x (0 - exit): " );

        if ( scanf( "%x", &x ) != 1 || x == 0 ) break;

        int n;

        printf( "Enter a negative or positive value of n (0 - exit): " );
        if ( scanf( "%d", &n ) != 1 || n == 0 ) break;

        printf( "\n%#x shifted %d is %#x\n", x, n, bitRotate( x, n ) );
    }

    return 0;
}

Its output might look like

Enter a hexadecimal value of x (0 - exit): 0xabcdef00
Enter a negative or positive value of n (0 - exit): 16

0xabcdef00 shifted 16 is 0xef00abcd

Enter a hexadecimal value of x (0 - exit): 0xabcdef00
Enter a negative or positive value of n (0 - exit): -16

0xabcdef00 shifted -16 is 0xef00abcd

Enter a hexadecimal value of x (0 - exit): 0xabcdef00
Enter a negative or positive value of n (0 - exit): 32

0xabcdef00 shifted 16 is 0xabcdef00

Enter a hexadecimal value of x (0 - exit): 0

Answer 3

This code shows you if the value has the sign bit or not:

#include <stdio.h>
#include <stdlib.h>

int main(void) {
    const int arr[7] = {0x80000000u, -2, -1, 0, 1, 2, 0x7fffffffu};
    unsigned int i, j, signVal, uiSiz;

    uiSiz = sizeof(unsigned int);

    for (i = 0, j = 7; i < 7; i++) {
        signVal = ((unsigned int) arr[i] & (0x1u << ((8 * uiSiz) - 1))) >> ((8 * uiSiz) - 1);
        fprintf(stdout, "arr[%u]: %i, uiSiz: %u, signVal: %u;\n", i, arr[i], uiSiz, signVal);
    }

    return 0;
}

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