Thread: C Read an integer and number of octades

I' m trying to solve a problem:

The program should read a positive integer and a number for the octades (n).

The program should show the outcome of the value by rotating the given number by n octades to the right. We have to suppose that the length for integers is 32 bits and that the user enters for the number of octades a value of 1-3. For example, if the user enters 553799974 (bin: 00100001|00000010|01010001|00100110) the program should display: 639697489 (bin: 00100110|00100001|00000010|010 10001).

I wrote this code for the case of 1 octade:

Code:

#include <stdio.h>
int main()
{


    unsigned int num;
    unsigned int temp;
    int n;


    //printf("Enter number: ");
    //scanf("%d",&num);
    num = 553799974;
    //printf("Enter oct: ");
    //scanf("%d",&n);
    n=1;
    
    if (n == 1){
    
    printf("Num is: %d\nand Temp is: %d\n\n",num, 0xFF);
    
    temp = num & 0xFF; //00000000000000000000000011111111 (under the 32-bit integer).
    
    temp <<= 8;
    
    temp += num >> 8;
    
}


    
    printf("Now\n\nNum is: %d\nand Temp is: %d",num, temp);
    
    return 0;
}

Where is my mistake? It returns 2173009 back and not the 639697489

OK I managed to solve it myself. Sorry for any inconvenience..

[...]

Originally Posted by akis

We have to suppose that the length for integers is 32 bits and that the user enters for the number of octades a value of 1-3.

[...]

What about including <stdint.h> and using types like uint32_t ?