I need to reverse the 5 digit number input by the user in C. Code has no syntax error (no error squiggles) but still code doesn't run

Question

There might be some error in the logic used by me. Could someone please point that out to me?

//Reverse the number

#include<stdio.h>
int main()
{
    int m,a,b,c,d,e,n;

    printf("\nEnter a five digit number: ");
    scanf("%d",m);

    a = m/10000;                                        //1st digit     
    b = (m - a*10000)/1000;                             //2nd digit
    c = ((m-a*10000) - b*1000)/100;                     //3rd digit
    d = (((m-a*10000) - b*1000) - c*100)/10;            //4th digit
    e = ((((m-a*10000) - b*1000) - c*100) - d*10);      //5th digit

    n = e*10000 + d*1000 + c*100 + b*10 + a;            //reverse of the number
    
    printf("Reverse of the number given by you is %d\n", n);

    return 0;
}

Answer 1

As already pointed out in a comment your bug is here:

scanf("%d",m);  -->  scanf("%d", &m);
           ^                     ^^
         wrong                  correct

as scanf must be passed a pointer to the object where the converted value is to be stored.

Your calculation can be simplified by using % like

e = m % 10;      //5th digit
m = m / 10;
d = m % 10;      //4th digit
m = m / 10;
c = m % 10;      //3rd digit
m = m / 10;
b = m % 10;      //2nd digit
m = m / 10;
a = m % 10;      //1st digit

As you can see the calculation is the same every time so you could use a loop.

int factor = 10000;
n = 0;
while(m != 0)
{
    n = n + (m % 10) * factor;
    m = m / 10;
    factor = factor / 10;
}

Note: The loop above only works when the input has exactly 5 digits. A more generic solution could be:

n = 0;
while(m != 0)
{
    n = n * 10;
    n = n + (m % 10);
    m = m / 10;
}