C - reverse a number

Question

I am coding in C on linux, and I need to reverse a number. (EG: 12345 would turn into 54321), I was going to just convert it into a string using itoa and then reverse that,

Answer 1

you can use stack to do this,

struct node
{
  char character;
  struct node *next;
};

struct node *list_head,*neos;

main()
{
  list_head=NULL;
  char str[14];
  int number,i;
  scanf("%d",&number);     

  sprintf(str,"%d",number);  //here i convert number to string
 for(i=0;i<strlen(str);i++)  //until the end of the string
 {
   add_to_stack(str[i]);   //i take every character and put it in the stack
 }
 print_the_number();

}

attention here,in stack the item which is added last, it taken out first, that why it works..

void add_to_stack(char charac)
{
  neos=(struct node*)malloc(sizeof(struct node));
  neos->character=charac;
  neos->next=list_head;
  list_head=neos;
}

void print_the_number()
{
   struct node *ptr;
   ptr=list_head;
   while(ptr!=NULL)
   {
     printf("%c",ptr->character);
     ptr=ptr->next;
   }  
}

Answer 2

iota() is not a standard C function, but snprintf() serves the purpose just as well.

/* assume decimal conversion */
const char * my_itoa (int input, char *buffer, size_t buffersz) {
    if (snprintf(buffer, sz, "%d", input) < sz) return buffer;
    return 0;
}

Since the input cannot be negative, you can use an unsigned type:

unsigned long long irev (unsigned input) {
    unsigned long long output = 0;
    while (input) {
        output = 10 * output + input % 10;
        input /= 10;
    }
    return output;
}

Reversing the input may result in a value that no longer fits the input type, so the return result attempts to use a wider type. This may still fail if unsigned and unsigned long long have the same width. For such cases, it is probably easiest to use a string to represent the reversed value. Or, if the only goal is to print the number, you can just use a loop to print the digits in reverse order.

void print_irev (unsigned input) {
    if (input) {
        do {
            putchar('0' + input % 10);
            input /= 10;
        } while (input);
    } else {
        putchar('0');
    }
    putchar('\n');
}

Answer 3

int n;
scanf("%d",&n);
int rev=0,rem;
while(n>0)
{
    rem=n%10; //take out the remainder .. so it becomes 5 for 12345
    rev=rev*10+rem; //multiply the current number by 10 and add this remainder.
    n=n/10; //divide the number. So it becomes 1234.
}
printf("%d",rev);

Answer 4

#include<stdio.h>
main()
{
                 int rev=0,n;
                 scanf("%d",&n);
                 while(n)
                 {
                         rev=10*rev+n%10;
                         n/=10;
                 }
                 printf("result=%d",rev);
}

Answer 5

Do it without strings.

fkt()
   {
    int i = 12345;

    int n = 0;
    int x;
    char nr[10];
    char *p = &nr[0];

    while(i != 0)
    {
        x = i % 10;
        i = i/10;
        n = n * 10 + x;
        *p = x+'0';
        p++;
    }

    *p = 0;

    printf("%d   %s\n", n, nr);
    return 0;
}

Answer 6

If you really want to use strings, you can use sprintf to do what itoa does.

int k = 12345;
char str[40];
sprintf(str,"%d",k);

Then reverse the string and convert it back to int using atoi or sscanf.