Accept a number and print its alternate digits, - c

Question

int Rearrange(int a)
{
long int b,j,i=0,num=0,count=0,arr[100];
while(a>0)
{
b=a%10;a=a/10;
arr[i]=b;
i++;
count ++;
}
j=count;
for(i=0;i<=count/2;i++)
{
t=arr[i];
arr[i]=arr[count-i-1];
arr[count-i-1]=t;
count--;
}
for(i=0;i<j;i+=2)
{
num=num*10 + arr[i]%10;
}
return num;
}
I want to write a function in c rearrange which prints the alternate digits of a number it is given.
for example:
input:- 12345
output:- 135
Thank you

Answer 1

Why complicating a simple problem?
If you don't mind an alternative approach, please check the below code.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
int input = 0;
int len = 0;
int i = 0;
char sinput[64] = {0, };
printf("Enter the number :");
scanf("%d", &input);
sprintf(sinput, "%d", input);
len = strlen(sinput);
printf("Output : ");
for (i = 0; i < len; i+=2)
{
printf("%c\t", sinput[i]);
}
printf("\n");
return 0;
}
Sample i/o:
[sourav#braodsword temp]$ ./a.out
Enter the number :123456
Output : 1 3 5
[sourav#braodsword temp]$

Answer 2

Your for loops are faulty. Change those to:
for(i=0;i<=count/2;i++)
{
int t=arr[i];
arr[i]=arr[j]; /* Use j */
arr[j]=t; /* Use j */
/* count--; Dont decrement */
j--;
}
for(i=0;i<count;i+=2) /* Should be count */
{
num=num*10 + arr[i]%10;
}
Demo
There can be many alternate ways to solve, but I just want to show you how the approach in thought process can be implemented correctly.

Answer 3

In your code problem is with the first for loop.
Please check the below code.
int Rearrange(int a)
{
long int b = 0, j = 0, i = 0, num = 0, count = 0, arr[100];
while (a > 0)
{
b = a % 10; a = a/10;
arr[i] = b;
i++;
count++;
}
j = count;
for (i = 0; i < count/2; i++) // Condition is problematic
{
long int t = arr[i];
arr[i] = arr[count-i-1];
arr[count - i - 1] = t;
// count--; // this is problamatic.
}
for (i = 0; i < j; i += 2)
{
num = num * 10 + arr[i] % 10;
}
return num;
}

Answer 4

#include<stdio.h>
int main()
{
int n,arr[40];
scanf("%d",&n);
printf("%d",n);
int s=0,i=0;
while(n!=0)
{
arr[i]=n%10;
printf("%d",arr[i]);
n=n/10;
i++;
}
for(int j=i-1;j>=0;j-=2)
{
s =s*10+arr[j];
}
printf("\n%d",s);
return 0;
}

Answer 5

int alternatedigits(int n)
{
int a[10],i=0,count=0,sum=0;
while(n!=0)
{
a[i]=n%10;
i++;
count++;
n=n/10;
}
for(int i=count;i>=0;i++)
{
if(i%2==0)
{
sum=sum*10+a[i];
}
}
return sum;
}