Write a c program to reverse the any number without using loops?

Answer 1

#include<stdio.h>

#include<conio.h>

void main(){

int a[100],i,temp,j,n;

printf("\t \n enter the size of the array");

scanf("%d",&n);

printf("\n enter the numbers ");

for(i=0;i<n;i++){

scanf("%d",&a[i]);

}

j=n-1;

for(i=0;i<(n/2);i++){

temp=a[i];

a[i]=a[j];

a[j]=temp;

j--;

}

for(i=0;i<n;i++){

printf("\n %d",a[i]);

}

}

Answer 2

//Use recursion to reverse the number. You can do that without using loops.

#include

main()

{

int num,rev;

printf("\nEnter any number : ");

scanf("%d",&num);

rev = reverse(num);

printf("\nReverse is %d.\n", rev);

}

int sum = 0,r;

reverse(int num)

{

if(num)

{

r=num%10;

sum=sum*10+r;

reverse(num/10);

}

else

return sum;

return sum;

}

Answer 3

#include<iostream>

#include<string>

template<typename T>

void reverse(T a[], const size_t size)

{

int left=-1, right=size;

while((++left)<(--right))

{

T temp=a[left];

a[left]=a[right];

a[right]=temp;

}

}

int main()

{

using std::string;

using std::cout;

using std::endl;

const size_t size=5;

string a[size] = {"one","two","three","four","five"};

cout<<"Before:";

for(size_t i=0; i<size; ++i)

cout<<' '<<a[i];

cout<<'\n'<<endl;

reverse(a,size);

cout<<"After:";

for(size_t i=0; i<size; ++i)

cout<<' '<<a[i];

cout<<'\n'<<endl;

}

Output:

Before: one two three four five

After: five four three two one

Answer 4

/*

if your array is of int data type, then your loop will be..

Starting_Index_Of_array is having starting index of array as int.

Ending_Index_Of_array is having ending index of array as int.

*/

void reverseArray(int Starting_Index_Of_array, int Ending_Index_Of_array){

int array_start_index=Starting_Index_Of_array;

int array_last_Index =Ending_Index_Of_array;

while(array_start_index != array_last_Index)

{

int temp= ArrayName[array_last_Index];

for(int i=array_last_Index;i>=array_start_index;i--)

{

ArrayName[i] = ArrayName[i-1];

ArrayName[array_start_Index] = temp;

}

array_start_index++;

}

}