Reverse of a string

[Ajaycpp]

Hi,

If I am executing following code after reversing it is giving "1" but I need "0001".

#include <stdio.h>

int main()
{
int n=1000, reverse = 0;


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

printf("Reverse of entered number is = %d\n", reverse);

return 0;
}

can anyone help on this.

[condor]

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

#include <iostream> #include <algorithm> int main() { std::string myStr; int a; std::cout << "Enter a number: "; std::cin >> a; myStr = std::to_string(a); //reversing the string std::string strMy = std::string(myStr.rbegin(), myStr.rend()); //as string has the leading zeroes! std::cout << "As string: " << strMy << '\n'; //but as an int, obviously, has no leading zeroes! int b = std::stoi(strMy); std::cout << "As integer: " << b << '\n'; return 0; }

Example 1 --------------- Enter a number: 1234 As string: 4321 As integer: 4321 Example 2 --------------- Enter a number: 1000 As string: 0001 As integer: 1

[Thumper]

If you want to reverse a string, use a string, not an int.

see std::reverse
http://www.cplusplus.com/reference/algorithm/reverse/?kw=reverse

Or you can use std::ostream manipulators 'setfill' and 'setw' to add the 3 zeroes.
check out this:
http://ideone.com/f4Dk6a

[rich1]

In general, I'm not sure if it's possible to save number in variable with leading zeros.

when you surely know how many zeros precede the number, you can simply write
printf("%0Nd",var), where N is number of preceding zeros

when not, you can find out how many zeros are leading using while()

int zeros = 0; // zero counter
int d = 10; // variable used in modulo
while(number % d == 0)
{
zeros++;
d *= 10;
}

then do the reverse as you did, and at the end use for() to printf zeros we counted

for(int i = 0; i < zeros; i++)
{
printf("%d",0); // or you can print it as char using putchar()
}

now printf the reverse number

[Hiten Sharma]

According to your requirement , this is the best way.........

#include<iostream>
#include<conio>
#include<stdio>

void main()
{
   int i=0,temp;
   long num;
   char rev[10];
	cout<<"\n\nEnter Any Number - ";
   	cin>>num;
   while(num!=0)
   {
   	temp = num%10;
      rev[i] = temp + 48;
      num /= 10;
      i++;
   }
   rev[i]='\0';
   cout<<"\n\n\nReverse of the Number is - ";
   	puts(rev);
   getch();
}

Enter Any Number - 1000

Reverse of the Number is - 0001



Enter Any Number - 253

Reverse of the Number is - 352

[TwilightSpectre]

@Hiten Sharma: If the number is too long, which is possible, you will get out of bounds errors on your array. Instead, use someting like a std::string. Alternately, just do this (works not only for numbers):

#include <iostream>
#include <string>
#include <algorithm>

int main() {
    std::string num;

    std::cout << "Enter a number: ";
    std::cin >> num;

    std::reverse(num.begin(), num.end());

    std::cout << "Reverse of number is: " << num << std::endl;

    return 0;
}

Some other little things:
Don't use void main. Also, don't use conio.h and its either stdio.h or cstdio.

EDIT:
Just saw @Thumpers post, this is the same thing.

[Ajaycpp]

Hi Hiten Sharma,

can you explain what is happening exactly in the following line which is highlighted?

while(num!=0)
{
temp = num%10;
rev[i] = temp + 48;
num /= 10;
i++;
}

[Glandy]

In ascii the number 48 is the character '0'.
To store the number 2 for example in a char variable you have to look up the decimal-representation for the character '2' = 50. Now if temp is 2 for example and you add 48 ⇒ 50 ⇒ '2'.

[cronopio]

Hi, look at this modification to your original code..

#include <iostream>
#include <stdio.h>
#include <math.h>
using namespace std;

int main()
{
   int n=1000, reverse = 0;
   
   int len =  (int)log10(n) + 1;
   
   while (n != 0)
   {
     reverse = reverse * 10;
     reverse = reverse + n%10;
     n = n/10;
   }

   len -= (int)log10(reverse) + 1;
   
   printf("Reverse of entered number is = ");
   for (int i =0; i<len; i++ ) printf("0");   
   printf("%d\n", reverse);
   return 0;
}

[Hiten Sharma]

@cronopio

The code you provided is only applicable for the numbers which are perfect squares,cubes......so on.. of 10......What if the Number is something else like 1, 2001 or 589 ?

[cronopio]

@Hiten_Sharma

say if n = 1 or n = 2001 or n = 589??

[IWishIKnew]

well, I would say you need a for loop:

string whatever("Hello World!");

//display the string from end to beginning
for(string::const_iterator it = whatever.rbegin(); it != whatever.rend(); it++)
{
    cout<< *it;
}

//or...
{ //limit the scope of x...  you don't need to do this

    int x = whatever.size();
    do
    {
        x--;
        cout<< whatever[x];
    }
    while(x > 0);
}

If you have any questions about this, you should ask.