two letters = one number?

Expand Post »

I am wonder how do you make it that you have to letters = to one number with only typing it once? Here is my program. My problem is that the result printf("%d reversed is: %d\n",num , res); is 0 reversed is: (The number backwords).

(Toggle Plain Text)

#include <iostream>
#include <stdio.h>
#include "simpio.h"
#include "strlib.h"
#include "random.h"
using namespace std;

int main(void)
{
	int num, res=0, rem;
	printf("Please enter an integer: ");
	num = GetInteger();
	while(num > 0)
	
{
rem=num% 10;
res=res* 10 + rem;
num=num/10;
}
printf("%d reversed is: %d\n",num , res);
system("pause");
}

Similar Threads

convert lower case letters to uppercase and vice-versa in C++
Read a phrase and Display the amount of letters. in C++
Regex for password in Shell Scripting
Bash shell needs tweeking in Shell Scripting
help with counting letters part of program in C++

plike922

Reputation Points: 10

Solved Threads: 0

Light Poster

Offline

38 posts
since Jul 2008

Permalink

Jul 20th, 2008

Re: two letters = one number?

since this is a c program.
u cannot use iostream.h
and namespaces.
remove the 2 of them anf check it out.

adarshcu

Reputation Points: 19

Solved Threads: 25

Junior Poster

Offline

121 posts
since May 2008

Permalink

Jul 22nd, 2008

Re: two letters = one number?

You could save the numbers as you find them instead of printing them, say in a char array, and then print that array in reverse. On the other hand, it would probably be better to use a different algorithm.

What you need to do is find the most significant (highest and leftmost) digit first. To do this, you'll need to skip over the digits to the right first . . . .

Here's how you can do it.

c Syntax (Toggle Plain Text)
#include <stdio.h>
 
int main() {
    int num = 1234;
    int digit = 1;
 
    if(!num) printf("0\n");
    else {
        while(digit < (num / 10)) digit *= 10;
 
        do {
            printf("%d", (num / digit) % 10);
            digit /= 10;
        } while(digit);
 
        printf("\n");
    }
 
    return 0;
}
#include <stdio.h>

int main() {
    int num = 1234;
    int digit = 1;

    if(!num) printf("0\n");
    else {
        while(digit < (num / 10)) digit *= 10;

        do {
            printf("%d", (num / digit) % 10);
            digit /= 10;
        } while(digit);

        printf("\n");
    }

    return 0;
}

It's just a matter of finding the leftmost digit and starting there, instead of starting at the right . . . .

Last edited by dwks; Jul 22nd, 2008 at 6:44 pm.

dwks

Reputation Points: 185

Solved Threads: 28

Posting Whiz in Training

Offline

269 posts
since Nov 2005

This thread is solved

Either the thread starter or a moderator has marked this thread as solved. You can most likely trust the responses and answers given. There is most likely no reason for any further responses to be posted here. If you have a related question, please start a new thread in this forum instead.

This thread is more than three months old

No one has posted to this discussion for at least three months. Please let old threads die and do not reply to them unless you feel you have something new and valuable to contribute that absolutely must be added to make the discussion complete. Otherwise, please start a new thread in this forum instead.

Previous Thread in C Forum Timeline: Help Me With Sorting Algo

Next Thread in C Forum Timeline: urgent: modify code please

DaniWeb Message

Cancel Changes

User Name
Password