Thread: Not able to print the output of the conversion problem

Dear All,


I have written a program which converts an input decimal number into any number in a base between 2 and 36. Below is my program:

Code:

#include<stdio.h>
#include<stdlib.h>
int main()
{
    int newbase,n,i,ascii,ndigit=0,q,r,zero;
    char newrep[100];
    printf("Enter the number to be converted");
    scanf("%d",&n);
    printf("Enter the base between 2 and 36 to which the number is to be converted");
    scanf("%d",&newbase);
    zero=48;
    q=n;
    ndigit=0;
    while(q!=0)
    {
    r=q%newbase;
    ndigit+=1;
    ascii=zero+r;
    if(ascii>57)
    {
            ascii=ascii+7;
    }
    newrep[ndigit]=(char)ascii;
    q=q/newbase;
    }
    printf("Base %d representation of %d is\n",newbase,n);
    for(i=ndigit;i<1;i--)
    {
    printf("%s",newrep[ndigit]);
    }                              
    system("pause");
    return 0;       
}

I am getting correct values of newbase and ndigit but am unable to print the value of the converted output stored in newrep array. For example if I enter input number as 15 and base as 8 then I should get 17 as output. Kindly help me out to print the value of the output array. Thank you.

Your indentation could use some work. One of the first 3 styles mentioned here is good: Indent style - Wikipedia, the free encyclopedia.

When I compile your code, I get the following warning. If you don't, turn up the settings on your compiler.

Code:

$ make foo
gcc -Wall -g -std=c99  -lm -lpthread -lcurses -lefence  foo.c   -o foo
foo.c: In function ‘main’:
foo.c:29: warning: format ‘%s’ expects type ‘char *’, but argument 2 has type ‘int’

You need a string to pass to printf if you are using the %s modifier. Namely, a char array that is null terminated. More than likely though, you want to print individual characters with the %c modifier. A couple other things though.

Lastly, don't use magic numbers like 48 and 57 and 7, or confusing variable names like zero, which is set to 48. Not only are they ASCII specific, they're confusing. Use char literals like '0', '9' and 'A'. Better yet, use an array with all the digits in it. Try:

Code:

char digits[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
...
newrep[ndigit] = digits[r];

That way, you are not ASCII specific, and don't use awful magic numbers.

Hi,

Thank you for the reply and suggestions, i tried out the modifications that you suggested, but I am still unable to print the output array of the converted number. Below is my modified code:

Code:

#include<stdio.h>
#include<stdlib.h>
int main()
{
    int newbase,n,i,ascii,ndigit=0,q,r,zero;
    char newrep[100];
    char digits[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
    printf("Enter the number to be converted");
    scanf("%d",&n);
    printf("Enter the base between 2 and 36 to which the number is to be converted");
    scanf("%d",&newbase);
/*  zero=48;*/
    q=n;
    ndigit=0;
    while(q!=0)
    {
    r=q%newbase;
    ndigit+=1;
/*  ascii=zero+r;
    if(ascii>57)
    {
            ascii=ascii+7;
    }*/
    newrep[ndigit] = digits[r];
/*    newrep[ndigit]=(char)ascii; */
    q=q/newbase;
    }
    printf("Base %d representation of %d is\n",newbase,n);
    for(i=ndigit;i<1;i--)
    {
    printf("%c",newrep[ndigit]);
    }                              
    system("pause");
    return 0;       
}

Originally Posted by anduril462

Your indentation could use some work. One of the first 3 styles mentioned here is good: Indent style - Wikipedia, the free encyclopedia.

When I compile your code, I get the following warning. If you don't, turn up the settings on your compiler.

Code:

$ make foo
gcc -Wall -g -std=c99  -lm -lpthread -lcurses -lefence  foo.c   -o foo
foo.c: In function ‘main’:
foo.c:29: warning: format ‘%s’ expects type ‘char *’, but argument 2 has type ‘int’

You need a string to pass to printf if you are using the %s modifier. Namely, a char array that is null terminated. More than likely though, you want to print individual characters with the %c modifier. A couple other things though.

Lastly, don't use magic numbers like 48 and 57 and 7, or confusing variable names like zero, which is set to 48. Not only are they ASCII specific, they're confusing. Use char literals like '0', '9' and 'A'. Better yet, use an array with all the digits in it. Try:

Code:

char digits[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
...
newrep[ndigit] = digits[r];

That way, you are not ASCII specific, and don't use awful magic numbers.

Hi iMalc,


I have used the code provided by you, thank you for helping me out, but am still unable to see the output on my screen. For example, if I enter 15 as input number and 8 as the base, I should get 17 on my screen as the output, whereas I am not getting anything. Kindly help me out.

Code:

#include<stdio.h>
#include<stdlib.h>
int main()
{
    int newbase,n,i,ascii,ndigit=0,q,r,zero;
    char newrep[100];
    char digits[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
    printf("Enter the number to be converted");
    scanf("%d",&n);
    printf("Enter the base between 2 and 36 to which the number is to be converted");
    scanf("%d",&newbase);
/*  zero=48;*/
    q=n;
    ndigit=0;
    while(q!=0)
    {
    r=q%newbase;
    ndigit+=1;
/*  ascii=zero+r;
    if(ascii>57)
    {
            ascii=ascii+7;
    }*/
    newrep[ndigit] = digits[r];
/*    newrep[ndigit]=(char)ascii; */
    q=q/newbase;
    }
    printf("Base %d representation of %d is\n",newbase,n);
    i = ndigit;
    while (i < 1)
    {
        printf("%c",newrep[ndigit]);
        i--;
    }              
    system("pause");
    return 0;       
}

Thank you for your reply.

I have replaced the old for loop with the new while loop suggested by iMalc. I have used the debugger and according to it everything up to the while loops works perfectly. The problem is just in printing the output which is stored in newrep array. Looking forward to your suggestions.

Originally Posted by ledow

Sigh. The code that iMalc posted was an equivalent of your (BROKEN!) for loop, to show you what was wrong with it. It was not a copy/paste source for a fix. Of *course* the program is still broke. Because you still haven't worked out what's broke in order to fix it despite iMalc given you a HUGE clue.

Add this line above your last while loop and check what it gives:

Code:

printf("\ni = %d\n", i);

Now check again your while loop and voila

Hi,

Thanks for your reply. This line gives me the number of digits in my converted number. I am already able to print/retrieve this using ndigit variable. However I am unable to understand why my final printf statement is unable to print the value of the converted number, I have tried changing the output formats in my printf statement (%s, %c, %d etc.) but am still unable to print the output.

@ledow - thank you for your clarification. I hope you try to put your suggestions in a more legible format rather than throwing in self-concocted slangs. If you are such an ace, I hope you have seen the thing that is "fundamentally wrong" right away and can point it out. For your information, I am looking forward to learn from my mistakes rather than simply copy-pasting code and getting away with a solution.

Originally Posted by dpitz

Add this line above your last while loop and check what it gives:

Code:

printf("\ni = %d\n", i);

Now check again your while loop and voila

I think you didn't check your while loop after seeing the printed value of i ^^.
If you choose to convert 10 to base 2 (1010), you'll get i=4, since 1010 has 4 digits. What we're trying to say is that your while loop is

Code:

while(i < 1) { ... }

, what doesn't make sense, since i isn't smaller than 1.

I just would like to say that the fact that he hasn't found the error does not mean he didn't try or that he just copy-pasted the code. Some errors are so obvious that we don't see them.

Hope it's clear now.

Thanks dpitz. Your suggestion was right on target,I missed to spot the obvious, I have changed the statement to

Code:

while (i>1)

in my code and also found that I was printing with the wrong index values.

Instead of printing with index value i, I was printing with index value ndigit in my output loop. Here's my modified code that gives me the right answer.

Code:

#include<stdio.h>
#include<stdlib.h>
int main()
{
    int newbase,n,i,ascii,ndigit=0,q,r,zero;
    char newrep[100];
    char digits[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
    printf("Enter the number to be converted");
    scanf("%d",&n);
    printf("Enter the base between 2 and 36 to which the number is to be converted");
    scanf("%d",&newbase);
/*  zero=48;*/
    q=n;
    ndigit=0;
    while(q!=0)
    {
    r=q%newbase;
    ndigit+=1;
/*  ascii=zero+r;
    if(ascii>57)
    {
            ascii=ascii+7;
    }*/
    newrep[ndigit] = digits[r];
/*    newrep[ndigit]=(char)ascii; */
    q=q/newbase;
    }
    printf("Base %d representation of %d is\n",newbase,n);
    i = ndigit;
    printf("\ni = %d\n", i);
    while (i>=1)
    {
        printf("%c",newrep[i]);
        i--;
    }              
    system("pause");
    return 0;       
}

Originally Posted by dpitz

I think you didn't check your while loop after seeing the printed value of i ^^.
If you choose to convert 10 to base 2 (1010), you'll get i=4, since 1010 has 4 digits. What we're trying to say is that your while loop is

Code:

while(i < 1) { ... }

, what doesn't make sense, since i isn't smaller than 1.

I just would like to say that the fact that he hasn't found the error does not mean he didn't try or that he just copy-pasted the code. Some errors are so obvious that we don't see them.

Hope it's clear now.