CODEWITHSUNDEEP
cloopscodeblocksdigits
Ali Haroon
Ali Haroon

Reputation: 83

C - Position of digit in a number

void main()
{
    int i, j = 0, p, q, N;    // N is the integer and p is the position needed
    printf("Enter two positive integers: ");
    scanf("%d%d", &N, &p);  // Taking the values of N and p

    for(i = 1; i <= p; i++) {

        j = N / 10;
        q = j % 10;
        i++;
        j = q;
    }

    printf("The digit of %d in the position %d is %d", N, p, j);
}

Example Input:

Enter two positive integers:
123456789
3

Output:

The digit of 123456789 in the position 3 is 8

I actually have 2 problems with the above lines of code:

  1. if I use the method above the computer will start taking digits from right to left while usually, the count of digits starts from left to right
  2. the program is only repeating itself in the loop which means that it is not taking the new value of j to work with the second time it runs

Upvotes: 1

Views: 11283

Answers (5)

neverever
neverever

Reputation: 79

#include <stdio.h>

int getDgtLen (int n);

int main(void)
{
    // n is the integer and p is the position needed
    int i, q, n, p;

    // Prompt for taking the values of n and p
    do
    {
        printf("Enter two positive integers for number,  position: ");
        scanf("%d%d", &n, &p);
    }
    while (n < 0 || p <= 0);

    int nOrigin = n;
    int len = getDgtLen(n);

    if (p <= len)
    {
        for (i = 0; i < p; i++)
        {
            q = n % 10;
            n /= 10;
        }
        printf("The digit of %d in the position %d is %d\n", nOrigin, p, len - q + 1);
    }
    else
    {
        printf("position not found!\n");
    }
}

int getDgtLen (int n)
{
    int i = 0;
    while (n > 0)
    {
        n /= 10;
        i++;
    }
    return i;
}

Upvotes: 0

user3813674
user3813674

Reputation: 2683

There are a few ways you can do this. As some of the comment suggests, you can do either of the following:

Method 1: Convert the number into a string array, which can be accomplished by sprintf.

char str[256];
int len = sprintf(str,"%d",N);
//traverse str using len

Method 2: Since a number N with n digits in base b you can express any number up to pow(b,n)-1. To get the digit you can use the inverse function of pow, namely base-b logarithm. You may not get integer values, so you can use the floor and cast the result to int. The full program looks like this:

#include <stdio.h>
#include <math.h>

int getDigit(int num, int p)
{
   return (num / (int)pow(10, floor(log10(num)) - p)) % 10;
}

int main()
{
    int i,j=0,p,q,N;// N is the integer and p is the position needed
    printf("Enter two positive integers: ");
    scanf("%d%d",&N,&p);// Taking the values of N and p
    j = getDigit(N,p);
    //The result is j if you count from 0, j+1 if you count from 1
    printf("The digit of %d in the position %d is %d\n",N,p,j+1);

    return 0;
}

Upvotes: 0

user3528438
user3528438

Reputation: 2817

#include <stdio.h>

int main()
{
    int i=0;
    int j=0;
    int p=5;
    int n = 123456;
    int digits[12]={};

    j=n;
    while(j)
    {
        digits[i]=j%10;
        j = j/10;
        i++;
    }

     printf("The digit of %d in the position %d is %d",n,p,digits[i-p]);
     return 0;
}

The easiest way for me to design is to convert the integer to digits and look up.

Why this way:

To avoid pow, log10 because they need floating point support to run, and hardware floating point to run fast.

And to avoid two loops because those two loops does a lot of repeated calculation.

http://ideone.com/Vo9g6T

Upvotes: 0

AkaSh
AkaSh

Reputation: 526

Without using 2 loops :

int main()
{
    int i,ans,p,N,no_of_digits=0,temp;
    printf("Enter two positive integers: ");
    scanf("%d%d",&N,&p);
    temp=N;
    no_of_digits=(int)log10(N) + 1;
    while(i<=(no_of_digits-p))
    {
        ans=N%10;
        N=N/10;
        i++;
    }
    printf("The digit of %d in the position %d is %d\n",temp,p,ans);
}

http://ideone.com/kQFISY

Upvotes: 0

Rajeev Singh
Rajeev Singh

Reputation: 3332

First you count the number of digits in the integer,let say n.Then you can get the first digit by dividing the number by 10n-1.Then put this in a loop and decrease n by 1 & number N by N%=10n-1 

int main()
{
    int i,j=0,p,N;
    printf("Enter two positive integers: ");
    scanf("%d%d",&N,&p);

    int pow=1,tmp=N;
    //counting power of 10 to divide
    while(tmp>10)
    {
        pow*=10;
        tmp/=10;
    }
    tmp=N;
    for(i=1; i<=p;i++){
       j = tmp/pow;
       tmp%=pow;
       pow/=10;
    }
    printf("The digit of %d in the position %d is %d\n",N,p,j);
}

Upvotes: 1

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