C code to find number of digits in integer is not working as per.suggest modifications

Question

here is the link to the file(google docs link) containing my code -

#include<stdio.h>

int len(int);
void main()
{
    int n,p;
    printf("enter number");
    scanf("%d",&n);
    p=len(n);
    printf("length of entered number is %d",p);
}
int len(int num)
{
    int i,c,b;
    for(i=1;i<=50;i++){
        b=10*i;
        if(num<b) {
            return(i);
            break;
        }
    }
}

Please take some time to read [the help pages](http://stackoverflow.com/help), especially the sections named ["What topics can I ask about here?"](http://stackoverflow.com/help/on-topic) and ["What types of questions should I avoid asking?"](http://stackoverflow.com/help/dont-ask). Also please [read about how to ask good questions](http://stackoverflow.com/help/how-to-ask), and learn how to create a [Minimal, Complete, and Verifiable Example](http://stackoverflow.com/help/mcve). — Some programmer dude, Apr 19 '17 at 12:15
I also suggest you listen to your compiler and the warnings it gives you. And if it doesn't give you warnings then you need to enable more warnings. There are a few problems with the code you show, all detecable by the compiler. — Some programmer dude, Apr 19 '17 at 12:19
May be you are trying to do `b=pow(10, i)`. But this is not a good idea. See the answers for better implementation — kuro, Apr 19 '17 at 12:21
Notice that "number of digits in integer" and "length of entered number is" is potentially different when considering zero and negative numbers. — chux - Reinstate Monica, Apr 19 '17 at 15:02
That's some food for thought.I will keep that in mind. @chux — , Apr 28 '17 at 12:54

score 3 · Answer 1 · answered Apr 19 '17 at 12:19

3

Computing b=10*i; makes no logical sense. Did you want to raise 10 to a given power?

The normal way of solving this problem is to repeatedly divide by 10 until 0 is reached (using integer arithmetic), and count the number of divisions made. That is the number of digits in the original number. This method is also not vulnerable to integer overflow:

unsigned digits = 0;
for (; num /= 10; ++digits);
return digits;

answered Apr 19 '17 at 12:19

Bathsheba

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actualy correct method was to calculate pow(10,i) instead of 10*i. now the code is working fine. – Apr 19 '17 at 12:41
1

Don't use pow for this: i) it's expensive, ii) you risk overflowing an `int`, iii) can go off due to floating point. `pow(x, y)` is typically implemented as `exp (y log x)` in C. Repeated division by 10 is better in every respect. – Bathsheba Apr 19 '17 at 13:29
`for (; num /= 10; ++digits);` fails for `num==0` as zero is usually considered a one digit number.. Consider `do { ++digits; } while (num /= 10);` – chux - Reinstate Monica Apr 19 '17 at 14:52
@chux: I probably should have mentioned zero as a special case. To me, defining the number of digits necessary to hold 0 is down to personal choice (I like to think of 0 as requiring 0 digits to store, but then I guess mathematicians are strange folk!) – Bathsheba Apr 19 '17 at 14:53

sstefan · Answer 2 · 2017-04-19T15:14:46.133

1

int len(int num){
    int i = 0;

    if(num == 0)
        return 1;

    while(num!=0){
        num = num / 10;
        i++;
    }
    return i;
}

here we divide num by 10 until it reaches 0, and each time loop executes we increment the count i++

edited Apr 19 '17 at 15:14

answered Apr 19 '17 at 12:20

sstefan

385
4
15

thanks for the help.but my code is now working when i replaced 10*i with pow(10,i). – Apr 19 '17 at 12:51
@frontMatter Ok its working, but consider changing it. Your for loop only goes 50 times, it works but its not right. You should set something else as end condition. Also, int c...i didn't see where you used it – sstefan Apr 19 '17 at 12:58
i used 50 since it was a constraint for the problem i was working on. and int c, thanks for ponting out. – Apr 19 '17 at 13:07
1

Note that `len(-123)` returns 1. I'd expect 3. – chux - Reinstate Monica Apr 19 '17 at 14:57
my bad..changed it – sstefan Apr 19 '17 at 15:14

score -1 · Answer 3 · answered Apr 19 '17 at 12:31

-1

frontmatter i am not able to recognize your answer but i think this following code is helpful as per your title.

#include<stdio.h>
#include<conio.h>
void main()
{
    char a[50000];
    int i,c=0;
    printf("enter number:");
    fflush(stdin);
    gets(a);
    for(i=0;a[i]!='\0';i++)
    {
        if(a[i]>=0 || a[i]<=9)
        {
            c++;
        }
    }
    printf("\n %d digit number",c);
    getch();
}

answered Apr 19 '17 at 12:31

Team Work

35
1
7

1

`fflush(stdin);` = one cat potentially eaten due to undefined behaviour. See http://stackoverflow.com/questions/2979209/using-fflushstdin – Bathsheba Apr 19 '17 at 12:32
thanks for the help.but my code is now working when i replaced 10*i with pow(10,i) – Apr 19 '17 at 12:51

C code to find number of digits in integer is not working as per.suggest modifications

3 Answers3