Check if a string has only numbers in C?

Question

I'm trying to write a simple code to check if a string only has numbers in it. So far it's not working, any help would be appreciated.

#include <stdio.h>
#include <string.h>
#include <ctype.h>

int main()
{
    char numbers[10];
    int i, correctNum = 0;

    scanf("%s", numbers);

    for(i = 0 ; i <= numbers ; ++i)
    {
        if(isalpha(numbers[i]))
        {
            correctNum = 1;
            break;
        }
    }

    if(correctNum == 1)
    {
        printf("That number has a char in it. FIX IT.\n");
    }
    else
    {
        printf("All numbers. Good.\n");
    }
    return 0;
}

Define "not working"? You're not checking if everything is numeric, for starters. You're checking if there's a letter in it. You probably would rather want to use `isdigit()` — Sami Kuhmonen, Jan 21 '17 at 07:49
Welcome to Stack Overflow! It looks like you are asking for homework help. While we have no issues with that per se, please observe these [dos and don'ts](http://meta.stackoverflow.com/questions/334822/how-do-i-ask-and-answer-homework-questions/338845#338845), and edit your question accordingly. — Joe C, Jan 21 '17 at 08:13
Use a *positive* test with `isdigit` as you may encounter chars that are not alpha and not digits... — Jean-Baptiste Yunès, Jan 21 '17 at 08:19
`char numbers[10];` is fairly small, suggest 1) `char numbers[100];` 2) `isdigit()` and 3) `scanf("%99s", numbers);` — chux - Reinstate Monica, Jan 21 '17 at 09:11
@JoeC This is just a very small part of a significantly larger project I'm working on. I was never taught any of this in a school setting, I'm just trying to implement some extra nifty features. — SycoSins, Jan 21 '17 at 09:48
`i <= numbers` is an error and your compiler should have diagnosed it. Pay attention to compiler messages. — M.M, Jan 21 '17 at 10:37

RoadRunner · Answer 1 · 2017-01-21T10:27:20.030

Adding to the others answers, you can also use strtol to determine if a string has all numbers or not. It basically converts the string to an integer, and leaves out any non-integers. You can read the man page for more information on this function, and the extensive error checking you can do with it.

Also, you should use:

scanf("%9s", numbers);

Instead of:

scanf("%s", numbers);

To avoid buffer overflow.

Here is some example code:

#include <stdio.h>
#include <stdlib.h>

#define MAXNUM 10
#define BASE 10

int main(void) {
    char numbers[MAXNUM];
    char *endptr;
    int number;

    printf("Enter string: ");
    scanf("%9s", numbers);

    number = strtol(numbers, &endptr, BASE);

    if (*endptr != '\0' || endptr == numbers) {
        printf("'%s' contains non-numbers\n", numbers);
    } else {
        printf("'%s' gives %d, which has all numbers\n", numbers, number);
    }

    return 0;
}

Example input 1:

Enter string: 1234

Output:

'1234' gives 1234, which has all numbers

Example input 2:

Enter string: 1234hello

Output:

'1234hello' contains non-numbers

score 1 · Answer 2 · answered Oct 05 '17 at 19:27

You might consider using strspn:

#include <stdio.h>
#include <string.h>

int main(int argc, char* argv[]) {
    int i;
    for (i=1; i < argc; i++) {
        printf("%s %s\n",
            strlen(argv[i]) == strspn(argv[i], "0123456789") ? "digits" : "mixed",
            argv[i]
        );
    }
}

Demoed:

$ ./try foo 123 ba23a 123.4
mixed foo
digits 123
mixed ba23a
mixed 123.4

strspn returns the initial number of characters from the first argument that appear in the second argument. Super simple examples:

strspn("abba", "a");  // == 1
strspn("abba", "b");  // == 0
strspn("abba", "ab"); // == 2

score 0 · Answer 3 · answered Jan 21 '17 at 07:53

0

for(i = 0 ; i <= numbers ; ++i) //how is this supposed to work.

Run the loop from 0 to 1 less than the length of the input.

for(i = 0 ; i < strlen(numbers) ; ++i)

answered Jan 21 '17 at 07:53

Gaurav Sehgal

7,422
2
18
34

Why not also mention that the loop guard can be `numbers[i] != '\0`. This is more C like. – RoadRunner Jan 21 '17 at 12:21

Shreyas Patil · Answer 4 · 2017-01-21T08:30:42.687

0

#include <stdio.h>
#include <string.h>

int main()
{
    char numbers[10];
    int i, correctNum = 0;

    scanf("%s", numbers);

    for(i = 0 ; i < 10 ; i++)
    {
        if(numbers[i]<48||numbers[i]>57)
        {
            correctNum = 1;
            break;
        }
    }

    if(correctNum == 1)
    {
        printf("That number has a char in it. FIX IT.\n");
    }
    else
    {
        printf("All numbers. Good.\n");
    }
    return 0;
}

edited Jan 21 '17 at 08:30

answered Jan 21 '17 at 08:01

Shreyas Patil

1
2

your code is not correct. For loop condition should be i – GAURANG VYAS Jan 21 '17 at 08:07
Sorry my bad!!! I just edited his code and forgot to remove that =. Edited it. Thanks @GAURANG Rest is assuring. – Shreyas Patil Jan 21 '17 at 08:32
you didnot understand. strlen(numbers) will not be 10 always. 10 here just means the maximum length of the string to be entered. You will have to replace ' i<10 ' by ' i – GAURANG VYAS Jan 21 '17 at 08:44
Yeah i know @Gaurang. I was just trying to give him an answer. that's it. – Shreyas Patil Jan 22 '17 at 08:05

score -1 · Answer 5 · answered Jan 21 '17 at 07:52

-1

You have an error in the for loop - for(i = 0 ; i <= numbers ; ++i)

numbers is pointer and the comparison of it with integer is forbidden. Correct Code -

#include <stdio.h>
#include <string.h>
#include <ctype.h>

int main()
{
   char numbers[10];
   int i, correctNum = 0;

    scanf("%s", numbers);

    for(i = 0 ; i < strlen(numbers) ; ++i)
    {
       if(!(numbers[i]>='0' && numbers[i]<='9'))
       {
           correctNum = 1;
            break;
       }
    }

    if(correctNum == 1)
    {
         printf("That number has a char in it. FIX IT.\n");
     }
   else
    {
        printf("All numbers. Good.\n");
    }
    return 0;
 }

answered Jan 21 '17 at 07:52

GAURANG VYAS

689
5
16

It would be better to use `!isdigit(numbers[i])` instead of `!(numbers[i] >= ...` – linuxfan says Reinstate Monica Jan 21 '17 at 08:27
will that affect the efficiency significantly ? @linuxfan – GAURANG VYAS Jan 21 '17 at 08:40
how is isdigit() function implemented internally ? @linuxfan – GAURANG VYAS Jan 21 '17 at 08:46
1

it is not for efficiency, but for brevity, clarity and correctness. Anyway isdigit(), if not more efficient, surely is not less. – linuxfan says Reinstate Monica Jan 21 '17 at 08:47
I also said correctness. Don't assume that C programs always run on ascii systems. Read this: https://en.wikipedia.org/wiki/C_character_classification – linuxfan says Reinstate Monica Jan 21 '17 at 08:54
Thanks! It also works when the only code I change is "numbers" in the for loop to "strlen(numbers)". – SycoSins Jan 21 '17 at 09:50
1

Talking about efficiency: 1st of all you want to remove the call to `strlen` on a constant string out of the `for`-statement. – alk Jan 21 '17 at 10:22
1

Why not just `numbers[i] != '\0'` as the guard? – RoadRunner Jan 21 '17 at 10:24

Check if a string has only numbers in C?

5 Answers5

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