C programming - validating int input so it isn't a character

Question

Need to write a program to take input from a user in the form of an int and validate that what they entered is not a char from a-z.

Is there another way to do this other than:

if((num != 'a') && (num != 'b') && (num != 'c') && (num != 'd') etc.....) printf("You entered %d", num);

You should be care of the negative number which has - as a character ahead. — Tim
– Tim, Commented Mar 17, 2014 at 3:04
Your question is ambiguous. It appears that you might be able to use: int num; if (scanf("%d", &num) == 1 && !islower(num)) printf("You entered %d\n", num); (noting the newline in the output). If the input is inherently non-numeric, the scanf() will fail; if the input is numeric but in the range of lower-case letters (97..122 in codesets derived from ISO 8859), then it will not be accepted. If you're using int num = getchar(); to read a single character, the answer is somewhat different. — Jonathan Leffler
– Jonathan Leffler, Commented Mar 17, 2014 at 3:12

aghoribaba · Accepted Answer · 2014-03-17 03:04:06Z

4

Yes, use the isdigit() function provided in the ctype.h header file

answered Mar 17, 2014 at 3:04

aghoribaba

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1 Comment

hookenz Over a year ago

Dang, you bet me by 30secs

hookenz · Accepted Answer · 2014-03-17 03:11:13Z

1

Rather than trying to work out that it's not a character, work out that it is a digit and discard the rest using the isdigit function from ctype like below.

#include <ctype.h>

...

if (isdigit(num)) {
  printf("You entered %d\n", num);
}

But this only works on single characters which is quite useless when you read in strings. So instead you could instead use the function sscanf. Like this.

int num;
if (sscanf(numstr, "%d", &num)){
  printf("You entered %d\n", num);
} else {
  printf("Invalid input '%s'\n", numstr);
}

Another option is to use the atoi function. But as I recall it doesn't handle errors and I find quite inferior to sscanf.

edited Mar 17, 2014 at 3:11

answered Mar 17, 2014 at 3:05

hookenz

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Comments

Mio · Accepted Answer · 2014-03-17 03:05:18Z

1

You can use the isalpha() function in the "ctype.h" header. It returns true if it is a letter. So you could do something like:

if ( !isalpha() )
{
    // Do whatever
}

Here is a link to the documentation for more information.

answered Mar 17, 2014 at 3:05

Mio

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2 Comments

Saraph Over a year ago

Assuming he'd be fine with anything except numbers, which I kinda doubt, since he mentions int. isdigit() would be a better solution.

Mio Over a year ago

@Saraph I originally was going to go with using that, but I re-read the question and must have interpreted it wrong. You are correct.

Alaeddine · Accepted Answer · 2014-03-17 03:02:14Z

-1

#include <stdio.h>
int main()
{
    char c;
    printf("Enter a character: ");
    scanf("%c",&c);
    if( (c>='a'&& c<='z') || (c>='A' && c<='Z'))
       printf("%c is an alphabet.",c);
    else
       printf("%c is not an alphabet.",c);
    return 0;
}

answered Mar 17, 2014 at 3:02

Alaeddine

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5 Comments

KThund Over a year ago

The variable must be declared as an integer, not a char.

Saraph Over a year ago

This answer wasn't very different from the solution OP originally proposed.

Jonathan Leffler Over a year ago

Theoretically, your alphabetic check could capture non-alphabetic characters. If you were on an IBM mainframe that used EBCDIC, then there are non-alpha characters between the codes for 'a' and 'z' — and the difference 'z' - 'a' is not 25 in EBCDIC. Use the isalpha() macro from <ctype.h>.

Jonathan Leffler Over a year ago

@Saraph: this solution is a lot more succinct than 26 equality comparisons.

Saraph Over a year ago

@JonathanLeffler Okay, I'm assuming OP would eventually find out about <, > operators.

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