How to determine if user input is an number using a boolean function in C

Question

I am trying to figure out how to determine if the user input is a number and not a letter/symbol (A,a,!,@,#,$,%) in C, using a boolean function. The code I am using (if statement) only works for lower case and capitals letters. Below is my code for the if statement that works. Below that I will include the boolean function I am (unsuccessfully) attempting. Am I missing something in my boolean function?

    if((ch >= 'a' && ch <= 'z') || (ch >= 'A' && ch <= 'Z')){
    printf("Character '%c' does not represent a digit\n", ch);}


    _Bool isNumber(char ch) {
    printf("Character '%c' does not represent a digit\n", ch);
    return 0;}

Is it possible without a library function?

paxdiablo · Accepted Answer · 2019-05-12 13:52:52Z

1

The C standard library already has such a beast:

#include <ctype.h>
:
if (! isdigit(ch)) {
    printf("'%c' is not a digit\n", ch);
}

If, for some bizarre reason^(a) you cannot use the standard library function, simply roll your own:

bool isDigit(int ch) {              // Uppercase D to distinguish it.
    return ch >= '0' && ch <= '9';
}

You can also do it directly in the code rather than as a function:

if (ch <  '0' || ch >  '9') puts("Not a digit"); // or:
if (ch >= '0' && ch <= '9') puts("Is  a digit");

The digit characters in C are guaranteed by the standard to be contiguous, unlike all the other characters. That's why doing the same for alphas (like ch >= 'a' && ch <= 'z') is a bad idea.

Keep in mind that's for a single character being a digit. That appears to be what you want. To check if a character string is a valid integer, it'll need to be more complex. Basically, every character will need to be a digit, and it may have an optional sign at the front. Something like this would be a good start:

bool isInt(char *sstr) {
    unsigned char *str = sstr;
    if (*str == '+' || *str == '-')
        ++str;
    if (! isdigit(*str++))          // or your own isDigit if desired.
        return false;
    while (*str != '\0')
        if (! isdigit(*str++))      // ditto.
            return false;
    return true;
}

^(a) Though I suppose you may want to do it for educational purposes, so not necessarily that bizarre :-)

edited May 12, 2019 at 13:52

answered May 12, 2019 at 6:28

paxdiablo

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2 Comments

Markus Z Over a year ago

Thank you :) this works perfectly! Just out of curiosity, can this be done using a Boolean function without using another library function? I was told there is a "special" trick to accomplishing such a goal.Will mark as correct answer in 4 minutes.

paxdiablo Over a year ago

@MarkusZ, yes it can. Have added the boolean function to show how.

Tom Roush · Accepted Answer · 2019-05-12 06:29:18Z

0

If you want to stay true to your original logic, if(ch >= '0' && ch <= '9')) will be true if ch is any digit and false for any other character.

answered May 12, 2019 at 6:29

Tom Roush

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