Member Avatar for revenge2

hey guys how would i figure out if a string has a number in it without using any aditional libraries?

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  • Latest Post Latest Post by Hiroshe

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Two common needs here are: (1) detect whether a string contains any digit at all, and (2) verify the entire input is a valid number (e.g., only digits, or only 0/1 for binary). In C, prefer the standard library for portability and correctness.

C (portable and safe):

#include <ctype.h>
#include <stdbool.h>

bool has_digit(const char *s) {
    for (; *s; ++s)
        if (isdigit((unsigned char)*s)) return true;  // cast avoids UB on signed char
    return false;
}

bool is_all_decimal_digits(const char *s) {
    if (!*s) return false;  // reject empty
    for (; *s; ++s)
        if (!isdigit((unsigned char)*s)) return false;
    return true;
}

If you actually need to validate and convert a binary or decimal number, use strtol and ensure the whole string is consumed. This handles signs and overflow checks:

#include <stdlib.h>
#include <errno.h>

bool parse_binary_strict(const char *s, long *out) {
    errno = 0; char *end;
    long v = strtol(s, &end, 2);
    if (errno == ERANGE || s == end || *end != '\0') return false;
    *out = v; return true;
}

See isdigit and strtol.

Python:

  • Contains any digit: any(ch.isdigit() for ch in s)
  • All digits (ASCII/Unicode): s.isdecimal() is strict decimal; s.isdigit() also accepts other numeric digit characters.
  • Parse with error handling: int(s, 2) for binary or int(s, 10) for decimal inside try/except ValueError. Docs: str.isdigit, int.

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All 5 Replies

Member Avatar for Member #46692

store a char array = ["0123456789"]

Loop through said string to check if it contains any of those numbers. Boom job done.

Member Avatar for Hiroshe

store a char array = ["0123456789"]

Loop through said string to check if it contains any of those numbers. Boom job done.

Or loop through the string and check if any char is between the acsii value's of '0' and '9'.

Member Avatar for Tom Gunn

Or loop through the string and check if any char is between the acsii value's of '0' and '9'.

Or just compare against '0' and '9' because the values have to be contiguous and using the underlying values means your code is tied to ASCII and any other character sets that correspond directly to ASCII for the decimal digits:

int isDigit = (c >= '0' && c <= '9');

Do not write unportable code if the portable version is easier to read, write, and maintain. :)

Member Avatar for revenge2

ah thanks alot. My question was wrong ealier i was actually trying to fingure out if there were any characters in a character string. (the input is binary or decimal number). i wanted to check whether it contained any characters so that i could print out "invalid number" .

would this be correct for the above situation?

#include <stdio.h>

int main() {
char input[80];
printf("number: );
gets(input);

int i=0;
while (i != '\0') {
if ( input[i]<0 || input[i]>9) {
printf("invalid number!");
}
}

}
Member Avatar for Hiroshe

You for got a double quote in the first printf. :P
Change if ( input<0 || input>9) to if ( input<'0' || input>'9'). '9' is different than 9.
Use fgets() instead of gets(). It's safer.
Add a return 0; at the end of your main().
Change while (i != '\0') to while (input != '\0').
Remember to increment i at the end of your loop.

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