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;
}
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