How to format input to only accept integer values

Question

input value 123 -- this value is integer, and valid

input value 1b23a -- this value is invalid

How do I detect which values are valid and not?

Here is my code:

#include <stdio.h>
#include <conio.h>
void main()
{
    char str1[5],str2[5];
    int num,num1,i;
    num=0;
    clrscr();
    printf("Enter the Number ");
    scanf("%s",str1);
    for(i=0;str1[i]!='\0';i++)
    {
        if(str1[i]>=48&&str1[i]<=56)
            num=num1*10+(str[i]-48);
        else
        {
            printf("The value is invalid ");
        }
    }
    printf("This Number is %d",num);
    getch();
}

Answer 1

#include<stdio.h>
#include<conio.h>
void main()
{
char str1[5],str2[5];
int num,num1,i;
num=0;
clrscr();
printf("Enter the Number ");
scanf("%s",str1);
for(i=0;str1[i]!='\0';i++)
if(str1[i]>=48&&str1[i]<=56)
num=num1*10+(str[i]-48);
else
{
printf("The value is invalid ");
}
}
printf("This Number is %d",num);
getch();
}

Answer 2

Please see this answer regarding use of strtol(). It is a safe way to convert arbitrary input that should be a string representation of an integer, while also saving 'garbage' bytes for additional analysis.

Using it, your code would look something like this:

#include <stdio.h>
#include <errno.h>
#include <stdlib.h>
#ifdef LINUX_VERSION
#include <curses.h>
#else
#include <conio.h>
#endif

#define BUFF_SIZE 1024

int main(void)
{
    char str1[BUFF_SIZE], *garbage = NULL;
    long num = 0;

    printf("Enter the Number ");
    scanf("%s",str1);

    errno = 0;

    num = strtol(str1, &garbage, 0);
    if (errno) {
       printf("The number is invalid\n");
       return 1;
    }

    printf("You entered the number %ld\n", num);
    if (garbage != NULL) {
        printf("Additional garbage that was ignored is '%s'\n", garbage);
    }
    getch();
    return 0;
}

This doesn't fix everything that is questionable about what you posted, but it should help you get off to a better start.

Output is:

tpost@tpost-desktop:~$ ./t 
Enter the Number 1234abdc 
You entered the number 1234
Additional garbage that was ignored is 'abdc'

Compiled via:

gcc -Wall -DLINUX_VERSION -o t t.c -lcurses

I'm not sure what platform you are using, so additional fixes to the code may be needed.

Answer 3

One way is to use sscanf and check that there are no characters following the number. This is done most easily by adding a %c on the end and testing the return code, like this:

const char *yourString = ...;
int theValue, dummy;
if (sscanf(yourString, "%d%c", &theValue, &dummy) == 1) {
    // Was a pure number, parsed into 'theValue'
} else {
    // Either no number or had junk after it
}