Check for Character instead of Integer [duplicate] - c

Question

This question already has answers here:
Check if a value from scanf is a number?
(2 answers)
Closed 9 years ago.
my program adds numbers entered by the user. It runs and works great until a character is entered instead of an integer. Is there a simple way to make sure only integers are entered from the keyboard?
Here is my code.
#include<stdio.h>
#include <stdlib.h>
int main(int argc, char **argv)
{
int n, sum = 0, i, TotalOfNumbers;
printf("Enter the number of integers you want to add\n");
scanf("%d", &n);
printf("Enter %d integers\n",n);
for (i = 1; i <= n; i++)
{
scanf("%d",&TotalOfNumbers);
sum = sum + TotalOfNumbers;
}
printf("Sum of entered integers = %d\n",sum);
return 0;
}

Answer 1

You need to check the return value of scanf. If the input was a valid number, it will return 1. If the input was not a valid number, it will return something else. Here is your code modified to put the checks in.
#include<stdio.h>
#include <stdlib.h>
int get_number()
{
int num;
int ret;
ret = scanf("%d", &num);
if (ret != 1) {
printf("bad number\n");
exit(EXIT_FAILURE);
}
return num;
}
int main(int argc, char **argv)
{
int n, sum = 0, i, TotalOfNumbers;
printf("Enter the number of integers you want to add\n");
n = get_number();
printf("Enter %d integers\n",n);
for (i = 1; i <= n; i++)
{
TotalOfNumbers = get_number();
sum = sum + TotalOfNumbers;
}
printf("Sum of entered integers = %d\n",sum);
return 0;
}

Answer 2

Check the ferror state on the input stream
scanf("%d",&TotalOfNumbers);
if(!ferror(stdin)){
sum = sum + TotalOfNumbers;
}

Answer 3

In addition to posted answer, there options not general as posted, but quicker.
First if you want to skip some final set of characters.In following example all letters,! and + will be skiped
int n;
scanf("%*[a-zA-Z!+]%d",&n);
printf("\n%d",n);
for input
weweqewqQQWWW!!!!+++3332
the output is
3332
Next option is to use buffer wich allowed to read everything untill number is met, and then read the number. The disadvantage is that buffer size is limited
char buf[25];
int n;
scanf("%[^0-9]%d",buf,&n);
printf("\n%d",n);
For input
fgfuf#$#^^#^##4565
Output
4565