How to test input is sane

Question

Consider the following simple C program.

//C test

#include

int main()
{
   int a, b, c;

   printf(\"Enter two numbers to add\\n\");
   scan         


        

Answer 1

You can use the following macro

#define SCAN_ONEENTRY_WITHCHECK(FORM,X,COND) \
do {\
    char tmp;\
    while(((scanf(" "FORM"%c",X,&tmp)!=2 || !isspace(tmp)) && !scanf("%*[^\n]"))\
            || !(COND)) {\
        printf("Invalid input, please enter again: ");\
    }\
} while(0)

and you call it in this way in the main

int main()

{
    int a, b, c;

    printf("Input first integer, valid choice between 0 and 10: ");
    SCAN_ONEENTRY_WITHCHECK("%d",&a,(a>=0 && a<=10));

    printf("Input second integer, valid choice between 0 and 10: ");
    SCAN_ONEENTRY_WITHCHECK("%d",&b,(b>=0 && b<=10));

    c = a + b;
    printf("Sum of entered numbers = %d\n",c);
    return 0;

}

for more detail concerning this macro please refer to: Common macro to read input data and check its validity

Answer 2

You can use like:

if( scanf("%d%d",&a,&b) == 2)
{
   //two integer values has been read successfully
   //do your stuff here
}
else
{
   //Wrong input
}

Answer 3

you can test this one.

#include <stdio.h>

int main(void)
{
    int a, b, c;

    printf("Enter two numbers to add\n");
    scanf("%d%d",&a,&b);

    if(scanf("%d%d",&a,&b) == 2)
    {
        c = a + b;
        printf("Sum of entered numbers = %d\n",c);
    }
    return 0;
}

Answer 4

User input is evil. Parse per:

(optional whitespace)[decimal int][whitespace][decimal int](optional whitespace)

strtol() and family have better error handling than scanf().
Coda: Best to handle user input in a helper function. Break into 2 parts: I/O and parsing.

#include <ctype.h>
#include <errno.h>
#include <limits.h>
#include <stdio.h>
#include <stdlib.h>

// return 1 (success), -1 (EOF/IOError) or 0 (conversion failure)
int Readint(const char *prompt, int *dest, size_t n) {
  char buf[n * 21 * 2]; // big enough for `n` 64-bit int and then 2x

  fputs(prompt, stdout);  // do not use printf here to avoid UB
  fflush(stdout); // per @OP suggestion
  if (fgets(buf, sizeof buf, stdin) == NULL) {
    return -1;
  }
  const char *p = buf;

  while (n-- > 0) {
    char *endptr;
    errno = 0;
    long l = strtol(p, &endptr, 10);
    if (errno || (p == endptr) || (l < INT_MIN) || (l > INT_MAX)) {
      return 0;
    }
    *dest++ = (int) l;
    p = endptr;
  }

  // Trailing whitespace OK
  while (isspace((unsigned char) *p)) p++;
  // Still more text
  if (*p) return 0;
  return 1;
}

int main() {  // for testing
  int Result;
  do {
    int dest[2] = { -1 };
    Result = Readint("Enter two numbers to add\n", dest, 2);
    printf("%d %d %d\n", Result, dest[0], dest[1]);
  } while (Result >= 0);
  return 0;
}

Answer 5

Also you can do this to prevent anything after second number

int a,b;
char c;
if( scanf("%d%d%c", &a, &b, &c) == 3) {
    if (c == '\n') {
        puts("good");
    }
} else {
    puts("bad");
}
    return 0;
}

Answer 6

A simple way would be,

int a=0, b=0, c=0;

initialise them to 0

Additionally, the check suggested by Midhun is good to check if there are two inputs.