Check if input is integer type in C

Question

The catch is that I cannot use atoi or any other function like that (I'm pretty sure we're supposed to rely on mathematical operations).

 int num; 
 scanf("%d",&num);
 if(/* num is not integer */) {
  printf("enter integer");
  return;
 }

I've tried:

(num*2)/2 == num
num%1==0
if(scanf("%d",&num)!=1)

but none of these worked.

Any ideas?

Answer 1

There are several problems with using scanf with the %d conversion specifier to do this:

1. If the input string starts with a valid integer (such as "12abc"), then the "12" will be read from the input stream and converted and assigned to num, and scanf will return 1, so you'll indicate success when you (probably) shouldn't;

2. If the input string doesn't start with a digit, then scanf will not read any characters from the input stream, num will not be changed, and the return value will be 0;

3. You don't specify if you need to handle non-decimal formats, but this won't work if you have to handle integer values in octal or hexadecimal formats (0x1a). The %i conversion specifier handles decimal, octal, and hexadecimal formats, but you still have the first two problems.

First of all, you'll need to read the input as a string (preferably using fgets). If you aren't allowed to use atoi, you probably aren't allowed to use strtol either. So you'll need to examine each character in the string. The safe way to check for digit values is to use the isdigit library function (there are also the isodigit and isxdigit functions for checking octal and hexadecimal digits, respectively), such as

while (*input && isdigit(*input))
   input++;    

(if you're not even allowed to use isdigit, isodigit, or isxdigit, then slap your teacher/professor for making the assignment harder than it really needs to be).

If you need to be able to handle octal or hex formats, then it gets a little more complicated. The C convention is for octal formats to have a leading 0 digit and for hex formats to have a leading 0x. So, if the first non-whitespace character is a 0, you have to check the next character before you can know which non-decimal format to use.

The basic outline is

1. If the first non-whitespace character is not a '-', '+', '0', or non-zero decimal digit, then this is not a valid integer string;
2. If the first non-whitespace character is '-', then this is a negative value, otherwise we assume a positive value;
3. If the first character is '+', then this is a positive value;
4. If the first non-whitespace and non-sign character is a non-zero decimal digit, then the input is in decimal format, and you will use isdigit to check the remaining characters;
5. If the first non-whitespace and non-sign character is a '0', then the input is in either octal or hexadecimal format;
6. If the first non-whitespace and non-sign character was a '0' and the next character is a digit from '0' to '7', then the input is in octal format, and you will use isodigit to check the remaining characters;
7. If the first non-whitespace and non-sign character was a 0 and the second character is x or X, then the input is in hexadecimal format and you will use isxdigit to check the remaining characters;
8. If any of the remaining characters do not satisfy the check function specified above, then this is not a valid integer string.

Answer 2

This method works for everything (integers and even doubles) except zero (it calls it invalid):

The while loop is just for the repetitive user input. Basically it checks if the integer x/x = 1. If it does (as it would with a number), its an integer/double. If it doesn't, it obviously it isn't. Zero fails the test though.

#include <stdio.h> 
#include <math.h>

void main () {
    double x;
    int notDouble;
    int true = 1;
    while(true) {
        printf("Input an integer: \n");
        scanf("%lf", &x);
        if (x/x != 1) {
            notDouble = 1;
            fflush(stdin);
        }
        if (notDouble != 1) {
            printf("Input is valid\n");
        }
        else {
            printf("Input is invalid\n");
        }
        notDouble = 0;
    }
}

Answer 3

I developed this logic using gets and away from scanf hassle:

void readValidateInput() {

    char str[10] = { '\0' };

    readStdin: fgets(str, 10, stdin);
    //printf("fgets is returning %s\n", str);

    int numerical = 1;
    int i = 0;

    for (i = 0; i < 10; i++) {
        //printf("Digit at str[%d] is %c\n", i, str[i]);
        //printf("numerical = %d\n", numerical);
        if (isdigit(str[i]) == 0) {
            if (str[i] == '\n')break;
            numerical = 0;
            //printf("numerical changed= %d\n", numerical);
            break;
        }
    }
    if (!numerical) {
        printf("This is not a valid number of tasks, you need to enter at least 1 task\n");
        goto readStdin;
    }
    else if (str[i] == '\n') {
        str[i] = '\0';
        numOfTasks = atoi(str);
        //printf("Captured Number of tasks from stdin is %d\n", numOfTasks);
    }
}

Answer 4

num will always contain an integer because it's an int. The real problem with your code is that you don't check the scanf return value. scanf returns the number of successfully read items, so in this case it must return 1 for valid values. If not, an invalid integer value was entered and the num variable did probably not get changed (i.e. still has an arbitrary value because you didn't initialize it).

As of your comment, you only want to allow the user to enter an integer followed by the enter key. Unfortunately, this can't be simply achieved by scanf("%d\n"), but here's a trick to do it:

int num;
char term;
if(scanf("%d%c", &num, &term) != 2 || term != '\n')
    printf("failure\n");
else
    printf("valid integer followed by enter key\n");

Answer 5

First ask yourself how you would ever expect this code to NOT return an integer:

int num; 
scanf("%d",&num);

You specified the variable as type integer, then you scanf, but only for an integer (%d).

What else could it possibly contain at this point?

Answer 6

I looked over everyone's input above, which was very useful, and made a function which was appropriate for my own application. The function is really only evaluating that the user's input is not a "0", but it was good enough for my purpose. Hope this helps!

#include<stdio.h>

int iFunctErrorCheck(int iLowerBound, int iUpperBound){

int iUserInput=0;
while (iUserInput==0){
    scanf("%i", &iUserInput);
    if (iUserInput==0){
        printf("Please enter an integer (%i-%i).\n", iLowerBound, iUpperBound);
        getchar();
    }
    if ((iUserInput!=0) && (iUserInput<iLowerBound || iUserInput>iUpperBound)){
        printf("Please make a valid selection (%i-%i).\n", iLowerBound, iUpperBound);
        iUserInput=0;
    }
}
return iUserInput;
}