How do I force user to input a positive integer? - c

Question

Force user to input a positive integer and put user in a loop till they do.
So I want everything including characters not allowed just over > 0
I tried the following:
while (i < 0) do {
printf("please input a number that's positive");
scanf("%d", i);
}

Answer 1

For positive integer use the following code
int i;
do
{
printf("please input a number that's positive");
scanf("%d", &i);
}while (i < 0);
The c language provides no error checking for user input. The user is expected to enter the correct data type. For instance, if a user entered a character when an integer value was expected, the program may enter an infinite loop or abort abnormally.

Answer 2

Both of these while functions manage the numbers, the int k is the set integer which can only be set below 20, the first while loop makes a statement that calls for another scan if the number is greater than 20
and the second loop prints a k*k box.
Hope this helps.
int main ( )
{
int i, j,k;
printf("Please enter Box size:\n\n");
scanf("%d",&k);
while(k>20){
printf("Please enter a value below 20\n\n");
scanf("%d"),&k;
}
while(k<=20)
{
for (i = 0; i < k; i++)
{
printf("\n");
for (j = 0; j < k; j++)
{
printf("#");
}
}
return 0;
}
}

Answer 3

I would do this: declare char term and int wrong = 0.
do {
printf("Enter a number: ");
fflush(stdin);
if (scanf("%d%c", &n, &term) != 2 || term != '\n' || n <= 0) {
printf("Only positive numbers.\n");
wrong = 1;
}
else {
wrong = 0;
//do something here if correct;
}
} while (wrong);
The code above detects invalid input if the user entered a mixture of characters and numbers, or negative numbers (and zero).
However, it doesn't detect if the user entered trailing zeros in front followed by valid digits eg. 001or 00000738. If anyone else could figure this out, please share below. Thanks! :)

Answer 4

Here is another alternative of a function which takes in a char str[20] (of say, maybe 20 elements), analyses the string to check for positive integers, and returns a 0 or 1 accordingly. Lastly, convert that string to an integer using atoi().
int checkPositiveIntegers(char str[]) {
char *ptr = str;
if (*ptr == '-' || *ptr == '0') //checks for negative numbers or zero
return 1;
else {
do {
if (isdigit(*ptr) == 0) { //checks for non-digit at ptr location; isdigit() returns 0 if non-digit
return 1;
break;
}
ptr++;
} while (*ptr != '\0' && *ptr != '\n');
return 0; //returns 0 if positive integer
}
}
So the function only accepts positive numbers from 1 to 9,999,999,999,999,999,999 (up to 19 digits if char str[] holds 20 elements).
However, if you converted the string back to int n = atoi(str);, the maximum value it could reach would be 2,147,483,647 since n is declared as a signed integer. Play around with different datatypes for exploration.
Hope this helps! :)