Here is some C code trying simply to prevent the user from typing a character or an integer less than 0 or more than 23.
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
const char *input;
char *iPtr;
int count = 0;
int rows;
printf("Enter an integer: ");
scanf("%s", input);
rows = strtol(input, &iPtr, 0);
while( *iPtr != '\0') // Check if any character has been inserted
{
printf("Enter an integer between 1 and 23: ");
scanf("%s", input);
}
while(0 < rows && rows < 24) // check if the user input is within the boundaries
{
printf("Select an integer from 1 to 23: ");
scanf("%s", input);
}
while (count != rows)
{
/* Do some stuff */
}
return 0;
}
I made it halfway through and a small push up will be appreciated.
Use scanf("%d",&rows)
instead of scanf("%s",input)
This allow you to get direcly the integer value from stdin without need to convert to int.
If the user enter a string containing a non numeric characters then you have to clean your stdin before the next scanf("%d",&rows)
.
your code could look like this:
#include <stdio.h>
#include <stdlib.h>
int clean_stdin()
{
while (getchar()!='\n');
return 1;
}
int main(void)
{
int rows =0;
char c;
do
{
printf("\nEnter an integer from 1 to 23: ");
} while (((scanf("%d%c", &rows, &c)!=2 || c!='\n') && clean_stdin()) || rows<1 || rows>23);
return 0;
}
Explanation
1)
scanf("%d%c", &rows, &c)
This means expecting from the user input an integer and close to it a non numeric character.
Example1: If the user enter aaddk
and then ENTER
, the scanf will return 0. Nothing capted
Example2: If the user enter 45
and then ENTER
, the scanf will return 2 (2 elements are capted). Here %d
is capting 45
and %c
is capting \n
Example3: If the user enter 45aaadd
and then ENTER
, the scanf will return 2 (2 elements are capted). Here %d
is capting 45
and %c
is capting a
2)
(scanf("%d%c", &rows, &c)!=2 || c!='\n')
In the example1: this condition is TRUE
because scanf return 0
(!=2
)
In the example2: this condition is FALSE
because scanf return 2
and c == '\n'
In the example3: this condition is TRUE
because scanf return 2
and c == 'a' (!='\n')
3)
((scanf("%d%c", &rows, &c)!=2 || c!='\n') && clean_stdin())
clean_stdin()
is always TRUE
because the function return always 1
In the example1: The (scanf("%d%c", &rows, &c)!=2 || c!='\n')
is TRUE
so the condition after the &&
should be checked so the clean_stdin()
will be executed and the whole condition is TRUE
In the example2: The (scanf("%d%c", &rows, &c)!=2 || c!='\n')
is FALSE
so the condition after the &&
will not checked (because what ever its result is the whole condition will be FALSE
) so the clean_stdin()
will not be executed and the whole condition is FALSE
In the example3: The (scanf("%d%c", &rows, &c)!=2 || c!='\n')
is TRUE
so the condition after the &&
should be checked so the clean_stdin()
will be executed and the whole condition is TRUE
So you can remark that clean_stdin()
will be executed only if the user enter a string containing non numeric character.
And this condition ((scanf("%d%c", &rows, &c)!=2 || c!='\n') && clean_stdin())
will return FALSE
only if the user enter an integer
and nothing else
And if the condition ((scanf("%d%c", &rows, &c)!=2 || c!='\n') && clean_stdin())
is FALSE
and the integer
is between and 1
and 23
then the while
loop will break else the while
loop will continue