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Question

My program compiles correctly but i am having problem when i run it. The first scanf (width) works correctly but when i try with another scanf(height) i get segmentation fault 11. And can i do this program to work without using pointers . (Also i need limit checker function because i have to use it again and again in my program).

#include <stdio.h>
void limitChecker(int x, int y, int* input);
int main(void)
{
    int* x;
    int* y;
    printf("Enter the width of the windows. (3 - 5) : ");
    scanf("%d", x);
    limitChecker(3, 5, x);
    printf("width: %d \n", *x);
    printf("Enter the height of the windows. (2 - 4) : ");
    scanf("%d", y);
    limitChecker(2, 4, y);
    printf("Height: %d \n", *y);

}

void limitChecker(int x, int y, int* input)
{
    while(!(*input>=x && *input<=y))
    {
    printf("Please enter a value between (%d - %d): ",x,y);
    scanf("%d", input);
    }
}

Answer 1

You did not allocate memory to hold x and y .

Allocate them on the stack and then use the & address of operator to obtain a pointer to that memory.

#include <stdio.h>
int limitChecker(int x, int y, int input);
int main(void)
{
    int x;
    int y;
    printf("Enter the width of the windows. (3 - 5) : ");
    scanf("%d", &x);
    x = limitChecker(3, 5, x);
    printf("width: %d \n", x);
    printf("Enter the height of the windows. (2 - 4) : ");
    scanf("%d", &y);
    y = limitChecker(2, 4, y);
    printf("Height: %d \n", y);

}

int limitChecker(int x, int y, int input)
{
    while(!(input>=x && input<=y))
    {
    printf("Please enter a value between (%d - %d): ",x,y);
    scanf("%d", &input);
    }

    return input;
}

If you want x and y to be pointers then you have to assign them valid memory before you use them.

int * x = malloc(sizeof(int));
int * y = malloc(sizeof(int));

Answer 2

You need to use a reference to the variables used in the scanf().

For example, scanf("%d", &x);

The first parameter of scanf() is for the type of data, and the following parameter(s) are a list of pointers to where you would like the user input to be stored.

CORRECTED CODE:

#include <stdio.h>
void limitChecker(int x, int y, int* input);
int main(void)
{
    int x;
    int y;
    printf("Enter the width of the windows. (3 - 5) : ");
    scanf("%d", &x);
    limitChecker(3, 5, &x);
    printf("width: %d \n", x);
    printf("Enter the height of the windows. (2 - 4) : ");
    scanf("%d", &y);
    limitChecker(2, 4, &y);
    printf("Height: %d \n", y);

}

void limitChecker(int x, int y, int* input)
{
    while(!(*input>=x && *input<=y))
    {
    printf("Please enter a value between (%d - %d): ",x,y);
    scanf("%d", input);
    }
}

Answer 3

#include <stdio.h>

int limitChecker(int x, int y, int value){
    return x <= value && value <= y;
}

int inputInt(void){
    //x >= 0
    int x = 0;
    int ch;
    while('\n'!=(ch=getchar())){
        if('0'<=ch && ch <= '9')
            x = x * 10 + (ch - '0');
        else 
            break;
    }
    return x;
}

int main(void){
    int x, y;
    do{
        printf("Enter the width of the windows. (3 - 5) : ");
        x = inputInt();
    }while(!limitChecker(3, 5, x));
    printf("width: %d \n", x);
    do{
        printf("Enter the height of the windows. (2 - 4) : ");
        y = inputInt();
    }while(!limitChecker(2, 4, y));
    printf("Height: %d \n", y);
    return 0;
}