如何将一个char中的值分配给一个指针？

Question

In function input, I was trying to do an input validation that will only accept input from the given choices. I tried using only the variable opt to get the values from scanf but when I compare with if, it gives an error saying something about comparing a pointer to another type.

With this code, I kinda managed to make that error disappear and the program now runs but the program ends after I input the choice. Can anyone help me with this? Thanks :)

#include <stdio.h>
#include <ctype.h>

int add(int n1, int n2);
int subtract(int n1, int n2);
int multiply(int n1, int n2);
int divide(int n1, int n2);
void input(int *n1, int *n2, char *opt);


int main(void)
{
    int n1, n2, ret;
    char opt;

    start:
    input(&n1, &n2, &opt);

    switch(opt)
{
            case '1': 
                ret = add(n1, n2);
                printf("The sum is %d\n", ret);
                break;
            case '2':
                ret = subtract(n1, n2);
                printf("The difference is %d\n", ret);
                break;
            case '3': 
                ret = multiply(n1, n2);
                printf("The product is %d\n", ret); 
                break;              
            case '4': 
                ret = divide(n1, n2);
                printf("The quotient is %d\n", ret);
                break;
            case 'R':
                goto start;
                break;
            case 'E':
                printf("Goodbye!\n");
                return 0;
                break;
    }
    return 0;   
}


void input(int *n1, int *n2, char *opt)
{
    int valid;
    char choice;

    printf("Enter first number: \n");
    scanf("%d", n1);

    printf("Enter second number: \n");
    scanf("%d", n2);   

    getchar();
    valid = 0;
    while( valid == 0)
    {
        printf("Addition -> 1\nSubtraction -> 2\nMultiplication -> 3\nDivision -> 4\nReset -> R\nExit -> E\n");


        if ( scanf("%c", &choice) == ('1' || '2' || '3' || '4' || 'R' || 'E'))
        {
            valid = 1;
        }
        else
        {
            printf("Invalid input!\n");
        }
    }
    opt = &choice;

}


int add(n1, n2)
{
    int result;
    result = (n1+n2);
    return result;
}

int subtract(n1, n2)
{
    int result;
    result = (n1-n2);
    return result;
}

int divide(n1, n2)
{
    int result;
    result = (n1/n2);
    return result;
}

multiply(n1, n2)
{
    int result;
    result = (n1*n2);
    return result;
}

Answer 1

One of the problems with the program you show is that it will not give you the option entered. That's because arguments are passed by value in C, which means that arguments are copied and changing an argument will not change the original value.

This is noticeable when you do

opt = &choice;

This reassignment of the opt pointer will be lost once the function returns.

What you want in this specific case is to use the dereference operator to dereference the pointer you pass in, and assign the value of choice to the dereferenced pointer:

*opt = choice;

---

There are many other errors in your code, including one which you don't notice because it works but not in the way you expect it to.

Lets take the condition

scanf("%c", &choice) == ('1' || '2' || '3' || '4' || 'R' || 'E')

The scanf function returns the number of successfully parsed formats, or EOF . In your case it would return 1 if it read and parsed a character.

The expression ('1' || '2' || '3' || '4' || 'R' || 'E') is the one that doesn't work as expected. In C all non-zero values are considered "true", only zero is "false". If your system is using the ASCII alphabet (most likely) then the expression

('1' || '2' || '3' || '4' || 'R' || 'E')

is equivalent to

(49 || 50 || 51 || 52 || 82 || 69)

All of the sub-expressions are "true" leading to the while expression to be "true" which in C is equivalent to 1 .

So the expression

scanf("%c", &choice) == ('1' || '2' || '3' || '4' || 'R' || 'E')

is equivalent to

scanf("%c", &choice) == 1

which is what you should check for, but not in this way.

What you should do is to put a check inside the loop for the correct alternative, but remember what I said the result of ('1' || '2' || '3' || '4' || 'R' || 'E') would be.

Answer 2

I think the above answer explains why your code doesn't work very well, so here is a working version of your input() function and a bit shorter version of the operator functions:

void input(int *n1, int *n2, char *opt)
{
    int valid;
    char choice = '0';

    printf("Enter first number: \n");
    scanf("%d", n1);

    printf("Enter second number: \n");
    scanf("%d", n2);   

    getchar();
    valid = 0;
    while( valid == 0)
    {
        printf("Addition -> 1\nSubtraction -> 2\nMultiplication -> 3\nDivision -> 4\nReset -> R\nExit -> E\n");

        choice = getchar();
        switch (choice) {
            case ('1', '2', '3', '4', 'R', 'E'):
                valid = 1;
                break;
            default: printf("Invalid input!\n");
                break;
        }
    }
    *opt = choice;
}

and the operators you can just do

int add(n1, n2)
{
    return n1 + n2;
}