Getting integer value out from while loop doesn't work

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I am trying to fix the code for a program that should just use a while loop in a separate function to get the int numbers 1 or 2 and then I want them printed with printf(). I cant get it to work.
Here is the code now:

#include <stdio.h>

int get_1_or_2( void );

int main( void ) {

    printf("%d", get_1_or_2() );

    return 0;
}

int get_1_or_2( void )
{
    int answer = 0;

    while((answer < 1) || (answer > 2)) {
        printf("Enter 1 for Yes, 2 for No\n");
        scanf("%d", answer);
    }
    return answer;
}

I am starting to think that the loop cant write to the answer variable outside the loop. However I dont know for sure.

Please help me understand how to fix this code.

What I have tried:

I have tried to put the

return answer;

code in the while loop. I have also tried with a for loop. I also tried with a for loop with the

return answer;

code inside the loop.

Posted 5-Jun-21 21:54pm

Mieczyslaw1683

Updated 5-Jun-21 22:05pm

v2

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OriginalGriff · Accepted Answer · 2021-06-05T22:05:00

Solution 1

Look closely at your code, and pay attention to this line:

C

scanf("%d", answer);

Where is scanf supposed to put the value?

Remember, in C all parameters are passed by value, so a copy of the current value of answer will be handed to scanf - which will assume that it's an address to save the new value in: C library function - scanf() - Tutorialspoint[^]
You need to pass an address to scanf, not a value!
Try this:

C

scanf("%d", &answer);

Posted 5-Jun-21 22:05pm

OriginalGriff

Comments

Mieczyslaw1683 6-Jun-21 4:11am

Okey, now the program works as intended (with a & before). Weird that this site: https://www.tutorialspoint.com/c_standard_library/c_function_scanf.htm Is showing without the "&" sign before the char that the value should go into. Thanks for the reply.

OriginalGriff 6-Jun-21 4:34am

Not weird at all - it's example shows an array being passed, and the name of an array is a pointer to the first element of the array.

Assuming this:

int arr[10];

then this:

int *p = arr;

is equivelant to this:

int *p = &(arr[0]);

But a lot easier to read!

You are passing a single value variable, so you need to get the address.