Why does global pointers behave erratically in this function? - c

Question

I've tried to write a program that uses global pointers to track down local variables, but when I use the right syntaxes, it doesn't run.
When I alter the syntaxes, the program yields correct results. I don't fully understand the mechanism behind it, and hoping that more experienced coders could explain it. Thank you.
#include <stdio.h>
#include <math.h>
int *q;
int *u;
int *p;
void test () {
printf("Insert #1?\t"); scanf("%d", &u);
/*should be scanf("%d", u) for u is a pointer*/
printf("Insert #2?\t"); scanf("%d", &p);
/*should be scanf("%d", p) for p is a pointer*/
}
int main () {
test ();
printf("%d\n", u);
/*should be printf("%d\n", *u) since I am trying to dereference the value of u*/
printf("%d\n", p);
printf("%d\n", q);
}
The output should be 3 numbers, 2 first numbers are inputs from user, and the third one should be a nonsense number since the 3rd pointer is not pointing anywhere.
Update: It works!!!!! Thank you so much for your help :).

Answer 1

This has nothing to do with global variables. There are multiple problems here but the most pressing error is that you are attempting to use uninitialised pointers. This never works. You need to understand pointers before you can use them properly.
In particular, if you have a declaration like this:
int *x;
Then, no matter what you do, you first need to assign a valid pointer to x1.
In your case the error is compounded by the fact that you’re using scanf wrong. Luckily the solution here is very easy: Don’t declare pointers. Instead, declare int variables.
#include <stdio.h>
int q;
int u;
int p;
void test () {
printf("Insert #1?\t"); scanf("%d", &u);
printf("Insert #2?\t"); scanf("%d", &p);
}
int main () {
test();
printf("%d\n", u);
printf("%d\n", p);
printf("%d\n", q);
}
Ironically the comments in your original code already point out the issue, they just suggest a wrong solution. It’s a bit of a mystery where these comments are from. ;-)
1 Technically the only exception is for statically allocated pointers, which are correctly zero-initialised and can therefore be used in null pointer comparisons.

Answer 2

Your global variables are the pointers to the ints, not the ints themselves. When you read in the values, you change where in memory u points to, not the value of u.
You should replace it with either scanf("%d", u); or change the declaration to int u.
You need to be careful taking pointer values from the user. Otherwise, you could risk things like null pointer dereferencing. For instance, Entering 0 when you scanf("%d", &u);, it would then make the value of u be NULL.

Answer 3

As I said in my comment to the question, a pointer is a type that holds the address to another variable of some type, so int* holds an address for a variable of type int. In your case, &u is referring to the address of u, so you're trying to scan a number into a pointer which won't work. You should change it to only u so it will scan to the variable it points too, but it needs to point to a valid variable in the first place. Here is a valid example:
int someNumber = 5;
int* u = &someNumber; // u points to a valid address statically allocated
scanf("%d", u); // Scans the number inputed to someNumber because u points to it
getchar(); // Avoid buffer overflow
printf("%d\n", *u); // A * dereferences u - gets the value of the variable it's pointing to

Answer 4

Pointers need to point to something, such as another variable, or memory allocated dynamically with malloc().
You can't use scanf() to read directly into a pointer. You can use it to read into what the pointer points to.
So you can write something like this:
#include <stdio.h>
#include <math.h>
int q;
int *qp = &q;
int *up;
int p;
int *pp = &p;
void test () {
printf("Insert #1?\t"); scanf("%d", up);
printf("Insert #2?\t"); scanf("%d", pp);
}
int main () {
up = malloc(sizeof(*up));
test ();
printf("%d\n", *up);
printf("%d\n", p);
printf("%d\n", q);
}

Answer 5

Your pointers have no value assigned to them.