Printing pointed value from passed pointer - c

Question

I am confused as to why this code succesfully gives me back the right input.
#include <stdio.h>
#include <stdlib.h>
int bintest();
int main()
{
int* x = malloc(sizeof(int));
printf("\n\n*****\nEnter value of x\n*****\n");
scanf("\n%d", &x);
bintest(x);
}
int bintest(int* y)
{
printf("\ny: %d", y);
}
Why doesnt it work when I print **y, I assume pointer x is passed to bintest and stored in pointer y, which then should be called **y to print the value of the input?

Answer 1

You should start turning on compiler warnings and paying attention to them. Your short snippet yields 3! warnings.
$ gcc main.c -Wall -Wextra
main.c: In function ‘main’:
main.c:12:19: warning: format ‘%d’ expects argument of type ‘int *’, but argument 2 has type ‘int **’ [-Wformat=]
scanf("\n%d", &x);
~^ ~~
main.c: In function ‘bintest’:
main.c:18:23: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘int *’ [-Wformat=]
printf("\ny: %d", y);
~^ ~
%ls
main.c:19:5: warning: control reaches end of non-void function [-Wreturn-type]
}
^
You program has undefined behavior, and this can manifest itself by working as it should.
The correct code should be
scanf("\n%d", x);
...
printf("\ny: %d", *y);
And declare bintest as a void function if you don't want to return anything.
I am confused as to why this code succesfully gives me back the right input.
You're storing the read value in the pointer x which creates a memory leak since you throw away the malloc call. You could remove the malloc call and your program would behave the same. (Probably. It does have undefined behavior).

Answer 2

It's undefined behaviour.
It happens to work for following reason:
&x is the address of the x pointer. As the size of a pointer is usually at least the size of int, scanf("\n%d", &x); simply overwrites the pointer. Then you pass that pointer (whose value is now what the user typed in) to bintest
Your compiler should have given you at least two warnings such as:
'scanf' : format string '%d' requires an argument of type 'int *', but variadic argument 1 has type 'int **'
'printf' : format string '%d' requires an argument of type 'int', but variadic argument 1 has type 'int *'
These two warnings are rather errors and you should not ignore them.

Answer 3

You did NOT store the value of an integer in x IF that is what you wanted; you asked user for an input and overwrote the pointer x with user provided value. That pointer x now points to whatever the user gave.
Later you pass that x into that function and print that value: so you get what the user provided.
However your CANNOT de reference it. The x is pointing to a user provided value. What value is that? Do you know that it is some valid memory location.
Best read help documentation on scanf.
IF you really wanted to store an integer x with user provided value the code goes like
int x;
scanf(... &x).
Then you pass &x to that function.

Answer 4

Supporting what #Jabberwocky said,
test.c: In function ‘main’:
test.c:7:12: warning: format ‘%d’ expects argument of type ‘int *’, but argument 2 has type ‘int **’ [-Wformat=]
7 | scanf("\n%d", &x);
| ~^ ~~
| | |
| | int **
| int *
test.c: In function ‘bintest’:
test.c:11:16: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘int *’ [-Wformat=]
11 | printf("\ny: %d", y);
| ~^ ~
| | |
| int int *
| %ls
test.c:12:1: warning: control reaches end of non-void function [-Wreturn-type]
12 | }
| ^
This came up when I tried to compile the code with CLion IDE (gcc (GCC) 9.1.1 20190503 (Red Hat 9.1.1-1), C11) although even I don't understand it fully. I hope it helps and someone could explain it further.

Answer 5

If you want to scan a value to where x is pointing, then use x like this: scanf("\n%d", x). To print the value that you just placed, then pass in x like this: bintest(x);, but print *y like this: printf("%d", *y);
scanf("\n%d", &x); is the same as x = scan;
scanf("\n%d", x); is the same as *x = scan; which is what you want.
So in this example scanf("\n%d", &x); you give scanf &x, which is the address to x itself. Doing this means scanf will change the value of x, thus changing what x is pointing to.
The function scanf wants an address to where it puts its value. So to put it where x is pointing to, you should just pass in x.
The reason the values you have are the same is because you are simply passing in x like this: bintest(x); and printing y like this: printf("%d", y);. In this example, y is just a copy of x.
If you want to test out how these functions are working with pointers try using this code. I entered in 99 as an example.
#include <stdio.h>
#include <stdlib.h>
int bintest();
int main(){
int* x = malloc(sizeof(int));
printf("value of x: %d\n",x);
printf("address of x: %d\n",&x);
printf("value at %d: %d\n", x, *x);;
printf("\n\n*****\nEnter value of x\n*****\n");
scanf("\n%d", x);
printf("value of x: %d\n",x);
printf("address of x: %d\n",&x);
printf("value at %d: %d\n", x, *x);
bintest(x);
}
int bintest(int* y){
printf("\ny: %d", *y);
}
You should get an output similar to this:
value of x: 7018424
address of x: 6422316
value at 7018424: 7017768
*****
Enter value of x
*****
99
value of x: 7018424
address of x: 6422316
value at 7018424: 99
y: 99