I did an exercise on youtube and I ended up with the same solution he had but his code was able to get the address of what the pointer is pointing to but when I compiled mine and run it I get an error segmentation fault (core dump), am I somehow accessing memory that I am not supposed to access? How is it that his code was able to run but mine doesn't? I know the problem lies with printf("The address of num is %d\n", *ptr); but did the line work on youtube but on my compiler it doesn't.
#include <stdio.h>
int print_int_ptr();
int main(){
int num;
num = 10;
print_int_ptr(num);
return(0);
}
/* dont forget that in between the assignment of ptr = &num */
/* Take the address of an integer as a parameter, and print out both the
* address and the integer itself
*/
void print_int_ptr(int *ptr){
printf("The address of num is %d\n", *ptr);
printf("The value of num is %d\n", ptr);
}
In main(), num is an integer. You passes it to print_int_ptr(int *ptr).
int *ptr significates that the first argument for print_int_ptr() is a pointer, because int * declares a pointer. So as soon as you enter print_int_ptr(), ptr is considered as a pointer. But num = 10, so any reference to ptr will point out address 10, which, obviously, isn't right.
That's exactly what you're doing with printf("The address of num is %d\n", *ptr);. *ptr dereferences the pointer, it tries to access the value at address 10. But you can't access a value at an arbitrary address, unless you're lucky and the OS gave exactly this one to the process.