Okay, Pointer Primer :
First, in C, all variables are pass by value, including pointers. But a pointer happens to hold the address of another object, which can be used to change that object, but not the pointer you pass to the function.
typedef struct DATA { int x; };
Pass by value :
They are all pass by value, including the pointer. But the pointer allows you to change memory.
Here's a little quiz for you. What does the following code output? Does it compile? Does it run successfully? Why or why not?
DATA d = {1}; int i = 2; int* iptr = &i; IncrementLocalDataCopy(d); IncrementLocalDataPointerCopy(&d); IncrementLocalIntCopy(i); IncrementLocalIntPointerCopy(iptr); printf("d.x is %d , and i is %d\n" , d.x , i);
They are equal
<spoiler>
They equal 2
</spoiler>
Tips on working with pointers :
int i = 0; int* iptr = &i;/// & Means 'the address of' i i->x = i->x + 1;/// -> Derefences a pointer and accesses a field *iptr = 3;/// * Dereferences a pointer and returns a reference to i which is used to assign the value 3 iptr = 0;/// Does nothing to i *iptr = 4;/// Crashes if you're lucky with a segfault SIGSEGV accessing invalid memory address
And if you ever switch to C++, they introduce a second meaning for & when inside a function signature or variable declaration. It means 'a reference to' an object.
Example :
int i = 0; int& j = i;/// j now references i j = 1;/// Directly modifies i through its address printf("I should be 1, and the value is %d" , i);
There. Learn that, and then tell us why it wasn't working. 