Thread: Question from a Newbie

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First, make sure to post your code in [CODE] tags next time. Its difficult to read otherwise.

Code:

for ( i = 0; i < 20; i++ )

The loop will execute 20 times (0, 1, ..., 19, all inclusive). In the "for" loop we're calling a function called "AddOne", which we can assume will "add one" to the argument, which is "num". Since the initial value of "num" is 5, and we add one to it, twenty times, the result will be 5 + 1*20 = 25. Next,

Code:

AddOne( &num );

The "&" means to pass the address of the variable "num". This is usually done when you want to pass the variable to a function and have the function modify the variable. Next,

Code:

void AddOne( int *myVar ) {

Declaring a function, which expects to be passed the address of the argument "myVar". Finally,

Code:

(*myVar) ++;

When you want to access the contents of a pointer (i.e. the value that the address points to), you have to dereference it, this is done simply with "*", so "*myVar" says "go to the address that myVar points to and get the value stored there". The very first time "AddOne" is called, the value of "myVar" would be an address, something like "0xf123abc" or whatever. The value that this address points to, i.e. "*myVar", will be 5. The "++" is the postincrement operator. It "adds one" to the value, so that after this line, "myVar" will, of course, still have the same value (the address), while the value stored at that address (*myVar) will have changed to 6.

Modify the program so that the definition of "AddOne" is instead this

Code:

void AddOne( int myVar );

, and it is instead called this way

Code:

AddOne( num );

, and implemented this way

Code:

void AddOne( int myVar ) {
  myVar++;
}

Do you understand why the output is what it will be? (5)

Well I didnt say to remove the "for" loop, but the result should be the same. Did you run it? The result should be 5.

(almost) All parameters to functions are passed by value. This means that when you call "AddOne", a copy of "num" is passed to the function. So when, inside the "AddOne" function, it increments "myVar" by one ("myVar++"), the value of myVar is now 6. However, this doesnt have anything to do with the variable or value of "num" in main. "AddOne" modified a copy of "num"--"num" and "myVar" are two entirely different variables. So since "AddOne" doesnt have any effect on "num", "num" stays the same and it would output "5".

The same thing with pointers. When you pass "&num" to the previous versino of "AddOne" that accepted an "int *", a copy of the value is passed to "AddOne". In this case, the value is "&num", the address of "num". Now we have the variable "num" in main, and we have a copy of its address in "myVar" in "AddOne". Since we have the address, we can directly access and modify the value (via "*myVar"). The "(*myVar)++", again, says "go to the value pointed to by the address "myVar" and increment it by one". Since it modified the value at that address, it indirectly modified "num".

It would be difficult, or at least take a while to fully explain this concept. But it is very important, and you should try and look up a basic tutorial or explanation, preferably with graphical examples to understand it. One on this site is Pointers.

Let us know if you have questions