About variables

**ZaC** · 06-09-2009

I'm trying to understand how c variables are stored and deleted...
I tought that a variable created in a function was erased after the function stop to work.
But I think i'm wrong because of a code I'm studing like this:

Code:

...
item *function(int a, int b, int c){
    item *y;
    ...
    return y;
}

int main(){
    item *x;
    ...
    x=function(a,b,c);
    ...
    return 0;
}

Now "x" points to "y", but wasn't "y" erased after function execution?
I don't know if I explained well what I don't understand... I hope someone could help me.
Thanks in advance

**laserlight** · 06-09-2009

Originally Posted by ZaC

Now "x" points to "y", but wasn't "y" erased after function execution?

After the call of the function, x does not point to y. Rather, x points to what y pointed to.

**KBriggs** · 06-09-2009

y is just a pointer to a location in memory. So the label y is gone after the function finishes, but x will be pointing to the same place as y was

**MK27** · 06-09-2009

Originally Posted by ZaC

I'm trying to understand how c variables are stored and deleted...
Now "x" points to "y", but wasn't "y" erased after function execution?
I don't know if I explained well what I don't understand... I hope someone could help me.
Thanks in advance

If you are coming from some language that does not use either pointers or references, you have a leap to make.

ps. is that avatar an oil well with its head about to explode??

**ZaC** · 06-09-2009

Uhm... I see. Thank you very much!
I got rusty a lot in C programmin, it was so easy... but I can't remember this thing!
Thanks again

**sean** · 06-09-2009

As a side note (just mentioning this because it's what came to mind as I read the first part of you questions), variables do go out of scope, yes, but the actual value that was stored in memory is still there until something else overwrites it. So it's entirely possible that if a function is called many times, unless you properly initialize the variables - you're gonna get old data.

**Elysia** · 06-09-2009

Well, in the eyes of the language, that is not so. It is just undefined--anything can and will happen.
What you say is probably true of most modern operating systems such as Windows or Linux, but may not be true in other systems...

**cpjust** · 06-09-2009

It also depends a lot on what code you have in the ... sections.
If you have:

Code:

y = malloc( 10 );

then the memory that y holds should never get released until you call free() on that memory; but if you have:

Code:

item it;
y = &it;

then y is pointing to memory on the stack, which is released once the function returns. Although there's no guarantee that the stack memory will be erased (overwritten) unless you start putting more data onto the stack.

**ZaC** · 06-09-2009

Now I'm confused a little...
Reading the original code I posted, can I be sure x contains the data y pointed to? Or not? Or does it depends on... what?
Thanks again

PS: @ sean, nice note!

**tabstop** · 06-09-2009

In your original code, x will get a copy of the value of the pointer y, not what was pointed to by y.

**KBriggs** · 06-09-2009

function returns a pointer. In this case it returns a pointer to y, so x becomes a pointer to the same thing that y pointed to. y is gone, but whatever it pointed to is still in the program since x points to it instead.

**Elysia** · 06-09-2009

You can be certain that x has the same value as y.
Whatever data was stored in y will now be stored in x.
This is different from what it points to.

**sean** · 06-09-2009

edit: wow - lots of posts since the one I was answering!

x and y are both pointers. Be careful wherever you see the * and & operators - you might be talking about value, you might be talking about location.

Since the function returns y, and that is assigned to x, yes, they both contain the same data. But don't forget that pointers contain the location of another value. So we say that x points to the same place as y (like laserlight said).

In this case it returns a pointer to y

No, no, no... y is a pointer. The value of y is a memory address, but that value is return to x. A pointer to y does not exist in this code.

**ಠ_ಠ** · 06-09-2009

Originally Posted by MK27

ps. is that avatar an oil well with its head about to explode??

It's a person

**ZaC** · 06-09-2009

Ok, now, it's more clear! I should repeat a little bit more about pointers

The last question just to make it clear at all.
I understood that: variables are destroyed, but not their datas. So I can use the addresses of theese datas to treat them, isn't it? But how can I be sure those addresses will not be used in the rest of the program?

O.T.: @MK27: it's a person, sorry but i didn't read your ps before. I think oil is f*g all of us. And it's not the only one :/

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