Default parameters for a function in C --

Question

I'm a beginner, so apologies if I overstep any rules. Here's my question.

I am using a GCC compiler on Codeblocks and there is something peculiar I noticed with a particular snippet of code. I'm hoping someone could shed some light on this.

int main()
{
Tree *t;
//some operations on the tree
traverse();// No parameter is passed here.
...
}

void traverse(Tree *t)
{
..
}

In the following code, the function traverse() executes correctly. My question is why? I'm not sure about this, but if a function is not declared, its default type becomes int. Now, the compiler not only suppressed an error at the time of compilation, it also correctly supplied the parameter 't' to the function traverse().

Is this because of an intelligent compiler design?

So in general: the question I have is - what behavior does the compiler default to if it encounters a method that has not yet been declared? And more importantly, how does it "know" which parameter I wanted to pass?

For all you know, I could have had three instance of "Tree *": t1, t2 and t3. Which one would the compiler pass then?

I tried looking around on Google, but have yet to locate a definitive source.

This is incorrect and "works" only by chance. C does not have default parameters. — chill
– chill, Commented Nov 20, 2012 at 17:37

Graham Borland · Accepted Answer · 2012-11-20 17:36:10Z

3

The function is looking for its argument on the stack. The function doesn't know that the argument it's expecting isn't actually there.

By chance, the thing on the stack where it's looking for the argument is the local variable t in your main() function. If you had more local variables inside main(), then one of them would be misinterpreted as the function's argument, and things would go badly wrong.

So, it's working purely by chance.

answered Nov 20, 2012 at 17:36

Graham Borland

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5 Comments

Kanishk Over a year ago

Thanks. Yeah, that's what I suspected.

Kanishk Over a year ago

One question, though. This "stack" that you've referred to. What stack is it that stores the variables during computation? I mean, I know there's the stack for: 1) function calls(call stack)(clearly not in picture here), and 2) there's this stack for Assembly-level instructions. Is the #2 "stack" that you're referring to? The stack that is implemented when the Assembly code for the program is run?

ibid Over a year ago

The answer refers to the activation stack, also known as the call stack. It's an abstraction, not an actual stack in the memory (though some parts of it are in fact stored in memory in an actual stack). Each time you call a function (including the initial call to main), an activation record containing the local variables and some other information (such as the location of the call) is pushed on the activation stack.

ibid Over a year ago

On some machine architectures, particularly with a stupid compiler, the activation record is actually stored in the machine stack in its entirety. Modern machines and good compilers avoid using the actual stack as much as possible; it's quite possible your local variable t is stored in a machine register, which just happens to be the same register that the called function expects to find its parameter in.

Kanishk Over a year ago

Thanks, ibid. That clarified my question.

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