C help

austinium · 08-02-2009, 11:03 PM

Hi,

This may be something really simple, but i havent been able to figure out the reason why the statement is valid (apart from the fact that the standard probably says so

). What use could it possibly be of?

Code:

#include <stdio.h>

int main()
{
        int n;
        n;      /*Why is this a valid statement?*/
        return 0;
}

thanks

karamarisan · 08-02-2009, 11:27 PM

It's not necessarily simple - you have to think about it in more abstract terms. Don't look at code as a list of statements that each have to do something; look at each line as something to be evaluated, where the evaluation of functions (and sometimes other things) can change the state of the system. You may have heard the words 'substitution model' if you've taken basic CS classes; if not, don't worry about it. Consider this:

Code:

int ret = foo(0, bar(x), mything.baz);

1. Well, we're going to assign something to our new integer ret. What are we assigning? Check out the right side of the =.
2. Let's evaluate foo. To do that, we'll need to look at each of its arguments.
2-1. 0 is an int literal. No problem.
2-2. bar(x). Let's evaluate that. bar takes one argument.
2-2-1. x is a variable - somewhere else, it was defined as 62, so substitute 62 for x and continue.
2-2-1. bar(62). Dive into bar's function definition, and get a value. Let's say it comes back with 125.
2-2. Substitute 125 for bar(x). We're looking at foo(0, 125,...) right now.
2-3. mything.baz? Check out the baz field of the struct mything, and look, it's 15.
2. Great, now we've got foo(0, 125, 15). Plug all of those values into the appropriate places in foo's definition and evaluate that. Let's say that comes back as 70.
3. The right side is now a literal, so we can do the assignment. 70 is stored into ret.

A little long-winded, but do you see what I'm saying about evaluations? Whether calling foo() creates a window somewhere or not isn't important; all the system does is evaluate statements. If there were no 'int ret =', it'd be the same deal, and the return value would just be discarded. 'n;' evaluates to whatever's in n, and we move on to the next line. It doesn't do anything, but it fits the model we're working in. Call it a degenerate case.

I'm not entirely sure that will get across, so please let me know if I'm babbling.

austinium · 08-02-2009, 11:49 PM

thanks

If I am getting this right, then the compiler evaluates n; to whatever (garbage) value it contains, now we have no variable on the left side so the compiler does nothing else on this line moves to the next line of code.

karamarisan · 08-03-2009, 12:07 AM

Yep. If you get that there isn't really a difference from this perspective between 'evaluate n' and 'evaluate drawRectangle(0,0,50,50)', and if I am to be trusted (

), then you've got it.

sundialsvcs · 08-03-2009, 12:40 AM

To my way of thinking, "if 'n' is a simple variable, then I was not aware that any compiler would accept it. I therefore share austinium's sentiment: to me, "if such a statement is acceptable, then it's a bug in the compiler."

Very early compilers accepted many things that are regrettable today. There simply wasn't enough space to permit a "better" compiler to be written in those days.

Some programs nevertheless were written which took advantage of these peculiarities ... literally because they had to. Syntax checking existed in separate programs like lint because there was no room to put such tests into the compiler.

austinium · 08-03-2009, 02:29 AM

it compiles fine on cl(vc++ 6.0) and gcc 3.2.2

mrtiller · 08-03-2009, 05:03 PM

Quote:

Originally Posted by austinium

it compiles fine on cl(vc++ 6.0) and gcc 3.2.2

Also gcc (3.4.4 and possibly earlier) will warn you with
warning: statement with no effect
if you use the -Wall switch