Accessing an array across files in C

Question

I'm trying to access an array across files, like this;

int option[NUMBER_OF_OPTIONS];
...
addition(&option[0], num1, num2); 
...
printf("%d", option[0]);

thats the first(main) file

and the second is like this;

void addition(int * option, unsigned number1, unsigned number2)
{
int total = number1 + number2;
...
*option ++;
}

Something like that. Dont worry about the addition method.

The problem is that the printf method allways prints 0, as if *option ++; is never executed/read.

How do i fix this?

By the way, I get a warning in the "*option++;" file saying: warning: value computed is not used.

How do I solve this problem?

Thank you!

This code should not even compile. What compiler are you using? — Nikos C.
– Nikos C., Commented Jul 31, 2013 at 12:51

Nikos C. · Accepted Answer · 2013-07-31 19:51:25Z

5

This:

*option++;

doesn't do what you think it does. What it actually means is:

*(option++);

which first applies the increment operator to the option pointer and dereferences it afterwards. The effect is:

option++;
*option; // This is a statement with no effect, hence the warning.

You need this instead:

(*option)++;

edited Jul 31, 2013 at 19:51

answered Jul 31, 2013 at 12:47

Nikos C.

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8 Comments

Daniel Daranas Over a year ago

It is valid. It is the same as *(option ++);. And the compiler already gave a warning, suspecting that something was not right.

Grijesh Chauhan Over a year ago

@DanielDaranas because the effect of expression *option ++; is just option++;, an lhs variable missing. its should be like: x = *option ++;

Nikos C. Over a year ago

@DanielDaranas, AnttiHaapala: The correct term is "modifiable lvalue". I corrected the answer. Other than that, the code is not valid. A warning is not enough; the code should actually not compile at all.

This isn't my real name Over a year ago

You're wrong about *option++ being something that shouldn't compile. If you look closely, you'll see that at the point where option is being incremented, is inside the addition() function. In that scope, option is a pointer that was passed to the function.

Nikos C. Over a year ago

@ElchononEdelson You're right. A good real-life example of why shadowing a variable from an enclosing scope is a bad idea; people get confused :-)

|

nouney · Accepted Answer · 2013-07-31 12:48:16Z

2

++ has a higher priority than *. So *option ++; is the same as *(option ++);, which do nothing (that's why you get the warning).

Try this:

(*option) ++;

answered Jul 31, 2013 at 12:48

nouney

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Comments

Antti Haapala · Accepted Answer · 2013-07-31 12:49:18Z

1

* binds looser than ++, so *option ++ = *(option++); to modify the values in array, you need to write (*option)++; that is, the suffix increment has higher precedence than the dereference operator

answered Jul 31, 2013 at 12:49

Antti Haapala

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