Thread: static variables inside a function.

when the function f is called the second time,does the compiler ignore the declaration of a inside the function?

A more interesting question is will the compiler ignore the initialization on subsequent calls. And the answer is yes, provided you truly initialize rather than simply assign:

Code:

void foo()
{
    static int a = 0; /* Initialization happens once */
    static int b;

    b = 0; /* "Initialization" happens every call */

    ....
}

Originally Posted by CommonTater

Rookies? I've been at this for 6 years now and Prelude makes me look like a beginner!

You can't compete with a goddess tater.

so if i were to separate the assignment and initialization,then a will be assigned 0 and the o/p
would differ.

Here's a more complete example that shows the difference:

Code:

#include <stdio.h>

void foo()
{
    static int a = 0; /* Initialization happens once */
    static int b;

    b = 0; /* "Initialization" happens every call */

    printf("a=%d\tb=%d\n", a++, b++);
}

int main(void)
{
    foo();
    foo();
    foo();

    return 0;
}

Originally Posted by Prelude

Code:

printf("a=%d\tb=%d\n", a++, b++);

Now that we're on the topic of rookies: very very bad coding practice to show a beginner!

very very bad coding practice to show a beginner!

I'm not in the business of dumbing down my code to protect beginners from reality. They're going to see far worse expressions, so better to get things moving as quickly as possible. Though since you seem so adamant about it, why not offer up an explanation for the beginner's reading your post?

Dunno, it seems fine to me since the expressions are standalone and it is clear that the variables involved are not repeated or aliases. It feels akin to writing:

Code:

array[size++] = element;

which I would posit is not bad coding practice.

Originally Posted by laserlight

Dunno, it seems fine to me since the expressions are standalone and it is clear that the variables involved are not repeated or aliases. It feels akin to writing:

Code:

array[size++] = element;

which I would posit is not bad coding practice.

I would agree that it is not bad. However, it also runs afoul of coding guidelines that require any non-declarative statement to have only one observable effect other than an overt function call. That statement has two observable effects: it changes the value of array[size] and increments size.

Originally Posted by MWAAAHAAA

Now that we're on the topic of rookies: very very bad coding practice to show a beginner!

Why? It's perfectly legal and invokes no undefined behavior. If they don't know what the increment operator is, it's as good a time as any to go look it up.

Originally Posted by grumpy

However, it also runs afoul of coding guidelines that require any non-declarative statement to have only one observable effect other than an overt function call. That statement has two observable effects: it changes the value of array[size] and increments size.

Whose guidelines are those?


Quzah.