Need help: Explanation of a static storage class

I am going through a tutorial on C and just ran into static storage class on an int. Reading this code I would assume that the variable number is set to 7, print statement says number is 7. Than 7 turns to 8, we return to the top of the loop, "number is than set to 7 again", than we print number is 7. Can someone help me understand how 7 gets incremented and than holds that value even though when it loops through again number is defined again and set to 7.

Code:

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int main( int argc, char ** argv )

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Code:

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{
    for (int i =1; i <= 5; ++i) {
        static int number = 7;
        printf("number is %d\n" , number++);
    }
    return 0;
}

---

Output:

number is 7
number is 8
number is 9
number is 10
number is 11

Example

Code:

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void foo(void)
{
    static int var = 7;/* This runs only once*/
    printf("%d\n", var);
    var++;
}

int main(void)
{
    // call foo 4 times
    foo();
    foo();
    foo();
    foo();
 
    return 0;
}

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The output would be:
7
8
9
10
Why?
As you may know, local variables terminate their lifetime, when the function they live in terminates.
var, in our example, is a static variable though. As a result, it is "immortal". It won't die, no matter how many times the function the terminates. It is created once and dies once, when your program ends.
As a result, the value of the variable is remembered among function calls!
But, this line of code

Code:

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static int var = 7;

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runs only once, because the static variable is created one time only!


Notice, that static keyword can also been found for limiting the scope of a variable, making her have in -file scope. However, the static keyword has not the same effects on these situations.

//Also, Sam, welcome to the forum :D

static variables are initialized to a value ONLY once, but they are not constant, and can be changed. Between calls to a function, their value will remain in place.

Thank you for the explanation, and the welcome. That makes perfect since now. "runs only once, because the static variable is created one time only!"
I tested this out using and not using static, this was actually blowing my mind for a second. I appreciate the help.

Quote:

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Originally Posted by Sam Pertuit

Code:

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int main( int argc, char ** argv )

---

Code:

---

{
    for (int i =1; i <= 5; ++i) {
        static int number = 7;
        printf("number is %d\n" , number++);
    }
    return 0;
}

---

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The easiest way to think of it is that the line which says "static int number = 7" is executed exactly once, at compile time. Therefore, each time you enter the block where number is defined, it retains its previous value.

Note that there is also "static linkage" which does something different. For items with static linkage it also makes sense to define a static const member which cannot be changed.

Quote:

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Originally Posted by Sam Pertuit

Thank you for the explanation, and the welcome. That makes perfect since now. "runs only once, because the static variable is created one time only!"
I tested this out using and not using static, this was actually blowing my mind for a second. I appreciate the help.

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You are welcome :)

Thanks for the heads up on the "static linkage", I will be looking for this shortly.

Quote:

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Originally Posted by Sam Pertuit

Thanks for the heads up on the "static linkage", I will be looking for this shortly.

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I hope you read my last sentence on post #2, which is about linkage. :)