static variable programming Q

Question

Question:

Comments

15 20 15 15

I think something is fuzzy about the scope here

This one I get it I guess as when second time print1 is called, the statement will be skipped at runtime and so is assigning value 10 is skipped. Hence, 15 20. In print2, when second time is called, only static int x will get skipped but not x=10. Hence, 15 15

I have a doubt though, in this program, I typed print1(), print2(), print1(), print2() and to distinguish, in print2 I typed x=50.

Output is 15 55 20 55

when print1 is called again, declaration is skipped, then it should simply add 5 and show 60. How is it 20?

Is x different for both print1 and print2?

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Have a look at addresses

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yes it will print 15 55 20 55

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please copy your code post here ..

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It means that both x is declared once but in both functions. Both function have their own copy of one time declared x variable.

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#include<stdio.h>
void print1()
{
    static int x=10;
    x+=5;
    printf("x=%d and the address in print1 is %p \n",x,&x);
}
void print2()
{
    static int x;
    x=50;
    x+=5;
    printf("x=%d and the address in print2 is %p \n",x,&x);
}
int main() {
   print1();
   print2();
   print1();
   print2();
}

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see my comments its given by made easy

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But that doesn't explain the case in my code. Then it should be

15 55 60 55

Also, addresses are different for both.

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@joshi_nitish @Anu007 @hs_yadav @Ashwin Kulkarni

Even if variable name is same, with multiple static declarations in different functions, is the scope of variable limited to repeated calls for same function only?

Answer 1

I Think This will make you more clear....

Ans is ;- 15 20 15 15

Comments

ans is 15 20 15 15

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Solution :

(d)
• print 1( ): x = 10 + 5 = 15; since the variable is of static storage class, hence it will retain its value
between different function calls.
• print 1( ): x = 15 + 5 = 20; since it has retained its value 15.
• print 2( ): x is defined again inside the function and hence will print, x = x + 5 = 10 + 5 = 15. Again
when the function will be called, x = 10 + 5 = 15. Here second time also x = 10 will be there because
it is not initialized at the time of definition.
Hence output 15, 20, 15, 15.