[FIXED] What is the different effect of assigning to vs. initialising static variables in a function?

Issue

#include <stdio.h> 

int print1(void)    
{    
    static int x= 10;

    x+=5;    
    return x;    
}

int print2(void)     
{    
    static int x;

    x = 10;    
    x+= 5;    
    return x;
}

int main()    
{    
    int x;    
    x = print1();    
    x+= print1();    
    x+= print2();     
    x+= print2();    
    printf("%d", x);    
    return 0;    
}

My solution:

1. First in print1 statement the value of the will increment by 5 , x will be 15 .
2. Then upon calling print1 function for the 2nd time it will be 15+5=20, since it will hold the previous value.
3. Now for the function print2, it will again initialise the x value with 10, then increment it by 5. Value of x will be 15.
4. for the last statement value of x will be 15+5=20.
5. x=15+20+15+20=70

But the answer is 65. Please correct me where did I go wrong? I even wrote it in my compiler, it is giving output 65, but if change the position of x inside Print2 statement like static int x=10, it is giving different output?

Why this is happening?

Solution

The init of a static variable like static int x=10; only happens once per program execution. And that is not like a statement being executed, it is during setup of variables for the program.
The assignment statement x = 10; is executed each time the function is executed.
This means the two code versions you discuss are different and especially have different effects on variable values; which in turn explains the different output.

Answered By - Yunnosch

Answer Checked By - Mary Flores (PHPFixing Volunteer)