GATE CSE 2015 | Set 3 | Question: 54

Question

Consider the following C program:

#include<stdio.h>
int f1(void);
int f2(void);
int f3(void);
int x=10;
int main()
{
    int x=1;
    x += f1() + f2 () + f3() + f2();
    printf("%d", x);
    return 0;
}
int f1() { int x = 25; x++; return x;}
int f2() { static int x = 50; x++; return x;}
int f3() { x *= 10; return x;}

The output of the program is ______.

Comments

Associativity of operators is different from order of evaluation. In C, associativity and precedence of operators are defined, but not the order of evaluation.

 

Since each of the function operates on separate variables and the final computation is addition, it gives the same answer for any order of the function calls.

Answer 1

The variable $x$ is initialized to $1$. First and only call to $f1()$ returns $26$. First call to $f2()$ returns $51$. First and only call to $f3()$ returns $100$. Second call to $f2()$ returns $52$ (The value of local static variable $x$ in $f2()$ retains its previous value $51$ and is incremented by $1$).

$$x=1+26+51+100+52 = 230$$

Answer: $230$

Comments

Look....Static variable is always  executed only once 
the next time when you use the static variable it had became 51 so we'll use 51

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@Shaik Masthan Sir, If x=1 is in f3() code ,then f3() would return 10 right ? Because local value is preffered over global

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int f3() { x *= 10; return x;} - returns 100 - This is what mentioned in the original question. Here, x is global value.
int f3() { x=1; x *= 10; return x;} - returns 10 - here x is global value
int f3() { int x=1; x *= 10; return x;} - returns 10 - here x is referred local value

Answer 2

1+26+51+100+52

=230

Comments

In main, x is declared and initialized and x is declared globally in the program, that means global variable will get priority during execution ?

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Global has Least Priority.

Answer 3

About static variables: 

-Static variables are initialized only once. Subsequent calls to the function do not reinitialize them

 

x += f1() + f2 () + f3() + f2(); <=> x+= ((f1()+f2())+f3())+f4();  [ + operator is left to right associative]

=> f1() call: retruns 26
   f2() call: returns 51 ---- here local x is static int , stored in data segment 
   f3() call: returns 100
   f2() call: retuurn 52 ---- x is not initialized again, x = 51; x++; =>x=52
=> x+=26+51+100+52;
then "+ x"
=> x = 1+229;

=> x = 230

Order of function call is not determined:
for example, the compiler may choose the order:
f1 f2 f3 f2 
or f3 f1 f2 f2
or ....

 

About "int f(void);":

- void => function takes no arguments

 

More Details about Static variables:

-A normal local Variable(without static) is strored inside the stack => when function is done executing it is removed from the stack = > the variable is also removed

-An initialized Static variable is stored in Data segment  => Present in data segment till the end of program execution

-An uninitialized Static variable is stored in BSS

-Although a local static variable is not stored on the stack, its scope is still limited to the function in which it is declared. Only that function can access it by name.

Answer 4

Answer : 1+26+51+100+52=230

Answer 5

The difference between int x and static int x is the location in memory is dynamic and static respectively throughout the program depending upon the call.
Inside main function the value of x =1,
f1()=26
f2() first call = 51 now the value of x=50

f3()= 50*2 =100

f2() second call =52 because it has already return 51 now the value of x is assigned 51 after previous call going inside the function.
Therefore x=1+26+51+100+52 = 230

Comments

$f(3) = 50 * 2 ???$

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i think , as they are considering , is not given in f(3) , so best and simple way is x= 10 * 10 = 100