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Question

What is the output???

Comments

@ Shaikh, got it, I didnt thought about part1.....It should be undefined behaviour in both.....Thanks

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@ Shaikh, answer could be 302 also right????

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brother, don't solve the questions which shows undefined behaviour, those type of questions never ask in GATE

Answer 1

when static scoping is used :

x is global with value 10;

Part2(&x)

               another x is created with local scope and static : x=15;

               *b=*b *x; value of global x = 150;

                 Part1(&x) // this is calling to part1 by passing local x address

                                *a+=x++; // here += is RL assosiative so first x++ will be evaluated then +=   and this is static scoping so first we will search in local x then global x so no local x to part1 is there then go to global x and increment it, now global  x=151.

                               *a+= 150;

                               *a=15+150=165; local x=165;

                               print(165);

             //part2 print(local x = 165)

Part1(&x):// global x=151

                *a+=151; global x=152

                 *a=*a+151; 152+151=303 ; global x=303

                 print(303);

When Dynamic scoping is used:  https://courses.cs.washington.edu/courses/cse341/03wi/imperative/scoping.html

global x=10;

Part2(&x):

              another x is created with local scope and static : x=15;

              *b=*b *x; value of global x = 150;

               Part1(&x) // this is calling to part1 by passing local x address

                             *a+=x++; // this is dynamic scoping so first value it will search in local scope then in its caller function so at part2 it will get x=15;

                              *a+=15; now post increment x=16

                              *a=16+15=31;  local x=31

                              print(31);

               print(31)

Part1(&x):// global x=150

               *a+=150; here x will find x in global so x=151

               *a=151+150=301

               print(301);

 

 

with Static scoping :

165

165

303

with dynamic scoping :

31

31

301

Comments

@ Dharmendra,     how could you evaluate *a=*a + x++     .......while either side of + could be evaluated first, that would result in different answer 302/303

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*a+=x++;

in this statement += is right-to-left associative so first right portion will be evaluated then += will be evaluated.

http://www.difranco.net/compsci/C_Operator_Precedence_Table.htm

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@Dharmendra Lodhi

#include<stdio.h>

#include<conio.h>

void main(){

int i=10,x=10;

int *a;

a=&i;

*a+=x++;

printf("%d%d",i,x);

}

what will be output according to ur comment ???

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*a+=x++;

*a+=10; // now x=11

*a=*a+10=10+10=20

Ans should be: 2011

Answer 2

In static scoping value is directed from main () .while in dynamic scoping value is called from local function

Answer 3

In static scoping 165,165,303

In Dynamic Scoping 31,31,301

Answer 4

I m giving solution only using static scope,in which if there is any free variable then value will be taken by this free variable only from static area where all global variable and static variable are declared,.

Try to understand my solution ,let me know where u getting problem