Test series question of ace academy

Question

#include<stdio.h>

#define ADD(a,b)(a+b)

#define SQUARE(x)(x*x)

int main()

{

int x=2;

int y=3;

int z = ADD(SQUARE(x++),y);

printf("%d\n",z);

return 0;

}

What is the output of the above code snippet?

Comments

is ans 7 ?

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answer is 9

 

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#define is a preprocessor statement.

Square(x++) is replaced with (x++ * x++) - updating one variable morethan once before one sequence point -- undefined behavior in C.

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ah got it .  square (x) will be treated as x++ * x++  so it will  give 6  and after add it with it y it will give 9. bt as it is not acceptable in C so it will be undefined behavior.

Answer 1

9

 This question is easy the only catch is in int z = ADD(SQUARE(x++),y); in this line .

Firstly it will take the value of 2 as it is and then increment so the value will be 2*3.

z=(2*3)+3 = 9.

Comments

@Abha Gupta   @pinaksh10      see shaikh mastan sir's comment ur flow of execution is correct , bt it will undefined behaviour.

Answer 2

#include<stdio.h>

#define ADD(a,b)(a+b)

#define SQUARE(x)(x*x)

int main()

{

int x=2;

int y=3;

int z = ADD(SQUARE(x++),y);        // here is the cache which is 1st time Square takes 2 and 2nd time give 3 means SQUARE(old x*updated x)= SQUARE(2*3) gives 6  then add(6+3)=9

printf("%d\n",z);

return 0;

}

So the answer is 9