Ace programming

Question

#define swap(a,b)

a=a+b;b=a-b;a=a-b;

void main()

{

int x=5,y=10;

swap(x,y);

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

}

Will the output be 10 5 or 5 10?

Answer 1

i hope you mistakenly split the preprocessor statements in multiple lines without using \

what i mean, you typed

#define swap(a,b)

a=a+b;b=a-b;a=a-b;
 

 instead of 

#define swap(a,b) \

a=a+b;b=a-b;a=a-b;
 

 

output is x and y are swapped... why swapped?

due to preprocessor just simply replace what you need

then your code look like as 

 

void main()

{

int x=5,y=10;

x=x+y;

y=x-y;

x=x-y;

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

}

 

therefore output is 10,5 

Comments

I didn't get what u told.....ans should be 5 10.

---

I didn't get what u told.....ans should be 5 10.

 

void main() 

{ 

int x=5,y=10; 

x=x+y; ===> x=5+10 ===> x = 15

y=x-y; ===> y=15-10 ===> y = 5

x=x-y; ===> x=15-5 ===> x = 10

printf("%d,%d",x,y);  ===> 10, 5

}