Question

Three is 3 questions please answer them all

Question 1 int main(void) { int x = 10, y = 20; if (x == y); printf ("\n%d %d",x,y); } Output as you would see on the compiler:

$Question 2 int main(void) { int x = 3, y = 5; if (x == 3) printf (\n%d,x); else ; printf(%d, y); return 0; Output as you$

$Question 3 int main(void) { int x = 3; float y = 3.0; if (x == y) printf (\nx and y are equal); else printf(\nx and y are$

Answer 1

Answers:

Question 1 : 1020

Question 2 : 35

Question 3 : x and y are equal

Explanation:

Question 1:

{ 1 #include <stdio.h> 2 int main(void) 3 4 int x=10,y=20; 5 if(x==y); 6 printf("\n%d%d",x,y); 7 8 } L.l. Result $gcc -o main *.c $main 1020

In this program, compiler will consider the if - statement as bodyless. It will expand the program as bellow,

int main(void)

{

int x=10,y=20;

If(x==y);

{

}

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

}

Question 2:

1 #include <stdio.h> 2 int main(void) 3- { 4 int x=3, y=5; 5 if(x==3) 6 printf("\n%d",x); 7 else; 8 printf("%d",y); 9 return 0; 10 11. } L.lı Result 13 $gcc -o main *.c $main 35

here, else; consider as empty statement. Because it ends with semicolon(;). That is why the next printf("\n%d",y); is executed.

Question 3:

1 #include <stdio.h> 2 3 int main(void) 4- { 5 int X=3; 6 float y=3.0; 7 if(x==y) 8 printf("\nx and y are equal" D; 9 else 10 printf("\nx and y are not equal");/ 11. } L.lı Result 130 $gcc -o main *.c $main x and y are equal

Here, we compare if(3==3.0) [if(x==y)] . This condition is satisfied. Because compiler will convert the int (x=3) into float (x=3.0) then compared (3.0==3.0).since int and float cannot be compared. Hence it prints " x and y are equal".

Thank you!