Outputs Programming

What will be the output of the program? #include<stdio.h> int main() { int a=0, b=1, c=3; *((a) ? &b : &a) = a ? b : c; printf("%d, %d, %d\n", a, b, c); return 0; }

Answer: Option C

Explanation:

Step 1: int a=0, b=1, c=3; here variable a, b, and c are declared as integer type and initialized to 0, 1, 3 respectively.

Step 2: *((a) ? &b : &a) = a ? b : c; The right side of the expression(a?b:c) becomes (0?1:3). Hence it return the value '3'.

The left side of the expression *((a) ? &b : &a) becomes *((0) ? &b : &a). Hence this contains the address of the variable a *(&a).

Step 3: *((a) ? &b : &a) = a ? b : c; Finally this statement becomes *(&a)=3. Hence the variable a has the value '3'.

Step 4: printf("%d, %d, %d\n", a, b, c); It prints "3, 1, 3".

12.	What will be the output of the program? #include<stdio.h> int main() { int a = 300, b, c; if(a >= 400) b = 300; c = 200; printf("%d, %d, %d\n", a, b, c); return 0; }

Answer: Option C

Explanation:

Step 1: int a = 300, b, c; here variable a is initialized to '300', variable b and c are declared, but not initialized.
Step 2: if(a >= 400) means if(300 >= 400). Hence this condition will be failed.
Step 3: c = 200; here variable c is initialized to '200'.
Step 4: printf("%d, %d, %d\n", a, b, c); It prints "300, garbage value, 200". because variable b is not initialized.

16.

What will be the output of the program? #include<stdio.h> int main() { int x = 10, y = 20; if(!(!x) && x) printf("x = %d\n", x); else printf("y = %d\n", y); return 0; }

A. y =20 B. x = 0 C. x = 10 D. x = 1 Answer: Option C Explanation: The logical not operator takes expression and evaluates to true if the expression is false and evaluates to false if the expression is true. In other words it reverses the value of the expression. Step 1: if(!(!x) && x) Step 2: if(!(!10) && 10) Step 3: if(!(0) && 10) Step 3: if(1 && 10) Step 4: if(TRUE) here the if condition is satisfied. Hence it prints x = 10. 20.  What will be the output of the program? #include<stdio.h> int main() { int x, y, z; x=y=z=1; z = ++x || ++y && ++z; printf("x=%d, y=%d, z=%d\n", x, y, z); return 0; } A. x=2, y=1, z=1 B. x=2, y=2, z=1 C. x=2, y=2, z=2 D. x=1, y=2, z=1 Answer: Option A Explanation: Step 1: x=y=z=1; here the variables x ,y, z are initialized to value '1'. Step 2: z = ++x || ++y && ++z; becomes z = ( (++x) || (++y && ++z) ). Here ++x becomes 2. So there is no need to check the other side because ||(Logical OR) condition is satisfied.(z = (2 || ++y && ++z)). There is no need to process ++y && ++z. Hence it returns '1'. So the value of variable z is '1' Step 3: printf("x=%d, y=%d, z=%d\n", x, y, z); It prints "x=2, y=1, z=1". here x is increemented in previous step. y and z are not increemented.