What will be the output of the program?

#include<stdio.h>
int main()
{
    int x = 3;
    float y = 3.0;
    if(x == y)
        printf("x and y are equal");
    else
        printf("x and y are not equal");
    return 0;
}

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".

switch(i) has the variable i it has the value '1'(one).

Then case 1: statements got executed. so, it prints "Hi". The break; statement make the program to be exited from switch-case statement.

switch-case do not execute any statements outside these blocks case and default

Hence the output is "Hi".

Step 1: Initially the value of variable i is '5'.
Loop 1: while(i-- >= 0) here i = 5, this statement becomes while(5-- >= 0) Hence the while condition is satisfied and it prints '4'. (variable 'i' is decremented by '1'(one) in previous while condition)
Loop 2: while(i-- >= 0) here i = 4, this statement becomes while(4-- >= 0) Hence the while condition is satisfied and it prints '3'. (variable 'i' is decremented by '1'(one) in previous while condition)
Loop 3: while(i-- >= 0) here i = 3, this statement becomes while(3-- >= 0) Hence the while condition is satisfied and it prints '2'. (variable 'i' is decremented by '1'(one) in previous while condition)
Loop 4: while(i-- >= 0) here i = 2, this statement becomes while(2-- >= 0) Hence the while condition is satisfied and it prints '1'. (variable 'i' is decremented by '1'(one) in previous while condition)
Loop 5: while(i-- >= 0) here i = 1, this statement becomes while(1-- >= 0) Hence the while condition is satisfied and it prints '0'. (variable 'i' is decremented by '1'(one) in previous while condition)
Loop 6: while(i-- >= 0) here i = 0, this statement becomes while(0-- >= 0) Hence the while condition is satisfied and it prints '-1'. (variable 'i' is decremented by '1'(one) in previous while condition)
Loop 7: while(i-- >= 0) here i = -1, this statement becomes while(-1-- >= 0) Hence the while condition is not satisfied and loop exits.
The output of first while loop is 4,3,2,1,0,-1

Step 2: Then the value of variable i is initialized to '5' Then it prints a new line character(\n).
See the above Loop 1 to Loop 7 .
The output of second while loop is 4,3,2,1,0,-1

Step 3: The third while loop, while(i-- >= 0) here i = -1(because the variable 'i' is decremented to '-1' by previous while loop and it never initialized.). This statement becomes while(-1-- >= 0) Hence the while condition is not satisfied and loop exits.

Hence the output of the program is
4,3,2,1,0,-1
4,3,2,1,0,-1

Step 1: int i = 0; here variable i is an integer type and initialized to '0'.
Step 2: for(; i<=5; i++); variable i=0 is already assigned in previous step. The semi-colon at the end of this for loop tells, "there is no more statement is inside the loop".

Loop 1: here i=0, the condition in for(; 0<=5; i++) loop satisfies and then i is incremented by '1'(one)
Loop 2: here i=1, the condition in for(; 1<=5; i++) loop satisfies and then i is incremented by '1'(one)
Loop 3: here i=2, the condition in for(; 2<=5; i++) loop satisfies and then i is incremented by '1'(one)
Loop 4: here i=3, the condition in for(; 3<=5; i++) loop satisfies and then i is increemented by '1'(one)
Loop 5: here i=4, the condition in for(; 4<=5; i++) loop satisfies and then i is incremented by '1'(one)
Loop 6: here i=5, the condition in for(; 5<=5; i++) loop satisfies and then i is incremented by '1'(one)
Loop 7: here i=6, the condition in for(; 6<=5; i++) loop fails and then i is not incremented.

Step 3: printf("%d", i); here the value of i is 6. Hence the output is '6'.

In the very begining of switch-case statement default statement is encountered. So, it prints "This is default".

In default statement there is no break; statement is included. So it prints the case 1 statements. "This is case 1".

Then the break; statement is encountered. Hence the program exits from the switch-case block.

Initially variables a = 500, b = 100 and c is not assigned.

Step 1: if(!a >= 400)
Step 2: if(!500 >= 400)
Step 3: if(0 >= 400)
Step 4: if(FALSE) Hence the if condition is failed.
Step 5: So, variable c is assigned to a value '200'.
Step 6: printf("b = %d c = %d\n", b, c); It prints value of b and c.
Hence the output is "b = 100 c = 200"

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.

Step 1: char str[]="C-program"; here variable str contains "C-program".
Step 2: int a = 5; here variable a contains "5".
Step 3: printf(a >10?"Ps\n":"%s\n", str); this statement can be written as

if(a > 10)
{
    printf("Ps\n");
}
else
{
    printf("%s\n", str);
}

Here we are checking a > 10 means 5 > 10. Hence this condition will be failed. So it prints variable str.

Hence the output is "C-program".