What will be the output of the program?

#include<stdio.h>
int main()
{
    unsigned int i = 65535; /* Assume 2 byte integer*/
    while(i++ != 0)
        printf("%d",++i);
    printf("\n");
    return 0;
}

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.

Here unsigned int size is 2 bytes. It varies from 0,1,2,3, ... to 65535.

Step 1:unsigned int i = 65536; here variable i becomes '0'(zero). because unsigned int varies from 0 to 65535.

Step 2: while(i != 0) this statement becomes while(0 != 0). Hence the while(FALSE) condition is not satisfied. So, the inside the statements of while loop will not get executed.

Hence there is no output.

Note: Don't forget that the size of int should be 2 bytes. If you run the above program in GCC it may run infinite loop, because in Linux platform the size of the integer is 4 bytes.

for(i<=5 && i>=-1; ++i; i>0) so expression i<=5 && i>=-1 initializes for loop. expression ++i is the loop condition. expression i>0 is the increment expression.

In for( i <= 5 && i >= -1; ++i; i>0) expression i<=5 && i>=-1 evaluates to one.

Loop condition always get evaluated to true. Also at this point it increases i by one.

An increment_expression i>0 has no effect on value of i.so for loop get executed till the limit of integer (ie. 65535)

printf() returns the number of charecters printed on the console.

Step 1: if(ch = printf("")) here printf() does not print anything, so it returns '0'(zero).
Step 2: if(ch = 0) here variable ch has the value '0'(zero).
Step 3: if(0) Hence the if condition is not satisfied. So it prints the else statements.
Hence the output is "It doesn't matters".

Note: Compiler shows a warning "possibly incorrect assinment".

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.

if(0.7 > a) here a is a float variable and 0.7 is a double constant. The double constant 0.7 is greater than the float variable a. Hence the if condition is satisfied and it prints 'Hi'
Example:

#include<stdio.h>
int main()
{
    float a=0.7;
    printf("%.10f %.10f\n",0.7, a);
    return 0;
}

Output:
0.7000000000 0.6999999881

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'.

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

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".