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;
}

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.

Function atoi() converts the string to integer.
Function atof() converts the string to float.

result1 = result1+atoi(i);
Here result1 = 10 + atoi(55.555);
result1 = 10 + 55;
result1 = 65;

result2 = result2+atof(i);
Here result2 = 11.111 + atof(55.555);
result2 = 11.111 + 55.555000;
result2 = 66.666000;
So the output is "65, 66.666000" .

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

Step 1:unsigned int i = 65535;

Step 2:
Loop 1: while(i++ != 0) this statement becomes while(65535 != 0). Hence the while(TRUE) condition is satisfied. Then the printf("%d", ++i); prints '1'(variable 'i' is already incremented by '1' in while statement and now incremented by '1' in printf statement) Loop 2: while(i++ != 0) this statement becomes while(1 != 0). Hence the while(TRUE) condition is satisfied. Then the printf("%d", ++i); prints '3'(variable 'i' is already incremented by '1' in while statement and now incremented by '1' in printf statement)
....
....

The while loop will never stops executing, because variable i will never become '0'(zero). Hence it is an 'Infinite loop'.

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".