The while(j loop will get executed 255 times. The size short int(2 byte wide) does not affect the while() loop.

Bitwise operators:
& is a Bitwise AND operator.

Logical operators:
&& is a Logical AND operator.
|| is a Logical OR operator.
! is a NOT operator.

So, ‘&‘ is not a Logical operator.

Simply called as BODMAS (Brackets, Order, Division, Multiplication, Addition and Subtraction).

Mnemonics are often used to help students remember the rules, but the rules taught by the use of acronyms can be misleading. In the United States the acronym PEMDAS is common. It stands for Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. In other English speaking countries, Parentheses may be called Brackets, or symbols of inclusion and Exponentiation may be called either Indices, Powers or Orders, and since multiplication and division are of equal precedence, M and D are often interchanged, leading to such acronyms as BEDMAS, BIDMAS, BODMAS, BERDMAS, PERDMAS, and BPODMAS.

The switch/case statement in the c language is defined by the language specification to use an int value, so you can not use a float value.

switch( expression )
{
    case constant-expression1:    statements 1;
    case constant-expression2:    statements 2;    
    case constant-expression3:    statements3 ;
    ...
    ...
    default : statements 4;
}

The value of the ‘expression‘ in a switch-case statement must be an integer, char, short, long. Float and double are not allowed.

Step 1: int i = 0; here variable i is an integer type and initialized to ‘0’.
Step 2: for(; 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(; 0i is incremented by ‘1’(one)
Loop 2: here i=1, the condition in for(; 1i is incremented by ‘1’(one)
Loop 3: here i=2, the condition in for(; 2i is incremented by ‘1’(one)
Loop 4: here i=3, the condition in for(; 3i is increemented by ‘1’(one)
Loop 5: here i=4, the condition in for(; 4i is incremented by ‘1’(one)
Loop 6: here i=5, the condition in for(; 5i is incremented by ‘1’(one)
Loop 7: here i=6, the condition in for(; 6i is not incremented.

Step 3: printf(“%d”, i); here the value of i is 6. Hence the output is ‘6’.

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”.

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”

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’.

Step 1: int x = 3; here variable x is an integer type and initialized to ‘3’.
Step 2: float y = 3.0; here variable y is an float type and initialized to ‘3.0’
Step 3: if(x == y) here we are comparing if(3 == 3.0) hence this condition is satisfied.
Hence it prints “x and y are equal”.

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

In for( i = -1; ++i; i>0) expression 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”.

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.

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
int main()
{
    float a=0.7;
    printf("%.10f %.10f\n",0.7, a);
    return 0;
}

Output:
0.7000000000 0.6999999881

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”.

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: 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

The keyword continue cannot be used in switch case. It must be used in for or while or do while loop. If there is any looping statement in switch case then we can use continue.

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.

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.

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: 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.

“Expression syntax” error occur in this line if(i = 5) && if(j = 10).

It should be like if((i == 5) && (j == 10)).

if(i % 2 = j % 3) This statement generates “LValue required error”. There is no variable on the left side of the expression to assign (j % 3).

This program will result in error “Statement missing ;”

printf(“x is less than y\n”) here ; should be added to the end of this statement.