What will be the output of the program?

#include<stdio.h>
int main()
{
    int i=4, j=-1, k=0, w, x, y, z;
    w = i || j || k;
    x = i && j && k;
    y = i || j &&k;
    z = i && j || k;
    printf("%d, %d, %d, %d\n", w, x, y, z);
    return 0;
}

True, we can find the size of short integer and long integer using the sizeof() operator.
Example:

#include<stdio.h>
int main()
{
    short int i = 10;
    long int j = 10;
    printf("short int is %d bytes.,\nlong int is %d bytes.",
            sizeof(i),sizeof(j));
    return 0;
}

Output:
short int is 2 bytes.
long int is 4 bytes.

True,
float = 4 bytes.
double = 8 bytes.

False, The range of double is -1.7e+308 to 1.7e+308.

True, we can use long double; if double range is not enough.

double = 8 bytes.
long double = 10 bytes.

False, The range of float is -3.4e+38 to 3.4e+38.

Step 1: char ch; ch = 'A'; here variable ch is declared as an character type an initialized to 'A'.

Step 2: printf("The letter is"); It prints "The letter is".

Step 3: printf("%c", ch >= 'A' && ch <= 'Z' ? ch + 'a' - 'A':ch);

The ASCII value of 'A' is 65 and 'a' is 97.

Here

=> ('A' >= 'A' && 'A' <= 'Z') ? (A + 'a' - 'A'):('A')

=> (TRUE && TRUE) ? (65 + 97 - 65) : ('A')

=> (TRUE) ? (97): ('A')

In printf the format specifier is '%c'. Hence prints 97 as 'a'.

Step 4: printf("Now the letter is"); It prints "Now the letter is".

Step 5: printf("%c\n", ch >= 'A' && ch <= 'Z' ? ch : ch + 'a' - 'A');

Here => ('A' >= 'A' && 'A' <= 'Z') ? ('A') : (A + 'a' - 'A')

=> (TRUE && TRUE) ? ('A') :(65 + 97 - 65)

=> (TRUE) ? ('A') : (97)

It prints 'A'

Hence the output is

The letter is a
Now the letter is A

Step 1: int x=12, y=7, z; here variable x, y and z are declared as an integer and variable x and y are initialized to 12, 7 respectively.

Step 2: z = x!=4 || y == 2;
becomes z = 12!=4 || 7 == 2;
then z = (condition true) || (condition false); Hence it returns 1. So the value of z=1.

Step 3: printf("z=%d\n", z); Hence the output of the program is "z=1".

Step 1: int i=-3, j=2, k=0, m; here variable i, j, k, m are declared as an integer type and variable i, j, k are initialized to -3, 2, 0 respectively.

Step 2: m = ++i && ++j || ++k;
becomes m = (-2 && 3) || ++k;
becomes m = TRUE || ++k;.
(++k) is not executed because (-2 && 3) alone return TRUE.
Hence this statement becomes TRUE. So it returns '1'(one). Hence m=1.

Step 3: printf("%d, %d, %d, %d\n", i, j, k, m); In the previous step the value of i,j are increemented by '1'(one).

Hence the output is "-2, 3, 0, 1".

Because, comma operator used in the expression i (1, 2, 3, 4, 5). The comma operator has left-right associativity. The left operand is always evaluated first, and the result of evaluation is discarded before the right operand is evaluated. In this expression 5 is the right most operand, hence after evaluating expression (1, 2, 3, 4, 5) the result is 5, which on adding to i results into 7.

Step 1: int k, num=30; here variable k and num are declared as an integer type and variable num is initialized to '30'.
Step 2: k = (num>5 ? (num <=10 ? 100 : 200): 500); This statement does not affect the output of the program. Because we are going to print the variable num in the next statement. So, we skip this statement.
Step 3: printf("%d\n", num); It prints the value of variable num '30'
Step 3: Hence the output of the program is '30'