What will be the output of the program?

#include<stdio.h>
typedef unsigned long int uli;
typedef uli u;

int main()
{
    uli a;
    u b = -1;
    a = -1;
    printf("%lu, %lu", a, b);
    return 0;
}

Step 1: int check(int); This prototype tells the compiler that the function check() accepts one integer parameter and returns an integer value.

Step 2: int l=45, c; The variable i and c are declared as an integer type and i is initialized to 45.

The function check(i) return 100 if the given value of variable i is >=(greater than or equal to) 45, else it will return 10.

Step 3: c = check(i); becomes c = check(45); The function check() return 100 and it get stored in the variable c.(c = 100)

Step 4: printf("%d\n", c); It prints the value of variable c.

Hence the output of the program is '100'.

Step 1: for(;;) this statement will genereate infinite loop.
Step 2: printf("%d\n", i++); this statement will print the value of variable i and increement i by 1(one).
Step 3: if(i>10) here, if the variable i value is greater than 10, then the for loop breaks.

Hence the output of the program is
1
2
3
4
5
6
7
8
9
10

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

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

Output:
0.7000000000 0.6999999881

The integer value 2 is represented as 00000000 00000010 in binary system.

Negative numbers are represented in 2's complement method.

1's complement of 00000000 00000010 is 11111111 11111101 (Change all 0s to 1 and 1s to 0).

2's complement of 00000000 00000010 is 11111111 11111110 (Add 1 to 1's complement to obtain the 2's complement value).

Therefore, in binary we represent -2 as: 11111111 11111110.

After left shifting it by 2 bits we obtain: 11111111 11111000, and it is equal to "fff8" in hexadecimal system.

In the program, printf() returns the number of charecters printed on the console i = printf("How r u\n"); This line prints "How r u" with a new line character and returns the length of string printed then assign it to variable i.
So i = 8 (length of '\n' is 1).

i = printf("%d\n", i); In the previous step the value of i is 8. So it prints "8" with a new line character and returns the length of string printed then assign it to variable i. So i = 2 (length of '\n' is 1).

printf("%d\n", i); In the previous step the value of i is 2. So it prints "2".

Yes, this program will compile and run successfully and prints 20.

The macro #ifdef NOTE evaluates the given expression to 1. If satisfied it executes the #ifdef block statements. Here #ifdef condition fails because the Macro NOTE is nowhere declared.

Hence the #else block gets executed, the variable a is declared and assigned a value of 20.

printf("%d\n", a); It prints the value of variable a 20.

True, We can prevent the same file from getting included again by using a preprocessor directive called #ifndef (short for "if not defined") to determine whether we've already defined a preprocessor symbol called XSTRING_H. If we have already defined this symbol, the compiler will ignore the rest of the file until it sees a #endif (which in this case is at the end of the file).

#ifndef XSTRING_H

#define XSTRING_H defines the same preprocessor symbol,

Finally, the last line of the file, #endif

Yes, the program prints "CuriousTab" and runs infinitely.

The macro INFINITELOOP while(1) replaces the text 'INFINITELOOP' by 'while(1)'

In the main function, while(1) satisfies the while condition and it prints "CuriousTab". Then it comes to while(1) and the loop runs infinitely.

The macro MIN(x, y) (x<y)? x : y; returns the smallest value from the given two numbers.

Step 1: int x=3, y=4, z; The variable x, y, z are declared as an integer type and the variable x, y are initialized to value 3, 4 respectively.

Step 2: z = MIN(x+y/2, y-1); becomes,

=> z = (x+y/2 < y-1)? x+y/2 : y - 1;

=> z = (3+4/2 < 4-1)? 3+4/2 : 4 - 1;

=> z = (3+2 < 4-1)? 3+2 : 4 - 1;

=> z = (5 < 3)? 5 : 3;

The macro return the number 3 and it is stored in the variable z.

Step 3: if(z > 0) becomes if(3 > 0) here the if condition is satisfied. It executes the if block statements.

Step 4: printf("%d\n", z);. It prints the value of variable z.

Hence the output of the program is 3