Online C Programming Test - C Programming Test - Random

Step 1: int fun(int); This is prototype of function fun(). It tells the compiler that the function fun() accept one integer parameter and returns an integer value.

Step 2: int i=3; The variable i is declared as an integer type and initialized to value 3.

Step 3: fun(i=fun(fun(i)));. The function fun(i) increements the value of i by 1(one) and return it.

Lets go step by step,

=> fun(i) becomes fun(3) is called and it returns 4.

=> i = fun(fun(i)) becomes i = fun(4) is called and it returns 5 and stored in variable i.(i=5)

=> fun(i=fun(fun(i))); becomes fun(5); is called and it return 6 and nowhere the return value is stored.

Step 4: printf("%d\n", i); It prints the value of variable i.(5)

Hence the output is '5'.

Yes, It is possible to allocate a block of memory (of arbitrary size) at run-time, using the standard library's malloc function, and treat it as an array.

The system will allocate 2 bytes for the union.

The statements u.ch[0]=3; u.ch[1]=2; store data in memory as given below.

Step 1: static int a[20]; here variable a is declared as an integer type and static. If a variable is declared as static and it will be automatically initialized to value '0'(zero).

Step 2: int i = 0; here vaiable i is declared as an integer type and initialized to '0'(zero).
Step 3: a[i] = i ; becomes a[0] = 0;
Step 4: printf("%d, %d, %d\n", a[0], a[1], i);
Here a[0] = 0, a[1] = 0(because all staic variables are initialized to '0') and i = 0.
Step 4: Hence the output is "0, 0, 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.

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.

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'.

The compiler will report the error "Constant expression required" in the line case P: . Because, variable names cannot be used with case statements.

The case statements will accept only constant expression.

scanf is a function that reads data with specified format from a given string stream source.
scanf("%d",&number);

atoi() convert string to integer.
var number;
number = atoi("string");

Step 1: int y=128; The variable 'y' is declared as an integer type and initialized to value "128".

Step 2: const int x=y; The constant variable 'x' is declared as an integer and it is initialized with the variable 'y' value.

Step 3: printf("%d\n", x); It prints the value of variable 'x'.

Hence the output of the program is "128"