There is an error in this line i>=45 ? return(*p): return(*q);. We cannot use

 returnkeyword in the terenary operators.

Step 1: int fun(int); Here we declare the prototype of the function fun().

Step 2: int i = fun(10); The variable i is declared as an integer type and 

the result of the fun(10) will be stored in the variable i.

Step 3: int fun(int i){ return (i++); } Inside the fun() we are returning a value

 return(i++). It returns 10. because i++ is the post-increement operator.

Step 4: Then the control back to the main function and the value 10 is assigned

 to variable i.

Step 5: printf("%dn", --i); Here --i denoted pre-increement. Hence it prints the value 9.

No answer description available for this question. 

Step 1: int no=5; The variable no is declared as integer type and initialized to 5.

Step 2: reverse(no); becomes reverse(5); It calls the function reverse() with '5' as parameter.

The function reverse accept an integer number 5 and it returns '0'(zero) if(5 == 0) if the given 

number is '0'(zero) or else printf("%d,", no); it prints that number 5 and calls the function reverse(5);.

The function runs infinetely because the there is a post-decrement operator is used. It will not decrease

 the value of 'n' before calling the reverse() function. So, it calls reverse(5) infinitely.

Note: If we use pre-decrement operator like reverse(--n), then the output will be 5, 4, 3, 2, 1. Because 

before calling the function, it decrements the value of 'n'.

Turbo C (WINDOW ): The return value of the function is taken from the Accumulator _AX=1990.

But it may not work as expected in GCC compiler (Linux).

Step 1: int i=5, j=2; Here variable i and j are declared as an integer type and 

initialized to 5 and 2 respectively.

Step 2: fun(&i, &j); Here the function fun() is called with two parameters &i and 

&j (The & denotes call by reference. So the address of the variable i and j are passed. )

Step 3: void fun(int *i, int *j) This function is called by reference, so we have to use

 * before the parameters.

Step 4: *i = *i**i; Here *i denotes the value of the variable i. We are multiplying 5*5 and 

storing the result 25 in same variable i.

Step 5: *j = *j**j; Here *j denotes the value of the variable j. We are multiplying 2*2 and 

storing the result 4 in same variable j.

Step 6: Then the function void fun(int *i, int *j) return back the control back to main()function.

Step 7: printf("%d, %d", i, j); It prints the value of variable i and j.

Hence the output is 25, 4.

Step 1: int i; The variable i is declared as an integer type.

Step 1: int fun(); This prototype tells the compiler that the function fun() does not accept any arguments and it returns an integer value.

Step 1: while(i) The value of i is not initialized so this while condition is failed. So, it does not execute the while block.

Step 1: printf("Hellon"); It prints "Hello".

Hence the output of the program is "Hello".