# Quark C Questions: C Questions Part II

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Explanation:

The structure yy is nested within structure xx. Hence, the elements are of yy are to be accessed through the instance of structure xx, which needs an instance of yy to be known. If the instance is created after defining the structure the compiler will not know about the instance relative to xx. Hence for nested structure yy you have to declare member.

main () { printf ( “\nab” ); printf ( “\bsi” ); printf ( “\rha” ); }

Answer: Hai

Explanation:

\n-newline

\b-backspace

\r-linefeed

main () { int i = 5; printf ( “%d%d%d%d%d%d” i + +, i--, + + i, -i, i); }

Answer: 45545

Explanation:

The arguments in a function call are pushed into the stack from left to right. The evaluation is by popping out from the stack. And the evaluation is from right to left, hence the result.

#define square (x) x * x

main () { int i; i = 64/square (4); printf ( “%d” i); }

Answer: 64

Explanation: The macro call square (4) will substituted by 4 * 4 so the expression becomes i = 64/4 * 4. Since/and * has equal priority the expression will be evaluated as (64/4) * 4 i.e., 16 * 4 = 64

main () { char * p = “hai friends” * p1; p1 = p; while ( * p! = ‘\0’ ) + + * p + +; printf ( “%s %s” p, p1); }

Answer: Ibj! gsjfoet

Explanation: + + * p + + will be parse in the given order

* p that is value at the location currently pointed by p will be taken

+ + * p the retrieved value will be incremented

when; is encountered the location will be incremented that is p + + will be executed

Hence, in the while loop initial value pointed by p is h ‘which is changed to i’ by executing + + * p and pointer moves to point, a ‘which is similarly changed to b’ and so on. Similarly blank space is converted to! ‘Thus, we obtain value in p becomes ibj! gsjfoet and since p reaches \0’ and p1 points to p thus p1doesnot print anything.

#include

#define a 10

main () { #define a 50 printf ( “%d” a); }

Answer: 50

Explanation:

The preprocessor directives can be redefined anywhere in the program. So the most recently assigned value will be taken.

#define clrscr () 100

main () { clrscr (); printf ( “%d\n” clrscr () ); }

Answer: 100

Explanation:

Preprocessor executes as a seperate pass before the execution of the compiler. So textual replacement of clrscr () to 100 occurs. The input program to compiler looks like this:

main () { 100; printf ( “%d\n” 100); }

Note: 100; is an executable statement but with no action. So it doesn't give any problem

main () { printf ( “%p” main); }

Answer:

Some address will be printed.

Explanation:

Function names are just addresses (just like array names are addresses).

main () is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They are printed as hexadecimal numbers.

1. main () { clrscr (); } clrscr (); Answer: No output/error Explanation: The first clrscr () occurs inside a function. So it becomes a function call. In the second clrscr (); is a function declaration (because it is not inside any function).

2. enum colors { BLACK, BLUE, GREEN} main () { printf ( “%d. %d. %d” BLACK, BLUE, GREEN); return (1); } Answer: 0. 1. 2 Explanation: Enum assigns numbers starting from 0, if not explicitly defined.

3. void main () { char far * farther, * farthest; printf ( “%d. %d” sizeof (farther), sizeof (farthest) ); } Answer: 4. 2 Explanation: The second pointer is of char type and not a far pointer

4. main () { int i = 400, j = 300; printf ( “%d. %d” ); } Answer: 400. 300 Explanation: Printf takes the values of the first two assignments of the program. Any number of printf's may be given. All of them take only the first two values. If more number of assignments given in the program, then printf will take garbage values.

5. main () { char * p; p = “Hello” printf ( “%c\n” * & * p); } Answer: H Explanation: * is a dereference operator & is a reference operator. They can be applied any number of times provided it is meaningful. Here p points to the first character in the string “Hello” * p dereferences it and so its value is H. Again & references it to an address and * dereferences it to the value H.

6. main () { int i = 1; while (i ⇐ 5) { printf ( “%d” i); if (i>2) goto here; i + +; } } fun () { here: Printf ( “PP” ); } Answer: Compiler error: Undefined label ‘here’ in function main Explanation: Labels have functions scope, in other words The scope of the labels is limited to functions. The label ‘here’ is available in function fun () Hence it is not visible in function main.

7. main () { static char names[5][20] = { “pascal” “ada” “cobol” “fortran” “perl” }; int i; char * t; t = names[3]; names[3] = names[4]; names[4] = t; for (i = 0; i ⇐ 4; i + + ) printf ( “%s” names[i]); } Answer: Compiler error: Lvalue required in function main Explanation: Array names are pointer constants. So it cannot be modified.

8. void main () { int i = 5; printf ( “%d” i + + + + + i); } Answer: Output Cannot be predicted exactly. Explanation: Side effects are involved in the evaluation of i

9. void main () { int i = 5; printf ( “%d” i + + + + + i); } Answer: Compiler Error Explanation: The expression i + + + + + i is parsed as i + + + + + i which is an illegal combination of operators.

10. #include main () { int i = 1, j = 2; switch (i) { case 1: Printf ( “GOOD” ); break; case j: Printf ( “BAD” ); break; } }

Answer: Compiler Error: Constant expression required in function main.

Explanation: The case statement can have only constant expressions (this implies that we cannot use variable names directly so an error).

Note: Enumerated types can be used in case statements.