Pointers Questions and Answers

Q: Can you combine the following two statements into one? char *p; p = (char*) malloc(100);

Correct answer: char *p = (char*)malloc(100);A declaration may carry an initialiser, so the declaration and the assignment combine directly: char *p = (char*)malloc(100);.The other forms misplace the asterisks. (char)malloc(100) casts the returned pointer to a single character, and the last option is not valid syntax at all. It is worth adding that in C the cast is unnecessary - void * converts to any object pointer type implicitly, and omitting the cast can expose a missing stdlib.h that the cast would hide.

Q: In which header file is the NULL macro defined?

Correct answer: stdio.h and stddef.hNULL is defined in several headers, among them stdio.h and stddef.h, and also stdlib.h, string.h, time.h and others.The duplication is deliberate. A programmer using any of those libraries is likely to need NULL, and the standard permits a macro to be defined identically in more than one header, so each defines it rather than forcing an extra include. stddef.h is the canonical home, holding it alongside size_t and ptrdiff_t.

Q: How many bytes are occupied by near, far and huge pointers (DOS)?

Correct answer: near=2 far=4 huge=4These are the pointer types of the segmented 16-bit DOS memory model. A near pointer holds a 16-bit offset within the current segment and occupies 2 bytes. A far pointer holds both segment and offset and occupies 4.A huge pointer is also 4 bytes but is normalised on every arithmetic operation, so it can address across segment boundaries correctly where a far pointer wraps within its segment. The distinction disappeared with flat 32-bit addressing, and the keywords are of historical interest only.

Q: If a variable is a pointer to a structure, then which of the following operator is used to access data members of the structure through the pointer variable?

Correct answer: ->The arrow operator accesses a member through a pointer to a structure: p->member is by definition equivalent to (*p).member.The parentheses in that expansion are essential, which is why the shorthand exists - . binds more tightly than *, so *p.member would be parsed as *(p.member) and rejected. The dot operator is for a structure object itself, not for a pointer to one.

Q: What would be the equivalent pointer expression for referring the array element a[i][j][k][l]

Correct answer: *(*(*(*(a+i)+j)+k)+l)Each subscript is defined as a dereference of a sum: x[y] means *(x+y). Applying that rule four times, from the outside in, gives one dereference for every subscript.\[ a[i][j][k][l] \equiv *(*(*(*(a+i)+j)+k)+l) \]The options without the asterisks compute addresses but never read through them. Note that the pointer arithmetic is scaled by the size of the sub-array at each level, which is why the array's dimensions must be known to the compiler.

Q: The operator used to get value at address stored in a pointer variable is

Correct answer: *The unary * is the indirection, or dereference, operator: it yields the object a pointer points to. Its counterpart & is the address-of operator, going the other way.The asterisk plays two distinct roles that are easy to conflate. In a declaration it forms a pointer type; in an expression it dereferences. The && and || operators are logical AND and OR and have nothing to do with pointers.

Q: What will be the output of the program ? #include <stdio.h> int main() { static char *s[] = {"black", "white", "pink", "violet"}; char **ptr[] = {s+3, s+2, s+1, s}, ***p; p = ptr; ++p; printf("%s", **p+1); return 0; }

Correct answer: inkWork outward. ptr holds {s+3, s+2, s+1, s}, and p = ptr then ++p leaves p pointing at ptr[1], which is s+2.So *p is s+2, and **p is s[2], the pointer to "pink". Finally **p+1 advances that char pointer one position, so printf starts from the second character and prints ink. The trap is applying the +1 at the wrong level - it binds to the char pointer, not to p.

Q: What will be the output of the program ? #include <stdio.h> int main() { int i=3, *j, k; j = &i; printf("%d\n", i**j*i+*j); return 0; }

Correct answer: 30The expression i**j*i+*j looks alarming but tokenises unambiguously as i * (*j) * i + (*j): the lexer takes the longest valid token at each point, so ** becomes * followed by unary *.With i = 3 and *j = 3:\[ 3 \times 3 \times 3 + 3 = 27 + 3 = 30 \]Multiplication binds tighter than addition, so no parentheses are needed to get this grouping.

Q: What will be the output of the program ? #include <stdio.h> int main() { int x=30, *y, *z; y=&x; /* Assume address of x is 500 and integer is 4 byte size */ z=y; *y++=*z++; x++; printf("x=%d, y=%d, z=%d\n", x, y, z); return 0; }

Correct answer: x=31, y=504, z=504*y++ = *z++ assigns *z to *y - writing 30 back over itself - and then increments both pointers, since ++ in the postfix position binds to the pointer rather than to the dereferenced value.Pointer arithmetic is scaled by the pointed-to type, so each advances by sizeof(int), that is 4 bytes: from 500 to 504. x++ then makes x 31. Both pointers now point one past x and must not be dereferenced.

31.

Can you combine the following two statements into one?

char *p;
p = (char*) malloc(100);

A.char p = *malloc(100);
B.char *p = (char) malloc(100);
C.char *p = (char*)malloc(100);
D.char *p = (char *)(malloc*)(100);

Answer: C

Correct answer: char *p = (char*)malloc(100);

A declaration may carry an initialiser, so the declaration and the assignment combine directly: char *p = (char*)malloc(100);.

The other forms misplace the asterisks. (char)malloc(100) casts the returned pointer to a single character, and the last option is not valid syntax at all. It is worth adding that in C the cast is unnecessary - void * converts to any object pointer type implicitly, and omitting the cast can expose a missing stdlib.h that the cast would hide.

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32.

In which header file is the NULL macro defined?

A.stdio.h
B.stddef.h
C.stdio.h and stddef.h
D.math.h

Answer: C

Correct answer: stdio.h and stddef.h

NULL is defined in several headers, among them stdio.h and stddef.h, and also stdlib.h, string.h, time.h and others.

The duplication is deliberate. A programmer using any of those libraries is likely to need NULL, and the standard permits a macro to be defined identically in more than one header, so each defines it rather than forcing an extra include. stddef.h is the canonical home, holding it alongside size_t and ptrdiff_t.

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33.

How many bytes are occupied by near, far and huge pointers (DOS)?

A.near=2 far=4 huge=4
B.near=4 far=8 huge=8
C.near=2 far=4 huge=8
D.near=4 far=4 huge=8

Answer: A

Correct answer: near=2 far=4 huge=4

These are the pointer types of the segmented 16-bit DOS memory model. A near pointer holds a 16-bit offset within the current segment and occupies 2 bytes. A far pointer holds both segment and offset and occupies 4.

A huge pointer is also 4 bytes but is normalised on every arithmetic operation, so it can address across segment boundaries correctly where a far pointer wraps within its segment. The distinction disappeared with flat 32-bit addressing, and the keywords are of historical interest only.

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34.

If a variable is a pointer to a structure, then which of the following operator is used to access data members of the structure through the pointer variable?

A..
B.&
C.*
D.->

Answer: D

Correct answer: ->

The arrow operator accesses a member through a pointer to a structure: p->member is by definition equivalent to (*p).member.

The parentheses in that expansion are essential, which is why the shorthand exists - . binds more tightly than *, so *p.member would be parsed as *(p.member) and rejected. The dot operator is for a structure object itself, not for a pointer to one.

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35.

What would be the equivalent pointer expression for referring the array element a[i][j][k][l]

A.((((a+i)+j)+k)+l)
B.*(*(*(*(a+i)+j)+k)+l)
C.(((a+i)+j)+k+l)
D.((a+i)+j+k+l)

Answer: B

Correct answer: *(*(*(*(a+i)+j)+k)+l)

Each subscript is defined as a dereference of a sum: x[y] means *(x+y). Applying that rule four times, from the outside in, gives one dereference for every subscript.

\[ a[i][j][k][l] \equiv *(*(*(*(a+i)+j)+k)+l) \]

The options without the asterisks compute addresses but never read through them. Note that the pointer arithmetic is scaled by the size of the sub-array at each level, which is why the array's dimensions must be known to the compiler.

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36.

A pointer is

A.A keyword used to create variables
B.A variable that stores address of an instruction
C.A variable that stores address of other variable
D.All of the above

Answer: C

Correct answer: A variable that stores address of other variable

A pointer is a variable whose value is the address of another object. It has a type - the type of what it points to - which determines both how the pointed-to bytes are interpreted and how pointer arithmetic is scaled.

It is not a keyword; there is no pointer keyword in C, the asterisk in a declaration doing the work. And while a function pointer holds the address of code, the general definition is in terms of objects, so the option naming instructions describes a special case rather than the concept.

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37.

The operator used to get value at address stored in a pointer variable is

A.*
B.&
C.&&
D.||

Answer: A

Correct answer: *

The unary * is the indirection, or dereference, operator: it yields the object a pointer points to. Its counterpart & is the address-of operator, going the other way.

The asterisk plays two distinct roles that are easy to conflate. In a declaration it forms a pointer type; in an expression it dereferences. The && and || operators are logical AND and OR and have nothing to do with pointers.

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38.

What will be the output of the program ?

#include <stdio.h>

int main()
{
    static char *s[] = {"black", "white", "pink", "violet"};
    char **ptr[] = {s+3, s+2, s+1, s}, ***p;
    p = ptr;
    ++p;
    printf("%s", **p+1);
    return 0;
}

A.ink
B.ack
C.ite
D.let

Answer: A

Correct answer: ink

Work outward. ptr holds {s+3, s+2, s+1, s}, and p = ptr then ++p leaves p pointing at ptr[1], which is s+2.

So *p is s+2, and **p is s[2], the pointer to "pink". Finally **p+1 advances that char pointer one position, so printf starts from the second character and prints ink. The trap is applying the +1 at the wrong level - it binds to the char pointer, not to p.

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39.

What will be the output of the program ?

#include <stdio.h>

int main()
{
    int i=3, *j, k;
    j = &i;
    printf("%d\n", i**j*i+*j);
    return 0;
}

A.30
B.37
C.7
D.9

Answer: A

Correct answer: 30

The expression i**j*i+*j looks alarming but tokenises unambiguously as i * (*j) * i + (*j): the lexer takes the longest valid token at each point, so ** becomes * followed by unary *.

With i = 3 and *j = 3:

\[ 3 \times 3 \times 3 + 3 = 27 + 3 = 30 \]

Multiplication binds tighter than addition, so no parentheses are needed to get this grouping.

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40.

What will be the output of the program ?

#include <stdio.h>

int main()
{
    int x=30, *y, *z;
    y=&x; /* Assume address of x is 500 and integer is 4 byte size */
    z=y;
    *y++=*z++;
    x++;
    printf("x=%d, y=%d, z=%d\n", x, y, z);
    return 0;
}

A.x=31, y=502, z=502
B.x=31, y=500, z=500
C.x=31, y=498, z=498
D.x=31, y=504, z=504

Answer: D

Correct answer: x=31, y=504, z=504

*y++ = *z++ assigns *z to *y - writing 30 back over itself - and then increments both pointers, since ++ in the postfix position binds to the pointer rather than to the dereferenced value.

Pointer arithmetic is scaled by the pointed-to type, so each advances by sizeof(int), that is 4 bytes: from 500 to 504. x++ then makes x 31. Both pointers now point one past x and must not be dereferenced.

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