Pointers Questions and Answers

Q: What is the output of this C code? int main() { char *p = NULL; char *q = 0; if (p) printf(" p "); else printf("nullp"); if (q) printf("q\n"); else printf(" nullq\n"); }

Correct answer: nullp nullqBoth pointers are null - NULL and a literal zero in a pointer context mean the same thing - and a null pointer used in a condition tests false. So both if statements take their else branch, and the output is nullp nullq. The result does not depend on how NULL is defined, since the comparison is against the null pointer value rather than against the bit pattern.

Q: What is the output of this C code? void main() { int x = 0; int *ptr = &x; printf("%d\n", *ptr); }

Correct answer: 0ptr holds the address of x, so *ptr retrieves what is stored there, which is 0. There is nothing indeterminate about it - x was explicitly initialised, and the pointer was set from its address, so the dereference is entirely defined. Printing the address rather than the value would require %p and the pointer itself.

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: Which of the following is the correct syntax to send an array as a parameter to function?

Correct answer: func(&array);Of the four forms offered this is the only valid C that passes the array's address. &array yields a pointer to the whole array, which the function can receive as int (*)[N].It is worth noting that the idiomatic form, func(array), is not among the options: an array name in an expression decays to a pointer to its first element, so writing the bare name already passes the address. The other options are not valid syntax at all, or pass a single element rather than the array.

Q: Is this a correct way for NULL pointer assignment? int i=0; char *q=(char*)i;

Correct answer: NoThis does not produce a null pointer reliably. The rule requires a null pointer constant - a constant expression evaluating to zero - and i is a variable, so what is converted is a run-time integer value. The compiler is obliged to reinterpret it as an address rather than substitute the machine’s null representation, which need not be all-zero bits.

Q: What is the output of this C code? void foo(int*); int main() { int i = 10, *p = &i; foo(p++); } void foo(int *p) { printf("%d\n", *p); }

Correct answer: 10The postfix ++ yields the value the pointer had before the increment, so foo receives the original address of i and prints 10. The increment happens to p in main, which afterwards points one int past i - harmless here since it is never dereferenced. Prefix ++p would have passed the advanced pointer and the behaviour would then have been undefined.

Q: What will be the output of the program ? #include <stdio.h> int main() { printf("%c\n", 7["IndiaBIX"]); return 0; }

Correct answer: print "X" of IndiaBIXSubscripting is defined as x[y] meaning *(x+y), and addition is commutative, so 7["IndiaBIX"] means *(7 + "IndiaBIX") - exactly the same as "IndiaBIX"[7].Counting from zero, position 7 of IndiaBIX is X. The reversed form is legal but deliberately perverse, and it is worth knowing chiefly as evidence that array subscripting in C really is just pointer arithmetic with syntactic sugar.

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 will be the output of the program ? #include <stdio.h> void fun(void *p); int i; int main() { void *vptr; vptr = &i; fun(vptr); return 0; } void fun(void *p) { int **q; q = (int**)&p; printf("%d\n", **q); }

Correct answer: 0i is a global, so it has static storage duration and is zero-initialised before the program starts.Inside fun, &p is the address of the parameter, which itself holds &i. Casting that to int ** and applying ** therefore reads first the parameter, obtaining &i, and then i itself - giving 0. The double cast is a roundabout way of reaching a value that *(int*)p would have given directly.

71.

What is the output of this C code?

int main()
{
char *p = NULL;
char *q = 0;
if (p)
printf(" p ");
else
printf("nullp");
if (q)
printf("q\n");
else
printf(" nullq\n");
}

A.nullp nullq
B.Depends on the compiler
C.x nullq where x can be p or nullp depending on the value of NULL
D.p q

Answer: A

Correct answer: nullp nullq

Both pointers are null - NULL and a literal zero in a pointer context mean the same thing - and a null pointer used in a condition tests false. So both if statements take their else branch, and the output is nullp nullq. The result does not depend on how NULL is defined, since the comparison is against the null pointer value rather than against the bit pattern.

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72.

Which of the following changes can correct the program so that it prints “Geeks Quiz”?

#include <stdio.h> 
  
void myStrcat(char *a, char *b) 
{ 
    int m = strlen(a); 
    int n = strlen(b); 
    int i; 
    for (i = 0; i <= n; i++) 
       a[m+i]  = b[i]; 
} 
  
int main() 
{ 
    char *str1 = "Geeks "; 
    char *str2 = "Quiz"; 
    myStrcat(str1, str2); 
    printf("%s ", str1); 
    return 0; 
}

A.char *str1 = “Geeks “; can be changed to char str1[100] = “Geeks “;
B.char *str1 = “Geeks “; can be changed to char str1[100] = “Geeks “; and a line a[m+n-1] = ‘\0’ is added at the end of myStrcat
C.A line a[m+n-1] = ‘\0’ is added at the end of myStrcat
D.A line ‘a = (char *)malloc(sizeof(char)*(strlen(a) + strlen(b) + 1)) is added at the beginning of myStrcat()

Answer: A

Correct answer: char *str1 = “Geeks “; can be changed to char str1[100] = “Geeks “;

The fault is that str1 points at a string literal, which must not be modified, and there is no room after it in any case. Making it an array of 100 characters gives writable storage with space to spare. No terminator line is needed, because the copy loop runs to i <= n and so copies b’s own null byte along with its characters.

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73.

What is the output of this C code?

void main()
{
int x = 0;
int *ptr = &x;
printf("%d\n", *ptr);
}

A.Address of x
B.Junk value
C.Run time error
D.0

Answer: D

Correct answer: 0

ptr holds the address of x, so *ptr retrieves what is stored there, which is 0. There is nothing indeterminate about it - x was explicitly initialised, and the pointer was set from its address, so the dereference is entirely defined. Printing the address rather than the value would require %p and the pointer itself.

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74.

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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75.

Which of the following is the correct syntax to send an array as a parameter to function?

A.func(&array);
B.func(#array);
C.func(*array);
D.func(array[size]);

Answer: A

Correct answer: func(&array);

Of the four forms offered this is the only valid C that passes the array's address. &array yields a pointer to the whole array, which the function can receive as int (*)[N].

It is worth noting that the idiomatic form, func(array), is not among the options: an array name in an expression decays to a pointer to its first element, so writing the bare name already passes the address. The other options are not valid syntax at all, or pass a single element rather than the array.

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76.

Is this a correct way for NULL pointer assignment?

int i=0;
char *q=(char*)i;

A.Yes
B.No

Answer: B

Correct answer: No

This does not produce a null pointer reliably. The rule requires a null pointer constant - a constant expression evaluating to zero - and i is a variable, so what is converted is a run-time integer value. The compiler is obliged to reinterpret it as an address rather than substitute the machine’s null representation, which need not be all-zero bits.

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77.

What is the output of this C code?

void foo(int*);
int main()
{
int i = 10, *p = &i;
foo(p++);
}
void foo(int *p)
{
printf("%d\n", *p);
}

A.10
B.Some garbage value
C.Compile time error
D.Segmentation fault

Answer: A

Correct answer: 10

The postfix ++ yields the value the pointer had before the increment, so foo receives the original address of i and prints 10. The increment happens to p in main, which afterwards points one int past i - harmless here since it is never dereferenced. Prefix ++p would have passed the advanced pointer and the behaviour would then have been undefined.

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78.

What will be the output of the program ?

#include <stdio.h>

int main()
{
    printf("%c\n", 7["IndiaBIX"]);
    return 0;
}

A.Error: in printf
B.Nothing will print
C.print "X" of IndiaBIX
D.print "7"

Answer: C

Correct answer: print "X" of IndiaBIX

Subscripting is defined as x[y] meaning *(x+y), and addition is commutative, so 7["IndiaBIX"] means *(7 + "IndiaBIX") - exactly the same as "IndiaBIX"[7].

Counting from zero, position 7 of IndiaBIX is X. The reversed form is legal but deliberately perverse, and it is worth knowing chiefly as evidence that array subscripting in C really is just pointer arithmetic with syntactic sugar.

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79.

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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80.

What will be the output of the program ?

#include <stdio.h>

void fun(void *p);
int i;

int main()
{
    void *vptr;
    vptr = &i;
    fun(vptr);
    return 0;
}
void fun(void *p)
{
    int **q;
    q = (int**)&p;
    printf("%d\n", **q);
}

A.Error: cannot convert from void** to int**
B.Garbage value
C.No output
D.0

Answer: D

Correct answer: 0

i is a global, so it has static storage duration and is zero-initialised before the program starts.

Inside fun, &p is the address of the parameter, which itself holds &i. Casting that to int ** and applying ** therefore reads first the parameter, obtaining &i, and then i itself - giving 0. The double cast is a roundabout way of reaching a value that *(int*)p would have given directly.

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