Pointers Questions and Answers

Q: Which of the following statements correctly declare a function that receives a pointer to pointer to a pointer to a float and returns a pointer to a pointer to a pointer to a pointer to a float?

Correct answer: float ****fun(float***);Read the declaration outward from the name. The parameter is a pointer to a pointer to a pointer to a float, so it is written float ***. The return type is one level deeper again, so it is float ****.Count the asterisks against the words: three for the parameter, four for the return. The other options mismatch one or both. Where such declarations become unreadable, a typedef for the intermediate types is the usual remedy.

Q: If p is a pointer, what does the expression *p give ?

Correct answer: The value stored at the address held by pThe asterisk is the dereference operator, which follows the pointer to the object it refers to. The address of p itself is obtained with &p, and p alone gives the address it holds.

Q: What is the output of this C code? int main() { int *ptr, a = 10; ptr = &a; *ptr += 1; printf("%d,%d/n", *ptr, a); }

Correct answer: 11,11ptr holds the address of a, so *ptr += 1 increments a itself - there is only one object, reached by two names. Both print as 11. Note also the /n in the format string, which is a forward slash and the letter n rather than an escape, so it appears literally instead of starting a new line.

Q: What will be the output of the program assuming that the array begins at the location 1002 and size of an integer is 4 bytes? #include <stdio.h> int main() { int a[3][4] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 }; printf("%u, %u, %u\n", a[0]+1, *(a[0]+1), *(*(a+0)+1)); return 0; }

Correct answer: 1006, 2, 2Take the three arguments in turn. a[0] is a pointer to the first row's first element, at 1002, and adding 1 advances it by sizeof(int), giving 1006.*(a[0]+1) dereferences that, yielding a[0][1], which is 2. And *(*(a+0)+1) is the same expression written with explicit pointer arithmetic - a+0 is a, *(a+0) is a[0] - so it too gives 2.

Q: What will be the output of the following C code? #include <stdio.h> void foo(int*); int main() { int i = 10; foo((&i)++); } void foo(int *p) { printf("%d\n", *p); }

Correct answer: Compile time error&i yields the address of i, but it is a value, not an object - there is no storage holding it that could be incremented.++ requires a modifiable lvalue, so (&i)++ is a constraint violation and the compiler rejects it. The contrast with the next case is instructive: if the address is first stored in a pointer variable, that variable is an object and can be incremented.

Q: What will be the output of the program ? #include <stdio.h> int main() { char *p; p="hello"; printf("%s\n", *&*&p); return 0; }

Correct answer: helloThe operators * and & are inverses, so each adjacent pair cancels. Reading *&*&p from the right: &p gives the address of p, * takes us back to p, & takes the address again, and the final * returns to p.The whole expression is therefore just p, and printf prints hello. Any even number of such pairs behaves the same way; the exercise exists to make the inverse relationship explicit.

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

Correct answer: 0.000000The four bytes holding the integer 10 are read as an IEEE single-precision float, and that bit pattern is a denormal of about \( 1.4 \times 10^{-44} \), which %f prints as 0.000000. The pointer types are incompatible, so a conforming compiler must issue a diagnostic; older compilers treat it as a warning and produce this output, which is what the question is asking for.

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

Correct answer: 0xbfd605e8 0xbfd605ecPointer arithmetic is in units of the pointed-to type, so incrementing an int * advances the address by sizeof(int), four bytes on a typical implementation. The second address is therefore four higher than the first, which the option with the last digit changing from 8 to c reflects. The particular addresses vary from run to run; the difference of four does not.

Q: What will be the output of the following C code? #include <stdio.h> int main() { int i = 97, *p = &i; foo(&i); printf("%d ", *p); } void foo(int *p) { int j = 2; p = &j; printf("%d ", *p); }

Correct answer: 2 97The pointer is passed by value, so foo receives a copy. Assigning p = &j inside foo changes only that copy, and the print there shows 2.Back in main, the original p still points to i, so the second print shows 97. This is the fundamental point about pointer arguments: passing a pointer lets a function modify what it points to, but not the caller's pointer itself. Changing the caller's pointer requires a pointer to pointer.

31.

Which of the following statements correctly declare a function that receives a pointer to pointer to a pointer to a float and returns a pointer to a pointer to a pointer to a pointer to a float?

A.float **fun(float***);
B.float *fun(float**);
C.float fun(float***);
D.float ****fun(float***);

Answer: D

Correct answer: float ****fun(float***);

Read the declaration outward from the name. The parameter is a pointer to a pointer to a pointer to a float, so it is written float ***. The return type is one level deeper again, so it is float ****.

Count the asterisks against the words: three for the parameter, four for the return. The other options mismatch one or both. Where such declarations become unreadable, a typedef for the intermediate types is the usual remedy.

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32.

If p is a pointer, what does the expression *p give ?

A.The address stored in p
B.The value stored at the address held by p
C.The address of p itself
D.The size of the type p points to

Answer: C

Correct answer: The value stored at the address held by p

The asterisk is the dereference operator, which follows the pointer to the object it refers to. The address of p itself is obtained with &p, and p alone gives the address it holds.

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33.

What is the output of this C code?

int main()
{
int *ptr, a = 10;
ptr = &a;
*ptr += 1;
printf("%d,%d/n", *ptr, a);
}

A.10,10
B.10,11
C.11,10
D.11,11

Answer: D

Correct answer: 11,11

ptr holds the address of a, so *ptr += 1 increments a itself - there is only one object, reached by two names. Both print as 11. Note also the /n in the format string, which is a forward slash and the letter n rather than an escape, so it appears literally instead of starting a new line.

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34.

What will be the output of the program assuming that the array begins at the location 1002 and size of an integer is 4 bytes?

#include <stdio.h>

int main()
{
    int a[3][4] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 };
    printf("%u, %u, %u\n", a[0]+1, *(a[0]+1), *(*(a+0)+1));
    return 0;
}

A.448, 4, 4
B.520, 2, 2
C.1006, 2, 2
D.Error

Answer: C

Correct answer: 1006, 2, 2

Take the three arguments in turn. a[0] is a pointer to the first row's first element, at 1002, and adding 1 advances it by sizeof(int), giving 1006.

*(a[0]+1) dereferences that, yielding a[0][1], which is 2. And *(*(a+0)+1) is the same expression written with explicit pointer arithmetic - a+0 is a, *(a+0) is a[0] - so it too gives 2.

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35.

What will be the output of the following C code?

#include <stdio.h>
    void foo(int*);
    int main()
    {
        int i = 10;
        foo((&i)++);
    }
    void foo(int *p)
    {
        printf("%d\n", *p);
    }

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

Answer: C

Correct answer: Compile time error

&i yields the address of i, but it is a value, not an object - there is no storage holding it that could be incremented.

++ requires a modifiable lvalue, so (&i)++ is a constraint violation and the compiler rejects it. The contrast with the next case is instructive: if the address is first stored in a pointer variable, that variable is an object and can be incremented.

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36.

What will be the output of the program ?

#include <stdio.h>

int main()
{
    char *p;
    p="hello";
    printf("%s\n", *&*&p);
    return 0;
}

A.llo
B.hello
C.ello
D.h

Answer: B

Correct answer: hello

The operators * and & are inverses, so each adjacent pair cancels. Reading *&*&p from the right: &p gives the address of p, * takes us back to p, & takes the address again, and the final * returns to p.

The whole expression is therefore just p, and printf prints hello. Any even number of such pairs behaves the same way; the exercise exists to make the inverse relationship explicit.

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37.

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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38.

What is the output of this C code?

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

A.10.000000
B.0.000000
C.Compile time error
D.Undefined behaviour

Answer: B

Correct answer: 0.000000

The four bytes holding the integer 10 are read as an IEEE single-precision float, and that bit pattern is a denormal of about \( 1.4 \times 10^{-44} \), which %f prints as 0.000000. The pointer types are incompatible, so a conforming compiler must issue a diagnostic; older compilers treat it as a warning and produce this output, which is what the question is asking for.

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39.

What is the output of this C code?

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

A.0xbfd605e8 0xbfd605ec
B.0xbfd605e8 0cbfd60520
C.0xbfd605e8 0xbfd605e9
D.Run time error

Answer: A

Correct answer: 0xbfd605e8 0xbfd605ec

Pointer arithmetic is in units of the pointed-to type, so incrementing an int * advances the address by sizeof(int), four bytes on a typical implementation. The second address is therefore four higher than the first, which the option with the last digit changing from 8 to c reflects. The particular addresses vary from run to run; the difference of four does not.

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40.

What will be the output of the following C code?

#include <stdio.h>
    int main()
    {
        int i = 97, *p = &i;
        foo(&i);
        printf("%d ", *p);
    }
    void foo(int *p)
    {
        int j = 2;
        p = &j;
        printf("%d ", *p);
    }

A.2 97
B.2 2
C.Compile time error
D.Segmentation fault/code crash

Answer: A

Correct answer: 2 97

The pointer is passed by value, so foo receives a copy. Assigning p = &j inside foo changes only that copy, and the print there shows 2.

Back in main, the original p still points to i, so the second print shows 97. This is the fundamental point about pointer arguments: passing a pointer lets a function modify what it points to, but not the caller's pointer itself. Changing the caller's pointer requires a pointer to pointer.

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