Pointers Questions and Answers

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 is the output of this C code? int main() { int i = 11; int *p = &i; foo(&p); printf("%d ", *p); } void foo(int *const *p) { int j = 10; *p = &j; printf("%d ", **p); }

Correct answer: Compile time errorint *const *p means a pointer to a constant pointer to int, so *p is const and cannot be assigned to. The line *p = &j therefore violates the constraint and the compiler rejects it. Reading such declarations from the name outward is the reliable method: p, pointer to, const, pointer to, int.

Q: What will be the output of the program ? #include <stdio.h> int main() { void *vp; char ch=74, *cp="JACK"; int j=65; vp=&ch; printf("%c", *(char*)vp); vp=&j; printf("%c", *(int*)vp); vp=cp; printf("%s", (char*)vp+2); return 0; }

Correct answer: JACKTake the three prints in turn. ch is 74, the code for J, so the first prints J. Then j is 65 and *(int*)vp yields 65, which %c prints as A.Finally vp is set to cp, and (char*)vp+2 advances two characters into "JACK", so %s prints CK. The three outputs concatenate to JACK - a coincidence engineered by the question rather than anything the code copies.

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

Correct answer: Compile time error&i is a value, not an object, so it cannot be incremented: the ++ operator needs a modifiable lvalue and the result of & is neither modifiable nor an lvalue. The next question shows the working version, in which a pointer variable holds the address and it is the variable that is incremented.

Q: Are the three declarations char **apple, char *apple[ ], and char apple[ ][ ] same?

Correct answer: FalseThe three declare different things. char **apple is a pointer to a pointer. char *apple[] is an array of pointers, which decays to char ** when passed as a parameter but is not the same object. And char apple[][] is not even legal, since all dimensions but the first must be given for the compiler to compute the element size.

Q: Is there any difference between the following two statements? char *p=0; char *t=NULL;

Correct answer: NoBoth give a null pointer. A constant expression with value zero used in a pointer context is a null pointer constant, and NULL is defined to be exactly such a constant, so the two initialisations are equivalent. Using NULL is preferred only because it documents the intent, not because it produces anything different.

Q: What is the output of this C code? void main() { int a[3] = {1, 2, 3}; int *p = a; int *r = &p; printf("%d", (**r)); }

Correct answer: Compile time errorp is an int *, so &p has type int **, and initialising an int * from it is a constraint violation - the levels of indirection do not match. The declaration should read int **r = &p;, which is exactly what the following question does. Some compilers downgrade this to a warning, but it is an error in standard C.

Q: Will the program compile in Turbo C? #include <stdio.h> int main() { int a=10, *j; void *k; j=k=&a; j++; k++; printf("%u %u\n", j, k); return 0; }

Correct answer: NoArithmetic on a void * is not allowed, because the compiler would have to know the size of the pointed-to type to advance the pointer, and void has none. So k++ is the offending line. The j++ above it is perfectly legal and advances by the size of an int. GCC permits void arithmetic as an extension, treating the size as one, but it is not standard C.

Q: Which of the following are correct syntaxes to send an array as a parameter to function:

Correct answer: func(array);An array name in an expression decays to a pointer to its first element, so passing the name is all that is required and the function receives that pointer. &array gives a pointer to the whole array, a different type; *array passes the first element rather than the array; and array[size] passes one element, and one that is out of bounds at that.

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.

61.

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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62.

What is the output of this C code?

int main()
{
int i = 11;
int *p = &i;
foo(&p);
printf("%d ", *p);
}
void foo(int *const *p)
{
int j = 10;
*p = &j;
printf("%d ", **p);
}

A.Compile time error
B.10 10
C.Undefined behaviour
D.10 11

Answer: A

Correct answer: Compile time error

int *const *p means a pointer to a constant pointer to int, so *p is const and cannot be assigned to. The line *p = &j therefore violates the constraint and the compiler rejects it. Reading such declarations from the name outward is the reliable method: p, pointer to, const, pointer to, int.

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63.

What will be the output of the program ?

#include <stdio.h>

int main()
{
    void *vp;
    char ch=74, *cp="JACK";
    int j=65;
    vp=&ch;
    printf("%c", *(char*)vp);
    vp=&j;
    printf("%c", *(int*)vp);
    vp=cp;
    printf("%s", (char*)vp+2);
    return 0;
}

A.JCK
B.J65K
C.JAK
D.JACK

Answer: D

Correct answer: JACK

Take the three prints in turn. ch is 74, the code for J, so the first prints J. Then j is 65 and *(int*)vp yields 65, which %c prints as A.

Finally vp is set to cp, and (char*)vp+2 advances two characters into "JACK", so %s prints CK. The three outputs concatenate to JACK - a coincidence engineered by the question rather than anything the code copies.

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64.

What is the output of this C code?

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 is a value, not an object, so it cannot be incremented: the ++ operator needs a modifiable lvalue and the result of & is neither modifiable nor an lvalue. The next question shows the working version, in which a pointer variable holds the address and it is the variable that is incremented.

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65.

Are the three declarations char **apple, char *apple[ ], and char apple[ ][ ] same?

A.True
B.False

Answer: B

Correct answer: False

The three declare different things. char **apple is a pointer to a pointer. char *apple[] is an array of pointers, which decays to char ** when passed as a parameter but is not the same object. And char apple[][] is not even legal, since all dimensions but the first must be given for the compiler to compute the element size.

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66.

Is there any difference between the following two statements?

char *p=0;
char *t=NULL;

A.Yes
B.No

Answer: B

Correct answer: No

Both give a null pointer. A constant expression with value zero used in a pointer context is a null pointer constant, and NULL is defined to be exactly such a constant, so the two initialisations are equivalent. Using NULL is preferred only because it documents the intent, not because it produces anything different.

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67.

What is the output of this C code?

void main()
{
int a[3] = {1, 2, 3};
int *p = a;
int *r = &p;
printf("%d", (**r));
}

A.1
B.Compile time error
C.Address of a
D.Junk value

Answer: B

Correct answer: Compile time error

p is an int *, so &p has type int **, and initialising an int * from it is a constraint violation - the levels of indirection do not match. The declaration should read int **r = &p;, which is exactly what the following question does. Some compilers downgrade this to a warning, but it is an error in standard C.

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68.

Will the program compile in Turbo C?

#include <stdio.h>
int main()
{
    int a=10, *j;
    void *k;
    j=k=&a;
    j++;
    k++;
    printf("%u %u\n", j, k);
    return 0;
}

A.Yes
B.No

Answer: B

Correct answer: No

Arithmetic on a void * is not allowed, because the compiler would have to know the size of the pointed-to type to advance the pointer, and void has none. So k++ is the offending line. The j++ above it is perfectly legal and advances by the size of an int. GCC permits void arithmetic as an extension, treating the size as one, but it is not standard C.

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69.

Which of the following are correct syntaxes to send an array as a parameter to function:

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

Answer: B

Correct answer: func(array);

An array name in an expression decays to a pointer to its first element, so passing the name is all that is required and the function receives that pointer. &array gives a pointer to the whole array, a different type; *array passes the first element rather than the array; and array[size] passes one element, and one that is out of bounds at that.

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70.

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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