Memory Allocation Questions and Answers

Q: Which of the following is/are true

Correct answer: All of the aboveAll three statements hold. calloc zeroes what it returns while malloc leaves arbitrary data; malloc followed by memset reproduces that effect; and both return void *.One difference the list does not mention is worth knowing: calloc multiplies its two arguments internally and is required to detect overflow in that product, whereas a hand-written malloc(m * n) can wrap silently and allocate far too little - a well-known source of security bugs.

Q: How many bytes of memory will the following code reserve? #include <stdio.h> #include <stdlib.h> int main() { int *p; p = (int *)malloc(256 * 256); if(p == NULL) printf("Allocation failed"); return 0; }

Hence 256*256 = 65536 is passed to malloc() function which can allocate upto 65535. So the memory allocation will be failed in 16 bit platform (Turbo C in DOS). If you compile the same program in 32 bit platform like Linux (GCC Compiler) it may allocate the required memory

Q: What is the output of this program? #include <stdio.h> #include <stdlib.h> int main() { int *numbers = (int*)calloc(4, sizeof(int)); numbers[0] = 2; free(numbers); printf("Stored integers are "); printf("numbers[%d] = %d ", 0, numbers[0]); return 0; }

Correct answer: Garbage valueThe block is freed and then read through the same pointer. Once free has been called the memory belongs to the allocator again and may be reused or unmapped, so numbers[0] reads whatever happens to be there.This is use after free, and it is undefined behaviour rather than merely unlucky - the program may print 2, may print rubbish, or may crash. It is particularly insidious because the old value often survives long enough to make the code look correct in testing.

Q: Point out the error in the following program. #include <stdio.h> #include <stdlib.h> int main() { char *ptr; *ptr = (char)malloc(30); strcpy(ptr, "RAM"); printf("%s", ptr); free(ptr); return 0; }

Correct answer: Error: in *ptr = (char)malloc(30);The line assigns to *ptr - a single character - rather than to ptr, and casts the returned address to (char), which throws away everything but its lowest byte.It should read ptr = malloc(30);. The strcpy that follows is then writing through a pointer that holds no valid address, so it is the line most likely to crash - but the cause lies in the allocation line above it, which is what the question asks for.

Q: The call realloc(NULL, size) is equivalent to ________.

Correct answer: malloc(size)The standard defines realloc() with a null first argument to behave exactly like malloc(). This lets a growing buffer be managed by a single realloc() call in a loop, with no special case for the first allocation.

Q: What does malloc() return if the requested memory cannot be allocated ?

Correct answer: A NULL pointerOn failure malloc() returns a null pointer, so every allocation should be tested before the pointer is used. Dereferencing the result without checking is one of the commonest causes of a crash in C.

11.

Which of the following is/are true

A.calloc() allocates the memory and also initializes the allocates memory to zero, while memory allocated using malloc() has random data.
B.malloc() and memset() can be used to get the same effect as calloc()
C.Both malloc() and calloc() return 'void *' pointer
D.All of the above

Answer: D

Correct answer: All of the above

All three statements hold. calloc zeroes what it returns while malloc leaves arbitrary data; malloc followed by memset reproduces that effect; and both return void *.

One difference the list does not mention is worth knowing: calloc multiplies its two arguments internally and is required to detect overflow in that product, whereas a hand-written malloc(m * n) can wrap silently and allocate far too little - a well-known source of security bugs.

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12.

How many bytes of memory will the following code reserve?

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int *p;
    p = (int *)malloc(256 * 256);
    if(p == NULL)
        printf("Allocation failed");
    return 0;
}

A.65536
B.Allocation failed
C.Error
D.No output

Answer: B

Hence 256*256 = 65536 is passed to malloc() function which can allocate upto 65535. So the memory allocation will be failed in 16 bit platform (Turbo C in DOS).

If you compile the same program in 32 bit platform like Linux (GCC Compiler) it may allocate the required memory

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13.

Which of the following is true?

A."ptr = calloc(m, n)" is equivalent to followingptr = malloc(m * n);
B."ptr = calloc(m, n)" is equivalent to followingptr = malloc(m * n); memset(ptr, 0, m * n);
C."ptr = calloc(m, n)" is equivalent to followingptr = malloc(m); memset(ptr, 0, m);
D."ptr = calloc(m, n)" is equivalent to followingptr = malloc(n); memset(ptr, 0, n);

Answer: B

Correct answer: "ptr = calloc(m, n)" is equivalent to following ptr = malloc(m * n); memset(ptr, 0, m * n);

calloc(m, n) does two things: it allocates \( m \times n \) bytes, and it sets every one of them to zero. Reproducing it with malloc therefore needs both steps - the allocation and the memset.

The option offering malloc(m * n) alone gets the size right but leaves the block holding arbitrary data. The two options using m or n alone allocate far too little, since calloc's two arguments are multiplied. That multiplication is also where calloc checks for overflow, which a hand-written product does not.

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14.

Why is calloc() function used for?

A.allocates the specified number of bytes
B.allocates the specified number of bytes and initializes them to zero
C.increases or decreases the size of the specified block of memory and reallocates it if needed
D.calls the specified block of memory for execution

Answer: B

Correct answer: allocates the specified number of bytes and initializes them to zero

calloc does two things where malloc does one: it allocates the storage and sets every byte of it to zero.

It also takes its size as a count and an element size rather than a single total, and it is required to detect overflow in the product of the two - a check a hand-written malloc(m * n) does not get. Resizing an existing block is realloc's work, described by the third option.

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15.

If malloc() successfully allocates memory it returns the number of bytes it has allocated.

#include <stdio.h>
#include <stdlib.h>
void fun(int *a)
{
    a = (int*)malloc(sizeof(int));
}
int main()
{
    int *p;
    fun(p);
    *p = 6;
    printf("%d\n",*p);
    return(0);
}

A.May not work
B.Works and prints 6
C.Compiler Error
D.Runtime error

Answer: A

The program is not valid. Try replacing "int *p;" with "int *p = NULL;" and it will try to dereference a null pointer. This is because fun() makes a copy of the pointer, so when malloc() is called, it is setting the copied pointer to the memory location, not p. p is pointing to random memory before and after the call to fun(), and when you dereference it, it will crash. If you want to add memory to a pointer from a function, you need to pass the address of the pointer (ie. double pointer).

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16.

What is the output of this program?

#include <stdio.h>
#include <stdlib.h>
int main()
{
   int *numbers = (int*)calloc(4, sizeof(int));
   numbers[0] = 2;
   free(numbers);
   printf("Stored integers are ");
   printf("numbers[%d] = %d ", 0, numbers[0]);
   return 0;
}

A.Garbage value
B.2
C.Compilation error

Answer: A

Correct answer: Garbage value

The block is freed and then read through the same pointer. Once free has been called the memory belongs to the allocator again and may be reused or unmapped, so numbers[0] reads whatever happens to be there.

This is use after free, and it is undefined behaviour rather than merely unlucky - the program may print 2, may print rubbish, or may crash. It is particularly insidious because the old value often survives long enough to make the code look correct in testing.

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17.

Point out the error in the following program.

#include <stdio.h>
#include <stdlib.h>
int main()
{
    char *ptr;
    *ptr = (char)malloc(30);
    strcpy(ptr, "RAM");
    printf("%s", ptr);
    free(ptr);
    return 0;
}

A.Error: in strcpy() statement
B.Error: in *ptr = (char)malloc(30);
C.Error: in free(ptr);
D.No Error

Answer: B

Correct answer: Error: in *ptr = (char)malloc(30);

The line assigns to *ptr - a single character - rather than to ptr, and casts the returned address to (char), which throws away everything but its lowest byte.

It should read ptr = malloc(30);. The strcpy that follows is then writing through a pointer that holds no valid address, so it is the line most likely to crash - but the cause lies in the allocation line above it, which is what the question asks for.

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18.

The call realloc(NULL, size) is equivalent to ________.

A.free(size)
B.malloc(size)
C.calloc(1, size)
D.an error

Answer: D

Correct answer: malloc(size)

The standard defines realloc() with a null first argument to behave exactly like malloc(). This lets a growing buffer be managed by a single realloc() call in a loop, with no special case for the first allocation.

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19.

What happens when realloc() cannot extend a block in place ?

A.It returns NULL and leaves the original block untouched forever
B.It allocates a new block, copies the contents and frees the old one
C.It truncates the block to the available size
D.It merges the block with the stack

Answer: C

Correct answer: It allocates a new block, copies the contents and frees the old one

When there is not enough adjacent free space, realloc() obtains a fresh block, copies the existing data across, releases the old block and returns the new address. This is why the returned pointer must always be captured rather than assumed unchanged.

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20.

What does malloc() return if the requested memory cannot be allocated ?

A.0 as an integer
B.A NULL pointer
C.A pointer to the last free block
D.It terminates the program

Answer: C

Correct answer: A NULL pointer

On failure malloc() returns a null pointer, so every allocation should be tested before the pointer is used. Dereferencing the result without checking is one of the commonest causes of a crash in C.

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