Memory Allocation Questions and Answers

Q: Consider the following three C functions : [PI] int * g (void) { int x= 10; return (&x); } [P2] int * g (void) { int * px; *px= 10; return px; } [P3] int *g (void) { int *px; px = (int *) malloc (sizeof(int)); *px= 10; return px; } Which of the above three functions are likely to cause problems with pointers?

Correct answer: Only P1 and P2P1 returns the address of a local variable. Its lifetime ends when the function returns, so the caller receives a pointer to storage that no longer belongs to it - a dangling pointer.P2 never assigns px an address at all, then writes through it, which is an uninitialised pointer and undefined behaviour. P3 is correct: it obtains storage from malloc, which has allocated lifetime and survives the return. The caller must free it, but that is a responsibility rather than a fault.

Q: Point out the correct statement which correctly free the memory pointed to by 's' and 'p' in the following program? #include <stdio.h> #include <stdlib.h> int main() { struct ex { int i; float j; char *s }; struct ex *p; p = (struct ex *)malloc(sizeof(struct ex)); p->s = (char*)malloc(20); return 0; }

Correct answer: free(p->s); , free(p);The order matters. p->s is reachable only through p, so freeing p first would destroy the only record of the inner block's address and leak it permanently.Releasing the inner allocation before the outer one is the general rule for any nested structure, and it means deallocation runs in the reverse order of allocation. The same reasoning applies to a two-dimensional array built as an array of row pointers: free every row, then free the array of pointers.

Q: What is the return type of malloc() or calloc()

Correct answer: void *Both return void *, a pointer to memory of no particular type. That is what allows one allocator to serve every type in the language: the caller assigns the result to a pointer of the type it wants, and the conversion happens implicitly.In C the cast is therefore unnecessary, and writing one can conceal a missing stdlib.h - without the prototype the compiler assumes a return type of int, and the cast silences the warning that would have exposed it. In C++ the cast is required, which is why it is so often seen.

Q: malloc() returns a NULL if it fails to allocate the requested memory.

Correct answer: TRUEWhen malloc cannot satisfy a request it returns a null pointer. That is the only way it reports failure - there is no error code and nothing is set beyond, on most systems, errno.Checking the return value is therefore the caller's responsibility, and omitting the check is a common fault: the program proceeds to dereference a null pointer and crashes at a line some distance from the real cause. Note that malloc(0) may legitimately return either null or a unique pointer.

Q: What will be the output of the program? #include <stdio.h> #include <string.h> int main() { char *s; char *fun(); s = fun(); printf("%s\n", s); return 0; } char *fun() { char buffer[30]; strcpy(buffer, "RAM"); return (buffer); }

Correct answer: Garbage valuefun returns the address of buffer, an automatic array whose lifetime ends when the function returns. The caller therefore holds a pointer to storage that has been reclaimed.Reading through it is undefined behaviour. The text may survive briefly, but the call to printf itself uses the stack and is quite likely to overwrite it, so garbage is the usual result. Returning a pointer to a local is one of the classic C faults; the fix is to allocate the buffer with malloc or to have the caller supply it.

Q: What is the problem with following code? #include <stdio.h> int main() { int *p = (int *)malloc(sizeof(int)); p = NULL; free(p); }

Correct answer: Memory LeakThe program allocates a block, then overwrites the only pointer to it with NULL. Nothing now holds the block's address, so it can never be freed - that is a memory leak.Calling free(NULL) is explicitly defined to do nothing, so the last line is harmless and there is no crash and no compiler error. Nor is there a dangling pointer, which would require the pointer still to refer to freed memory. The fault is that the block was lost before it could be released.

Q: Where does the uninitialized data gets stored in memory?

Correct answer: BSS- Block started by symbolThe BSS holds objects with static storage duration that have no explicit initialiser, and which are therefore zero-initialised before the program starts.Keeping them separate saves space in the executable: the file need only record how many bytes to reserve, and the loader clears them, rather than storing a long run of zeros on disk. Initialised globals go in the data segment proper, where their values must actually be stored, and the code segment holds the instructions.

61.

Consider the following three C functions :

[PI] int * g (void) 
{ 
  int x= 10; 
  return (&x); 
}  
  
[P2] int * g (void) 
{ 
  int * px; 
  *px= 10; 
  return px; 
} 
  
[P3] int *g (void) 
{ 
  int *px; 
  px = (int *) malloc (sizeof(int)); 
  *px= 10; 
  return px; 
}

Which of the above three functions are likely to cause problems with pointers?

A.Only P3
B.Only P1 and P3
C.Only P1 and P2
D.P1, P2 and P3

Answer: C

Correct answer: Only P1 and P2

P1 returns the address of a local variable. Its lifetime ends when the function returns, so the caller receives a pointer to storage that no longer belongs to it - a dangling pointer.

P2 never assigns px an address at all, then writes through it, which is an uninitialised pointer and undefined behaviour. P3 is correct: it obtains storage from malloc, which has allocated lifetime and survives the return. The caller must free it, but that is a responsibility rather than a fault.

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62.

What is the output of this program?

#include <stdio.h>
#include <stdlib.h>
int main()
{
  int i, numbers[1];
  numbers[0] = 15;
  free(numbers);
  printf("Stored integers are ");
  printf("numbers[%d] = %d ", 0, numbers[0]);
  return 0;
}

A.15
B.Compilation error
C.garbage value

Answer: A

he memory allocation function free() will not free or delete the content in the memory location, because free() can only free or delete the content in the memory, who's memory is allocated either by malloc() or calloc().

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63.

Point out the correct statement which correctly free the memory pointed to by 's' and 'p' in the following program?

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

int main()
{
    struct ex
    {
        int i;
        float j;
        char *s
    };
    struct ex *p;
    p = (struct ex *)malloc(sizeof(struct ex));
    p->s = (char*)malloc(20);
    return 0;
}

A.free(p); , free(p->s);
B.free(p->s); , free(p);
C.free(p->s);
D.free(p);

Answer: B

Correct answer: free(p->s); , free(p);

The order matters. p->s is reachable only through p, so freeing p first would destroy the only record of the inner block's address and leak it permanently.

Releasing the inner allocation before the outer one is the general rule for any nested structure, and it means deallocation runs in the reverse order of allocation. The same reasoning applies to a two-dimensional array built as an array of row pointers: free every row, then free the array of pointers.

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64.

What is the return type of malloc() or calloc()

A.void *
B.Pointer of allocated memory type
C.void **
D.int *

Answer: A

Correct answer: void *

Both return void *, a pointer to memory of no particular type. That is what allows one allocator to serve every type in the language: the caller assigns the result to a pointer of the type it wants, and the conversion happens implicitly.

In C the cast is therefore unnecessary, and writing one can conceal a missing stdlib.h - without the prototype the compiler assumes a return type of int, and the cast silences the warning that would have exposed it. In C++ the cast is required, which is why it is so often seen.

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65.

malloc() returns a NULL if it fails to allocate the requested memory.

A.True
B.False

Answer: A

Correct answer: TRUE

When malloc cannot satisfy a request it returns a null pointer. That is the only way it reports failure - there is no error code and nothing is set beyond, on most systems, errno.

Checking the return value is therefore the caller's responsibility, and omitting the check is a common fault: the program proceeds to dereference a null pointer and crashes at a line some distance from the real cause. Note that malloc(0) may legitimately return either null or a unique pointer.

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66.

What will be the output of the program?

#include <stdio.h>
#include <string.h>

int main()
{
    char *s;
    char *fun();
    s = fun();
    printf("%s\n", s);
    return 0;
}
char *fun()
{
    char buffer[30];
    strcpy(buffer, "RAM");
    return (buffer);
}

A.0xffff
B.Garbage value
C.0xffee
D.Error

Answer: B

Correct answer: Garbage value

fun returns the address of buffer, an automatic array whose lifetime ends when the function returns. The caller therefore holds a pointer to storage that has been reclaimed.

Reading through it is undefined behaviour. The text may survive briefly, but the call to printf itself uses the stack and is quite likely to overwrite it, so garbage is the usual result. Returning a pointer to a local is one of the classic C faults; the fix is to allocate the buffer with malloc or to have the caller supply it.

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67.

What is the output of this program?

#include <stdio.h>
void main()
{
   int *ptr = (int *)malloc(sizeof(int));
   *ptr = 10;
   free(ptr);
   p = 5;
   printf("%d", ptr);
}

A.Compilation error
B.5
C.Garbage value

Answer: B

free() will not deallocate the memory it just to delete all data's allocated to a variable (ptr). In this program, first integer value 10 is assigned to the pointer variable *ptr and then its data is deleted using free(ptr) and then new integer value 5 is assigned to the variable ptr and then 5 is outputted.

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68.

What is the problem with following code?

#include <stdio.h>
int main()
{
    int *p = (int *)malloc(sizeof(int));
 
    p = NULL;
 
    free(p);
}

A.Compiler Error: free can't be applied on NULL pointer
B.Memory Leak
C.Dangling Pointer
D.The program may crash as free() is called for NULL pointer

Answer: B

Correct answer: Memory Leak

The program allocates a block, then overwrites the only pointer to it with NULL. Nothing now holds the block's address, so it can never be freed - that is a memory leak.

Calling free(NULL) is explicitly defined to do nothing, so the last line is harmless and there is no crash and no compiler error. Nor is there a dangling pointer, which would require the pointer still to refer to freed memory. The fault is that the block was lost before it could be released.

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69.

Which statment is true about the given code ?

#include <stdio.h>
#include <stdlib.h>
int main()
{
    int *a[5];
    a = (int*) malloc(sizeof(int)*5);
    free(a);
    return 0;
}

A.Error: unable to allocate memory
B.Error: We cannot store address of allocated memory in a
C.Error: unable to free memory
D.No error

Answer: B

Correct answer: Error: We cannot store address of allocated memory in a

int *a[5] declares an array of five pointers. An array name is not a modifiable lvalue, so a = ... is rejected by the compiler - nothing can be assigned to it.

The declaration the code seems to intend is int *a;, a single pointer, which could then hold the allocated address. The distinction between an array of pointers and a pointer is exactly the kind of declaration that repays reading from the inside out.

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70.

Where does the uninitialized data gets stored in memory?

A.Code segment
B.Data segment
C.BSS- Block started by symbol
D.Heap

Answer: C

Correct answer: BSS- Block started by symbol

The BSS holds objects with static storage duration that have no explicit initialiser, and which are therefore zero-initialised before the program starts.

Keeping them separate saves space in the executable: the file need only record how many bytes to reserve, and the loader clears them, rather than storing a long run of zeros on disk. Initialised globals go in the data segment proper, where their values must actually be stored, and the code segment holds the instructions.

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