Difference between revisions of "Interview Preparation Pointers"
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Pointer 20 | Pointer 20 | ||
</pre> | </pre> | ||
+ | |||
+ | '''Malloc''' | ||
+ | The Malloc function dynamically allocates memory when required. The function allocates size of byte of memory and returns a pointer to the first byte of NULL. | ||
+ | <pre> | ||
+ | |||
+ | Syntax: | ||
+ | pointer = (type)malloc(size in bytes); | ||
+ | |||
+ | Code: | ||
+ | |||
+ | int* p; //Declare pointer | ||
+ | p = (int*)malloc(sizeof(int)); // Pointer equal to pointer type int that contain memory address space of int | ||
+ | *p = 5; // Finally pointer points to location containing value 5 | ||
+ | </pre> |
Revision as of 15:02, 17 December 2016
Pointers : A pointer is a variable who's value is address of some other variable i.e. it can be address of some memory location.
<varaible_type> *<name> eg : int *pointer_to_integer
The above example , we have declared a pointer to a variable (pointer_to_integer), the variable stores the address of an integer .
Implementation of Pointer :
Code: #include <stdio.h> int main(void) { int var = 20; int *p; p = &var; printf("Pointer %d\n",var); printf("Pointer %d\n",&var); // Prints the address of the varaible (var) printf("Pointer %d\n",p); printf("Pointer %d\n",*p); // Prints the value that (p) points to return 0; } Output : Pointer 20 Pointer 1809844068 Pointer 1809844068 Pointer 20
Malloc
The Malloc function dynamically allocates memory when required. The function allocates size of byte of memory and returns a pointer to the first byte of NULL.
Syntax: pointer = (type)malloc(size in bytes); Code: int* p; //Declare pointer p = (int*)malloc(sizeof(int)); // Pointer equal to pointer type int that contain memory address space of int *p = 5; // Finally pointer points to location containing value 5