In C, the **`memcpy()` function** copies `n` number of characters from one block of memory to another.

# Syntax
The general syntax for the `memcpy()` function is:

```C
void * memcpy(void *destination, const void *source, size_t n);
```

## Parameters
Let's look at the parameters that the `memcpy()` function takes as input:

* **Destination:** Pointer to the destination array where data will be copied.

* **Source:** Pointer to the source of the data to copy.

* **N:** The number of characters to copy.

> **Note:** 
>
>1. The [pointers](https://www.educative.io/answers/what-is-a-pointer-in-c) have been declared `void *` so that they may be used for any data type.
> 2. The `memcpy()` function does *not* check for the terminating null character `\0` and only copies `n` characters of a string.

## Return value
The function returns a pointer to the destination string. 

# Example
Let's look at a simple example of copying a string from one array to another using the `memcpy()` function.

In case of overlapping [memory segments](https://www.educative.io/answers/learning-about-segmentation-faults-in-cpp), the `memcpy()` function ceases to work properly, i.e. if you wish to copy characters within the same block of memory, the `memcpy()` function does not function as desired and therefore is not recommended for use. 

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

int main()
{
  char source[] = "World";   
  char destination[] = "Hello ";   
  /* Printing destination string before memcpy */
  printf("Original String: %s\n", destination);
  /* Copies contents of source to destination */
  memcpy (destination, source, sizeof(source)); 
  /* Printing destination string after memcpy */
  printf("Modified String: %s\n", destination);

  return 0;
}

C: Copying data using the memcpy() function in C

Syntax

Parameters

Return value

Example