I am little confused on the parameters for the memcpy function. If I have
int* arr = new int[5];
int* newarr = new int[6];
and I want to copy the elements in arr
into newarr
using memcopy
,
memcpy(parameter, parameter, parameter)
How do I do this?
(Copy Memory Block) In the C Programming Language, the memcpy function copies n characters from the object pointed to by s2 into the object pointed to by s1. It returns a pointer to the destination.
memcpy() function is an inbuilt function in C++ STL, which is defined in <cstring> header file. memcpy() function is used to copy blocks of memory. This function is used to copy the number of values from one memory location to another. The result of the function is a binary copy of the data.
C library function - memcpy() The C library function void *memcpy(void *dest, const void *src, size_t n) copies n characters from memory area src to memory area dest.
C strcpy() The strcpy() function copies the string pointed by source (including the null character) to the destination. The strcpy() function also returns the copied string.
The C library function void *memcpy (void *dest, const void *src, size_t n) copies n characters from memory area src to memory area dest. Following is the declaration for memcpy () function. dest − This is pointer to the destination array where the content is to be copied, type-casted to a pointer of type void*.
memcpy () is used to copy a block of memory from a location to another. It is declared in string.h Below is a sample C program to show working of memcpy (). 2) memcpy () leads to problems when source and destination addresses overlap. memmove () is another library function that handles overlapping well. Writing code in comment?
Following is the declaration for memcpy () function. dest − This is pointer to the destination array where the content is to be copied, type-casted to a pointer of type void*.
Following is the declaration for memcpy () function. dest − This is pointer to the destination array where the content is to be copied, type-casted to a pointer of type void*. src − This is pointer to the source of data to be copied, type-casted to a pointer of type void*.
So the order is memcpy(destination, source, number_of_bytes)
.
Therefore, you can place the old data at the beginning of newarr
with
memcpy(newarr, arr, 5 * sizeof *arr);
/* sizeof *arr == sizeof arr[0] == sizeof (int) */
or at the end with
memcpy(newarr+1, arr, 5 * sizeof *arr);
Because you know the data type of arr
and newarr
, pointer arithmetic works. But inside memcpy
it doesn't know the type, so it needs to know the number of bytes.
Another alternative is std::copy
or std::copy_n
.
std::copy_n(arr, 5, newarr);
For fundamental types like int
, the bitwise copy done by memcpy
will work fine. For actual class instances, you need to use std::copy
(or copy_n
) so that the class's customized assignment operator will be used.
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