C memmove() Function

January 18, 2021 9:10 pm IST

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C memmove() Function

The memmove() function copies a block of memory from a location to another.

This function copies the data first to an intermediate buffer, then from buffer to destination. So, memmove() is slightly slower approach than memcpy().

This function doesn't overlap the source. So, memmove() is safer than memcpy().

Example

C Compiler

#include <stdio.h> #include <string.h> int main() { char dest[] = "abcde"; char src[] = "123"; memmove(dest, src, 3); printf("New dest = %s", dest); return 0; }

Output

New dest = 123de

Syntax

void *memmove(void *dest, const void *src, size_t n)

Parameter Values

Value	Type	Explanation
dest	Required	Destination array where the content is to be copied.
src	Required	Source of data to be copied.
n	Required	Number of bytes to be copied.

Return Value

Value	Explanation
Address	Returns a pointer to the destination, which is dest array.

More Examples

In the following example, the destination string is "cde" and the source string is "abcde". So, the memmove() function will replace "cde" with "abc" and the result is ababc.

Example

C Compiler

#include <stdio.h> #include <string.h> int main() { char str1[] = "abcde"; memmove(&str1[2], &str1[0], 3); printf("New str1 = %s", str1); return 0; }

Output

New str1 = ababc

In the following example, memcpy() overlaps the source, but memove() doesn't.

Example

C Compiler

#include <stdio.h> #include <string.h> int main() { char str1[] = "abcde"; char str2[] = "abcde"; memmove(&str1[2], &str1[0], 3); memcpy(&str2[2], &str2[0], 3); printf("memmove's str1 = %s\n", str1); printf("memcpy's str2 = %s\n", str2); return 0; }