memmove function

I am trying to replicate the memmove function. I copied the following from online, and it's replicating the memmove function. But I am wondering, why we are checking source and destination memory addresses (if statement)? why that check is needed? is there any chance for overlap? If yes, how the destination and source string will overlap? memmove function has to copy the source string to the destination string in a non-destructive manner. Since both are stored in a different memory address how do both inrfere? where does the destruction happen?

void *ft_memmove(void *dst, const void *src, size_t len)

{

char *dst_aux;

char *src_aux;

char *dst_last;

char *src_last;

if (!dst && !src)

return (NULL);

dst_aux = (char *)dst;

src_aux = (char *)src;

if (dst_aux < src_aux)

while (len--)

*dst_aux++ = *src_aux++;

else

{

dst_last = dst_aux + (len - 1);

src_last = src_aux + (len - 1);

while (len--)

*dst_last-- = *src_last--;

}

return (dst);

}

memmove is a C standard library function used to copy a block of memory from one location to another. It's similar to the memcpy function, but it's designed to handle overlapping memory areas. In other words, memmove is used when the source and destination memory regions can overlap without causing issues.

The syntax for memmove is:

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

dest is a pointer to the destination memory buffer where the data will be copied.
src is a pointer to the source memory buffer from which the data will be copied.
n is the number of bytes to copy.

memmove ensures that the data is copied in a way that is safe even if the source and destination memory regions overlap. It achieves this by using a temporary buffer to perform the copy.

Here's an example of how you might use memmove:

#include
#include

int main() {
char str[] = "Hello, World!";
memmove(str + 7, str, 7);
printf("%s\n", str);
return 0;
}
In this example, memmove is used to move the string "World!" within the same character array str. The function safely handles the overlap between the source and destination, resulting in "Hello, World!" as the output. If you were to use memcpy in this case, it might not work correctly due to the overlap.

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Bhavesh Raval2y ago
Accepted answer
memmove is a C standard library function used to copy a block of memory from one location to another. It's similar to the memcpy function, but it's designed to handle overlapping memory areas. In other words, memmove is used when the source and destination memory regions can overlap without causing issues.

The syntax for memmove is:

void *memmove(void *dest, const void *src, size_t n);
- dest is a pointer to the destination memory buffer where the data will be copied.
- src is a pointer to the source memory buffer from which the data will be copied.
- n is the number of bytes to copy.
memmove ensures that the data is copied in a way that is safe even if the source and destination memory regions overlap. It achieves this by using a temporary buffer to perform the copy.

Here's an example of how you might use memmove:

#include
#include

int main() {
char str[] = "Hello, World!";
memmove(str + 7, str, 7);
printf("%s\n", str);
return 0;
}
In this example, memmove is used to move the string "World!" within the same character array str. The function safely handles the overlap between the source and destination, resulting in "Hello, World!" as the output. If you were to use memcpy in this case, it might not work correctly due to the overlap.