for (i = 0; i < n1; i++)
L[i] = arr[l + i];
Because I want to copy a large array,I heard that need to use memcpy.
2 Answers
memcpy as the name says, copy memory area. is a C standard function under string.h.
void *memcpy(void *dest, const void *src, size_t n);
description:
The memcpy() function copies n bytes from memory area src to memory
area dest. The memory areas must not overlap. Use memmove(3) if the
memory areas do overlap. The memcpy() function returns a pointer todest.
for more details goto man7: memcpy
so in your case the call would be:
memcpy(L, &arr[l], n1*sizeof(arr[l]));
sizeof one array elementis:
sizeof(arr[l])
make sure that (l+n1) doesnot exceeds the array boundaries! its you responsibility.
1 Comment
AdamF
@曾品翔, please accept my answer (by pushing the beside V ) if you find it useful. thanks.
memcopy(destination, source, length)
That would be in your case:
int len = sizeof(int) * n1;
memcopy(L, arr+l, len);
Note: You may have to fix the length calculation according to the type you are using. Moreover, you should also remember to add 1 to include the \0 character that terminates char arrays if you are dealing with strings.
lby using the address-of operator, as in&arr[l]. This could be used as the source argument for amemcpycall.