Why does `memmove` use` void * `as a parameter instead of` char * `?

Question

The definition of the c library function memmove

looks like this:

void* memmove(void *s1, const void *s2, size_t n)
{
    char *sc1;
    const char *sc2;

    sc1 = s1;
    sc2 = s2;
    ...
}

I am wondering why we need to use void*

and const void*

how type of parameters. Why not directly char*

and const char*

?

Update

int test_case[] = {1, 2, 3, 4, 5, 6, 7, 8, 10};

memmove(test_case+4, test_case+2, sizeof(int)*4);

Output: test_case = {1, 2, 3, 4, 3, 4, 5, 6, 10}

+3

c memmove

Fihop 03 Aug At 4:49 am

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2 answers

If char*

and are used const char*

, then when calling memmove

for other types, we should always use char*

.

Using void*

and const void*

, we can write shorter code and there is no casting overhead.

+6

cshu 03 Aug '15 at 5:21

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Yu hao · Accepted Answer · 2015-08-03T04:57:15+0000

void *

- general type of pointer. memmove

it is supposed to manipulate memory no matter what type of objects is in memory.

Likewise for memcpy

. Compare it to strcpy

one that uses parameters char *

as it must manipulate strings.