How to auto increment the destination pointer with memcpy()?-CodePudding

Basically I need to do multiple memcpy() and I don't want to increment the destination pointer dst ; after every memcpy() call.. just wondering how to save some coding time.

memcpy(dest, src1, 1);
dest  ;
memcpy(dest, src2, 2);
dest  ;
dest  ;

How to do multiple memcpy() that increment the destination pointer for you?

CodePudding user response：

Write your own function.

for example generic for any size elements:

void *generic(void *dest, const void *src, const size_t elemsize, size_t nelems)
{
    char *cptr = dest;

    memcpy(dest, src, elemsize * nelems);
    cptr  = nelems * elemsize;
    return cptr;
}

or as in your example (char size elements)

void *charSizeOnly(void *dest, const void *src, size_t nelems)
{
    char *cptr = dest;

    memcpy(dest, src, nelems);
    cptr  = nelems;
    return cptr;
}

example usage:

    mytype *dest;
    mytype *src;
    /* ... some code*/
    /*it will copy and update the dest by 30 elements of type `mytype`*/
    desyt = generic(dest, src, sizeof(*dest), 30);

CodePudding user response：

You could make a function:

void memcpy_auto_inc( void **pp_dest, void *src, size_t count )
{
    memcpy( *pp_dest, src, count );
    *pp_dest = ((char*)*pp_dest)   count;
}

Now, instead of writing

memcpy(dest, src1, 1);
dest  ;
memcpy(dest, src2, 2);
dest =2;

you can write:

memcpy_auto_inc( &dest, src1, 1);
memcpy_auto_inc( &dest, src2, 2);

However, I doubt that this is worth the effort.

If the name memcpy_auto_inc is too long for you, then you can also specify a shorter name.