Cast (void *) to any other in a safe way - c++

Question

Is this safe now?
#include <stdio.h>
#include <string.h>
long lee( void *v1)
{
long i1;
memcpy(&i1,v1,sizeof(long));
return i1;
}
//Added conversion to double:
double lee_d( void *v1)
{
double i1;
memcpy(&i1,v1,sizeof(double));
return i1;
}
void main()
{
long val=1777777;
long ii;
ii=lee((void *) &val);
double dd=377e-177;
dd=lee_d((void *) &dd);
ii=ii;
}
It works well under MS studio.
I do not use this pointer translations but when using curl libraries

Answer 1

The integral division will happen before the adding of 0.99999 therefore you are going to get an integer, which may be 0, 1, 2, add an arbitrary 0.99999 to it which achieves nothing, then cast it back to long.
size1 could still be 0 so val[0] might not be valid.
It is also difficult to know what you are trying to achieve.
To answer your question directly, it is not safe.

Answer 2

As far as lee() and your call to it goes, yes it's safe.

Answer 3

As has been said the (void *) are unneccessary & should be avoided until you know what you're doing. In the function declaration concerned it just means the function takes a pointer to something but it doesn't matter what the type of 'something' is.
E.g.
#include <stdio.h>
#include <stdlib.h>
#include <strings.h>
static void doit(int *); // Function Prototypes. VERY IMPORTANT.
static void doitAgain(void *);
int main(int argc, char *argv[]) {
int i = 0x12345;
doit(&i);
doitAgain(&i);
}
static void doit(int *b)
{
int *foo = malloc(sizeof(int));
memcpy(foo, b, sizeof(int));
fprintf(stdout, "foo is 0x%x\n", *foo);
}
static void doitAgain(void *b)
{
int *foo = malloc(sizeof(int));
memcpy(foo, b, sizeof(int));
fprintf(stdout, "foo is 0x%x\n", *foo);
}
And when we run it:
$ make
cc main.c -o main
$ ./main
foo is 0x12345
foo is 0x12345