C 编程 - 将 struct 成员变量分配给 char *

Question

I have a variable declared as const char * server = NULL;

This variable is assigned a value when the user passes the IP address of a server as a command line argument to the C program using the code below. For example, running ./cprogram -h 10.4.0.01

opt = 1;
while (opt < argc)
{
    if (argv[opt] == NULL    ||
        argv[opt][0] != '-'  ||
        argv[opt][2] != 0)
    {
        print_usage();
        return 0;
    }

    switch (argv[opt][1])
    {
        case 'h':
            opt++;
            if (opt >= argc)
            {
                print_usage();
                return 0;
            }
            server = argv[opt];
            break;
    }

I am removing the need to pass command line arguments from the program and putting the values in an .ini file.

I now read an .ini file at the start of the program and store the values in a struct.

struct lwm2m_object {
    char clientname[LG_BUF];
    char ipv4[LG_BUF];
    char server[LG_BUF];
};

In the correct way to assign the struct member variable server to the variable const char * server using this

server = &(lwm2m.server);

Answer 1

The assignment is not correct. The correct assignment is

server = lwm2m.server;

Answer 2

Your

server = &(lwm2m.server);

is doubly wrong . First, the string you intend to copy is lwm2m.server -- remember that an array decays to a pointer. So, better is

server = lwm2m.server;    // Still dangerous

But this merely copies the pointer to the data ( shallow copy ). If the lwm2m_object lwm2m goes out of scope, then the associated string lwm2m_object::server is freed and may be re-used. In other words, your copy in server then becomes a dangling pointer .

If you know that lwm2m survives server (so that there is no danger of getting a dangling pointer), then the shallow copy is okay (and preferable). Otherwise, you must make a deep copy, using strcpyn :

const char server[LG_BUF];
strcpyn(server, lwm2m.server, LG_BUF);