试图将一个空指针投射到一个结构

Question

I know this has been asked before but I never got an answer that suited me.. so here goes.. here is my project header file:

struct rss_s {
char * device_info;
char * device_model;
char * device_serial;
Radio_Types radio_type;
int power_48v;
int power_400hz;
int panel_lamps;
void * radio_info;
int sub_devices;
struct device_s {
    int fd[ FD_pair ];
    int frequency[ tuned ];
    int panel_lamp;
}sub_device[];
  };

and here is a struct for one of radios:

struct radio_614L8 {
loop_sw614L8 loop_sw_614L8;
mode_sw614L8 mode_sw_614L8;
int sw_band;
int sw_bfo;
int meter;
int dial_lamp;
  };

I initialized everything in main with the following:

static struct radio_G3713 G_3713;
static struct radio_614L8 C_614L8;
static struct radio_G1981 G_1981;
static struct radio_G3490 G_3490;
static struct radio_G4214 G_4214;

static struct rss_s radios[] = {
{ "COM/NAV #1", "G-3717", "81",  G3717, 0, 0, 0, & G_3713,  2, },
{ "ADF",        "614L8", "8384", C614L8,0, 0, 0, & C_614L8, 1, },
{ "ATC",        "G-1981", "336", G1981, 0, 0, 0, & G_1981,  1, },
{ "5in1",       "G-3490", "31",  G3490, 0, 0, 0, & G_3490,  4, },
{ "COM/NAV #2", "G-4214", "68",  G4214, 0, 0, 0, & G_4214,  1, }};

the forth column is a enum: so by using a pointer to "radio -> radio_info" I now know which radio sub-system I need to use.
the problem is I need to cast "void * radio_info" to a pointer type of "struct radio_C614L8" so I tried:
radios -> radio_info = ( struct radio_614L8 *) radios -> radio_info;
and got statement with no effect....

but all I get out of eclipse is an error message:
undefined reference to `init_C614L8'

ok I give up what am I doing wrong.. please help

Answer 1

I think, if understand correctly now, you need a temporary variable.

Where you would like to use a radio_G3713, you should do something like:

   struct radio_G3713 *radio = ((struct radio_G3713 *)(radios[i]->radio_info);
   radio->meter++;

Answer 2

radios -> radio_info = ( struct radio_614L8 *) radios -> radio_info;

does nothing because radios->radio_info is still type void *.

You need to put the result of your typecast in a variable of the right type

 struct radio_614L8 * r_614L8 = ( struct radio_614L8 *) radios -> radio_info;

for instance.

Answer 3

As I've stated as a comment, typecasting a variable creates only a temporary variable with the type you desire. You may not ever change the type of the variable you have already declared. For example, the following also won't work, and is truly analogous:

int a = 5;
a = (float) a;

Here, in the first line, a variable of type int is declared, assigned with the value 5 . Then with the second line, first the value 5 gets typecasted as float , which gives us a 5.0 . Then this 5.0 is to be assigned to a again, which happens to be an int , so it will get implicitly casted into an int first, into a 5 , then get assigned to a .

a may never be a float , a is destined to be an int ever since it got declared.

I don't really know what you are hoping to have in the end, but here's something:

By making a call like radios->radio_info , you refer to the radio_info inside the very first radio , which is the radio[0] , be aware of that. If you want to refer to the radio_info of a specific radio , then either use (radio + x)->radio_info or radio[x].radio_info , both are equivalent.

Second thing is, you first need to have a memory location that holds a struct radio_614L8 to be able to point to it. Two ways to do that:

// declaring one and then assigning its address to your pointer
struct radio_614L8 asd;
(radio + 0)->radio_info = &asd;

// or allocating memory for one and assigning the address of the memory to your pointer
(radio + 0)->radio_info = malloc( sizeof( struct radio_614L8 ) );

After that point, you can refer to your void pointer, as if it was a struct radio_614L8 pointer with the following manner, and change its sw_band , meter , whatever:

((struct radio_614L8 *) (radio + 0)->radio_info) -> meter = 2345;
((struct radio_614L8 *) (radio + 0)->radio_info) -> dial_lamp = 123;

I know, it's kind of long... But if you really want to keep the radio_info in your structure as a void * , then this is the way to go. What's happening there is:

(radio + 0)->radio_info;    // recognized as a void *
                            // has some allocated memory though...
(struct radio_614L8 *) (radio + 0)-> radio_info;
                            // recognized as a struct radio_614L8 * now
                            // allowing you to refer sw_bfo, etc.