questions on following statement

Hi All,

I was to trying understand an source file ,I came across following line

Code:

---

void *pParam;
*((char**)pParam) = "SCHD_MODULE";

---

Why is pParam typecasted to double char pointer?

Thanks

I'm thinking that void is already a pointer (NULL pointer), so *pParam would be a double pointer. Casted to char** since it's a char array (string) and then dereferenced once to give a char* pointer.

Equivalent:

Code:

---

char *pParam = NULL;
*pParam = "SCHD_MODULE"; // read only

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Quote:

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Originally Posted by Epy

I'm thinking that void is already a pointer (NULL pointer),

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I'm not sure what you're trying to say here, but it's wrong. A void pointer isn't a NULL pointer, it's also not initialized to NULL. I don't know where you got NULL from, but there was nothing set to NULL in this example. It's a generic pointer, not a NULL pointer.


Quzah.

I still didn't understand the need for typecasting it to double pointer....

I still didn't understand the need for typecasting it to double pointer....

Let's define out custom type:
typedef char mytype_t,;
Now we want to pass that variable to a function to modify it. In order to do that, we need to pass it by address, right?
So,

mtype_t mydata;
foo(&mydata);

And

Code:

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void foo(mytype_t * data)
{
    *data = 'H';
}

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Easy. Right?
Now let's redefine our type a little:

typedef const char* mytype_t;

And we once again pass it to a function:

mtype_t mydata;
foo(&mydata);

And

Code:

---

void foo(mytype_t * data)
{
    *data = "Hello World!"; // We probably should have malloc or something here, but this serves the example fine.
}

---

No problems! We've seen this stuff before.
Now let's substitute mydata_t for the type it's an alias for, and we get:

const char * mydata;
foo(&mydata);

And

Code:

---

void foo(char const* * data)
{
    *data = "Hello World!"; // We probably should have malloc or something here, but this serves the example fine.
}

---

There you go. A double pointer.

The reason for the casting is probably because the function takes an opaque pointer (ie void*). Thus it needs to convert it to its real type before doing something with it.