void* in c and pointers

Question

I am having trouble understanding the void* pointer in c. I've googled around but haven't really understood how to solve this specific problem: ``` typedef struct _Test{ char* c; }Test; void method(void* test){ Test t; t = *(Test*)test; t.c = "omg"; printf(t.c); //WORKS } int main(){ Test t; method(&t); printf(t.c); //NOT WORKING, prints nothing/random letters return 0;} ``` Why? Or rather, best way to fix/get around this issue?

Accepted Answer

You are changing the local object `t` inside `method()`, after copying `main()`'s object `t` into it. This doesn't change anything in `main()`'s object since you never copy in the other direction.

You should just access through the pointer and directly change the caller's object:

```
((Test *) test)->c = "omg";
```

or, you can make it a bit clearer by using a local pointer of the proper type, which might be what you were trying to do:

```
void method(void* test) {
    Test *t = test;
    t->c = "omg";
}
```

note that no cast is needed here, since `void *` automatically converts to `Test *` in C.

void* in c and pointers

Solution

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