转换void指针（在struct中工作时）

Question

Im very new (few weeks) to the whole C programming so please take it easy on me ;-)

Project details: I'm currently working on making a train travelplanner system, using a given trainplan to find the shortest route to a specific destination. This is done making a graph, a binary heap for sorting, and then the dijkstra algortihm. Hence the sorta messy structs (sorry! figured it was the best to get familiar with them).

My issue: I built a linkedlist, having data as void inside a struct (making it more general purpose for reuse), however when I want to cast the data while working with structs, I get a syntax error.

The struct:

typedef struct  linked_list{
    void *data;
    struct linked_list *next;
    struct linked_list *previous;
} linked_list;

And my, probably, glaring error:

if(current->((edge*)data)->to->distance > ...

This works perfectly fine if I change the data type to being edge in my struct (and not trying this stunt), but casting in this case apparently does not work.

If more details are needed, please let me know.

Answer 1

((edge*)current->data)->to->distance

Answer 2

The right operand of the -> operator must be the name of a member of a struct (or union); it cannot be a cast expression.

current is an expression of type linked_list (presumably).

current->data is an expression of type void* .

To convert that void* expression to an edge* expression, you need to apply a cast to the expression, not to the member name, so:

(edge*)current->data is an expression of type edge*