passing a function as a parameter
void dispatch_for(dispatch_queue_t *queue, long number, void (* work)(long)){
int loop = number;
int i;
task_t *ctask;
for(i = 0; i<loop;i++){
ctask = create_task((void *)work,number,"for_test");
dispatch_async(queue,ctask);
}
}
task_t *task_create(void (* work)(void *), void *param, char* name){
//do something
}
I'm getting work as a function and need to pass it to the function create_task开发者_高级运维..(1st parameter)
How should i pass it?name of the function without the parentheses is the pointer to that function :
void work(void) {
...;
}
void main(void) {
task_create(work, void, void);
}
Just use the identifier like any other parameter:
dispatch_for(queue, number, work);
The easiest thing would be a typedef to the wanted function type. So you can do
typedef void workfunc_t(void *);
workfunc_t sample_workfunc; // in order to ensure type-correctness
void workfunc_t(void *)
{
// ...
}
void dispatch_for(dispatch_queue_t *queue, long number, workfunc_t * work)
{
int loop = number;
int i;
task_t *ctask;
for(i = 0; i<loop;i++) {
ctask = create_task(work, number, "for_test");
dispatch_async(queue,ctask);
}
}
task_t *task_create(workfunc_t work, void *param, char* name){
//do something
}
The function work
doesn't have the same signature in dispatch_for
and task_create
(the argument is a pointer in one, a long in the other)
It seems strange you want to use the same function in both cases
Since you're using C, and work
is a type void (*)(long)
, but you want to make it a void (*)(void*)
, simply re-cast the type of work
(this is can be done the easiest by using a typedef)
//declare somewhere at a global level
typedef void (*task_create_func)(void*);
//at the point where you want to call task_create
task_create((task_create_func)work, PARAM1, PARM2);
Alternatively, if you don't want to deal with typedefs, you can simply do the cast using the desired pointer-type at the point-of-call like so:
task_create((void (*)(void*))work, PARAM1, PARAM2);
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