CODEWITHSUNDEEP
c++functionpointersfunction-pointers
Roland Soós
Roland Soós

Reputation: 3290

Function pointer as parameter

I try to call a function which passed as function pointer with no argument, but I can't make it work.

void *disconnectFunc;

void D::setDisconnectFunc(void (*func)){
    disconnectFunc = func;
}

void D::disconnected(){
    *disconnectFunc;
    connected = false;
}

Upvotes: 66

Views: 147103

Answers (3)

Nikolai Fetissov
Nikolai Fetissov

Reputation: 84239

Replace void *disconnectFunc; with void (*disconnectFunc)(); to declare function pointer type variable. Or even better use a typedef:

typedef void (*func_t)(); // pointer to function with no args and void return
...
func_t fptr; // variable of pointer to function
...
void D::setDisconnectFunc( func_t func )
{
    fptr = func;
}

void D::disconnected()
{
    fptr();
    connected = false;
}

Upvotes: 14

GManNickG
GManNickG

Reputation: 504253

The correct way to do this is:

typedef void (*callback_function)(void); // type for conciseness

callback_function disconnectFunc; // variable to store function pointer type

void D::setDisconnectFunc(callback_function pFunc)
{
    disconnectFunc = pFunc; // store
}

void D::disconnected()
{
    disconnectFunc(); // call
    connected = false;
}

Upvotes: 98

WhirlWind
WhirlWind

Reputation: 14112

You need to declare disconnectFunc as a function pointer, not a void pointer. You also need to call it as a function (with parentheses), and no "*" is needed.

Upvotes: 8

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