What if i want to have an array of pointers to a function and the size of the array is not known from the beginning? I'm just curious if there's a way to do that. Using new statement or maybe something else. Something looking similar to
void (* testArray[5])(void *) = new void ()(void *);
You could use a std::vector:
#include <vector>
typedef void (*FunPointer)(void *);
std::vector<FunPointer> pointers;
If you really want to use a static array, it would be better to do it using the FunPointer i defined in the snippet above:
FunPointer testArray[5];
testArray[0] = some_fun_pointer;
Though i would still go for the vector solution, taking into account that you don't know the size of the array during compilation time and that you are using C++ and not C.
With typedef, the new expression is trivial:
typedef void(*F)(void*);
int main () {
F *testArray = new F[5];
if(testArray[0]) testArray[0](0);
}
Without typedef, it is somewhat more difficult:
void x(void*) {}
int main () {
void (*(*testArray))(void*) = new (void(*[5])(void*));
testArray[3] = x;
if(testArray[3]) testArray[3](0);
}
for(i=0;i<length;i++)
A[i]=new node
or
#include <vector>
std::vector<someObj*> x;
x.resize(someSize);
