C++ Error 1 error C2664 passing array pointers - c++

Question

I don't understand how fix this problem, tried many things but no solution. Help on it would be much appreciated. Thanks.
Error 1 error C2664: 'void showAllBuses(const Bus *[],int)' : cannot convert argument 1 from 'Bus **' to 'const Bus *[]'
void showAllBuses(const Bus* pBuses[], int numBus) {
for (int i = 0; i < numBus; i++) {
cout << "Bus no ." << numBus << " details: " << endl;
cout << "Number: " << pBuses[i]->getNumber() << endl;
cout << "Driver name: " << pBuses[i]->getDriver().getName() << endl;
cout << "Driver experience(years): " << pBuses[i]->getDriver().getYearsDriving() << endl;
}
}
void listBusesWithYearsDriving(const Bus* pBuses[], int numBus, int drivingYears) {
for (int i = 0; i < numBus; i++) {
if (pBuses[i]->getDriver().getYearsDriving() >= drivingYears) {
cout << "Bus number: " << pBuses[i]->getNumber() << endl;
cout << "Driver name: " << pBuses[i]->getDriver().getName() << endl;
cout << "Driver experience: " << pBuses[i]->getDriver().getYearsDriving() << endl;
}
}
}
void removeDriver(Bus* pBuses[], int busPos) {
pBuses[busPos]->removeDriver();
}
void main() {
const int ASIZE = 4;
int drivingYears = 0;
Bus* buses = new Bus[ASIZE];
for (int i = 0; i < ASIZE; i++) {
addNewBus(&buses[i]);
cout << "Bus " << i << ": " << &buses[i] << endl;
}
showAllBuses(&buses, ASIZE);
cout << "Please enter a minimum years of experience to look for: " << endl;
cin >> drivingYears;
listBusesWithYearsDriving(&buses, ASIZE, drivingYears);
removeDriver(&buses, 0);
showAllBuses(&buses, ASIZE);
delete[] buses;
buses = nullptr;
cout << "\n\n";
system("pause");
}

Answer 1

For any type T, T* can be implicitly converted to const T*, but T** cannot be implicitly converted to const T**. Allowing this would make it possible to violate constness(1).
A T ** can be converted to const T* const *, though. Since your function does not modify the array in any way, you can simply change its parameter like that:
void showAllBuses(const Bus* const * pBuses, int numBus) {
Bear in mind that in a function parameter declaration, * and the outermost [] are synonyms.
(1) Here's the code:
const int C = 42;
int I = -42;
int *p = &I;
int *pp = &p;
const int **cp = pp; // error here, but if it was allowed:
*cp = &C; // no problem, *cp is `const int *`, but it's also `p`!
*p = 0; // p is &C!