I have a function whose parameter is a reference. I need to pass a pointer argument to this function. Is the below declaration valid for this situation? If not, what do I need to do?
function foo(int& myParam) {
//some code here
}
//function call
int *myArg;
foo(myArg);
To pass the pointer's value by reference, you must dereference the pointer. But in your situation, you never initialize the pointer, so dereferencing it will cause undefined behavior.
int* ptr;
int& ref = *ptr; // dereferencing garbage pointer
You can dereference the pointer and let the compiler turn the value into a reference, but you must have a valid pointer first, eg:
int *ptr = new int();
foo(*ptr)
delete ptr;
Or:
int i;
int *ptr = &i;
foo(*ptr)
if you want to use the value pointed by myArg, you can simply code:
foo(*myArg);
