A simple question for which I couldn't find the answer here.
What I understand is that while passing an argument to a function during call, e.g.
void myFunction(type myVariable)
{
}
void main()
{
myFunction(myVariable);
}
For simple datatypes like int
, float
, etc. the function is called by value.
But if myVariable
is an array, only the starting address is passed (even though our function is a call by value function).
If myVariable
is an object, also only the address of the object is passed rather than creating a copy and passing it.
So back to the question. Does C++ pass a object by reference or value?
Arguments are passed by value, unless the function signature specifies otherwise:
void foo(type arg)
, arg
is passed by value regardless of whether type
is a simple type, a pointer type or a class type,void foo(type& arg)
, arg
is passed by reference.In case of arrays, the value that is passed is a pointer to the first element of the array. If you know the size of the array at compile time, you can pass an array by reference as well: void foo(type (&arg)[10])
.