How to pass objects to functions in C++?
I am new to C++ programming, but I have experience in Java. I need guidance on how to pass objects to functions in C++.
Do I need to pass pointers, references, or non-pointer and non-reference values? I remember in Java there are no such issues since we pass just the variable that holds reference to the objects.
It would be great if you could also explain where to use each of those options.
Rules of thumb for C++11:
Pass by value , except when
const reference , const lvalue reference , const reference or not.) Passing by pointer is virtually never advised. Optional parameters are best expressed as a boost::optional , and aliasing is done fine by reference.
C++11's move semantics make passing and returning by value much more attractive even for complex objects.
Rules of thumb for C++03:
Pass arguments by const reference , except when
const reference NULL / 0 / nullptr instead; apply the previous rule to determine whether you should pass by a pointer to a const argument (here, "pass by value" is called "pass by copy", because passing by value always creates a copy in C++03)
There's more to this, but these few beginner's rules will get you quite far.
There are some differences in calling conventions in C++ and Java. In C++ there are technically speaking only two conventions: pass-by-value and pass-by-reference, with some literature including a third pass-by-pointer convention (that is actually pass-by-value of a pointer type). On top of that, you can add const-ness to the type of the argument, enhancing the semantics.
Pass by reference
Passing by reference means that the function will conceptually receive your object instance and not a copy of it. The reference is conceptually an alias to the object that was used in the calling context, and cannot be null. All operations performed inside the function apply to the object outside the function. This convention is not available in Java or C.
Pass by value (and pass-by-pointer)
The compiler will generate a copy of the object in the calling context and use that copy inside the function. All operations performed inside the function are done to the copy, not the external element. This is the convention for primitive types in Java.
An special version of it is passing a pointer (address-of the object) into a function. The function receives the pointer, and any and all operations applied to the pointer itself are applied to the copy (pointer), on the other hand, operations applied to the dereferenced pointer will apply to the object instance at that memory location, so the function can have side effects. The effect of using pass-by-value of a pointer to the object will allow the internal function to modify external values, as with pass-by-reference and will also allow for optional values (pass a null pointer).
This is the convention used in C when a function needs to modify an external variable, and the convention used in Java with reference types: the reference is copied, but the referred object is the same: changes to the reference/pointer are not visible outside the function, but changes to the pointed memory are.
Adding const to the equation
In C++ you can assign constant-ness to objects when defining variables, pointers and references at different levels. You can declare a variable to be constant, you can declare a reference to a constant instance, and you can define all pointers to constant objects, constant pointers to mutable objects and constant pointers to constant elements. Conversely in Java you can only define one level of constant-ness (final keyword): that of the variable (instance for primitive types, reference for reference types), but you cannot define a reference to an immutable element (unless the class itself is immutable).
This is extensively used in C++ calling conventions. When the objects are small you can pass the object by value. The compiler will generate a copy, but that copy is not an expensive operation. For any other type, if the function will not change the object, you can pass a reference to a constant instance (usually called constant reference) of the type. This will not copy the object, but pass it into the function. But at the same time the compiler will guarantee that the object is not changed inside the function.
Rules of thumb
This are some basic rules to follow:
There are other small deviations from these rules, the first of which is handling ownership of an object. When an object is dynamically allocated with new, it must be deallocated with delete (or the [] versions thereof). The object or function that is responsible for the destruction of the object is considered the owner of the resource. When a dynamically allocated object is created in a piece of code, but the ownership is transfered to a different element it is usually done with pass-by-pointer semantics, or if possible with smart pointers.
Side note
It is important to insist in the importance of the difference between C++ and Java references. In C++ references are conceptually the instance of the object, not an accessor to it. The simplest example is implementing a swap function:
// C++
class Type; // defined somewhere before, with the appropriate operations
void swap( Type & a, Type & b ) {
Type tmp = a;
a = b;
b = tmp;
}
int main() {
Type a, b;
Type old_a = a, old_b = b;
swap( a, b );
assert( a == old_b );
assert( b == old_a );
}
The swap function above changes both its arguments through the use of references. The closest code in Java:
public class C {
// ...
public static void swap( C a, C b ) {
C tmp = a;
a = b;
b = tmp;
}
public static void main( String args[] ) {
C a = new C();
C b = new C();
C old_a = a;
C old_b = b;
swap( a, b );
// a and b remain unchanged a==old_a, and b==old_b
}
}
The Java version of the code will modify the copies of the references internally, but will not modify the actual objects externally. Java references are C pointers without pointer arithmetic that get passed by value into functions.
There are several cases to consider.
Parameter modified ("out" and "in/out" parameters)
void modifies(T ¶m);
// vs
void modifies(T *param);
This case is mostly about style: do you want the code to look like call(obj) or call(&obj)? However, there are two points where the difference matters: the optional case, below, and you want to use a reference when overloading operators.
...and optional
void modifies(T *param=0); // default value optional, too
// vs
void modifies();
void modifies(T ¶m);
Parameter not modified
void uses(T const ¶m);
// vs
void uses(T param);
This is the interesting case. The rule of thumb is "cheap to copy" types are passed by value — these are generally small types (but not always) — while others are passed by const ref. However, if you need to make a copy within your function regardless, you should pass by value. (Yes, this exposes a bit of implementation detail. C'est le C++.)
...and optional
void uses(T const *param=0); // default value optional, too
// vs
void uses();
void uses(T const ¶m); // or optional(T param)
There's the least difference here between all situations, so choose whichever makes your life easiest.
Const by value is an implementation detail
void f(T);
void f(T const);
These declarations are actually the exact same function! When passing by value, const is purely an implementation detail. Try it out:
void f(int);
void f(int const) { /* implements above function, not an overload */ }
typedef void NC(int); // typedefing function types
typedef void C(int const);
NC *nc = &f; // nc is a function pointer
C *c = nc; // C and NC are identical types
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