What can I use instead of the arrow operator, `->`?
C++PointersC++ Problem Overview
What is the arrow operator (->) a synonym for?
C++ Solutions
Solution 1 - C++
The following two expressions are equivalent:
a->b
(*a).b
(subject to operator overloading, as Konrad mentions, but that's unusual).
Solution 2 - C++
a->b is generally a synonym for (*a).b. The parenthesises here are necessary because of the binding strength of the operators * and .: *a.b wouldn't work because . binds stronger and is executed first. This is thus equivalent to *(a.b).
Beware of overloading, though: Since both -> and * can be overloaded, their meaning can differ drastically.
Solution 3 - C++
The C++-language defines the arrow operator (->) as a synonym for dereferencing a pointer and then use the .-operator on that address.
For example:
If you have a an object, anObject, and a pointer, aPointer:
SomeClass anObject = new SomeClass();
SomeClass *aPointer = &anObject;
To be able to use one of the objects methods you dereference the pointer and do a method call on that address:
(*aPointer).method();
Which could be written with the arrow operator:
aPointer->method();
The main reason of the existents of the arrow operator is that it shortens the typing of a very common task and it also kind of easy to forgot the parentheses around the dereferencing of the pointer. If you forgot the parentheses the .-operator will bind stronger then *-operator and make our example execute as:
*(aPointer.method()); // Not our intention!
Some of the other answer have also mention both that C++ operators can be overload and that it is not that common.
Solution 4 - C++
In C++0x, the operator gets a second meaning, indicating the return type of a function or lambda expression
auto f() -> int; // "->" means "returns ..."
Solution 5 - C++
I mostly read it right-to-left and call "in"
foo->bar->baz = qux->croak
becomes:
"baz in bar in foo becomes croak in qux."
Solution 6 - C++
-> is used when accessing data which you've got a pointer to.
For example, you could create a pointer ptr to variable of type int intVar like this:
int* prt = &intVar;
You could then use a function, such as foo, on it only by dereferencing that pointer - to call the function on the variable which the pointer points to, rather than on the numeric value of the memory location of that variable:
(*ptr).foo();
Without the parentheses here, the compiler would understand this as *(ptr.foo()) due to operator precedence which isn't what we want.
This is actually just the same as typing
ptr->foo();
As the ->dereferences that pointer, and so calls the function foo() on the variable which the pointer is pointing to for us.
Similarly, we can use -> to access or set a member of a class:
myClass* ptr = &myClassMember;
ptr->myClassVar = 2;
Solution 7 - C++
You can use -> to define a function.
auto fun() -> int
{
return 100;
}
It's not a lambda. It's really a function. "->" indicates the return type of the function.