What is the arrow operator (->) a synonym for?
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7 Answers
The following two expressions are equivalent:
a->b
(*a).b
(subject to operator overloading, as Konrad mentions, but that's unusual).
5 Comments
Konrad Rudolph
Overloading issues are a lot less unusual than you think. Not long ago, STL implementors had no overloaded
-> operator for some iterator types so you had to use *.. Many libraries define them inconsistently. Becomes really annoying when you work with templates and don't know the precise type.
Sellorio
you can also do
a[0].b instead of (*a).b. But it wouldn't be as properly structured.
ATL_DEV
Boy, after many years of c# programming, going back to c++ is not only cognitively taxing, the c++ syntax is just ugly and yucky. I feel like taking a shower after using it. Programs written in c and c++ just encourage bad programming. Apple, pre-unix, struggled to make the language as pretty as Pascal.
Tim Seguine
@ATL_DEV I'd argue that a lot of the ugly stuff isn't considered idiomatic anymore, but unfortunately that doesn't mean you can afford to not be familiar with it as a practicing C++ programmer. Also the syntactically nice path is often not the semantically nice path, but that also has been getting better not worse. But then again I have C++ Stockholm Syndrome.
ATL_DEV
@TimSeguine If you ever want to see pretty code, then look at the documentation for inside the Macintosh. I think they invented CamelCase. Very descriptive variable names and elegantly formatted code. They managed to make their later C code almost as gorgeous as their earlier Pascal code.
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.
answered Oct 21, 2008 at 10:03
Konrad Rudolph
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2 Comments
vishless
By
binding strength you mean operator precedence? if not what is the difference between the two?Konrad Rudolph
@Vizkrig Yes, the two terms are used interchangeably (although “operator precedence” seems to be much more frequent, at least in recent years).
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.
2 Comments
musiphil
new SomeClass() returns a pointer (SomeClass *), not the SomeClass object. And you start with declaring anObject and aPointer but you're using p afterwards.
Code Man
overall this explanation is theoretically very apt, only the change of objects makes it a little convoluted. But the process is better described
In C++0x, the operator gets a second meaning, indicating the return type of a function or lambda expression
auto f() -> int; // "->" means "returns ..."
answered Nov 6, 2010 at 15:06
Johannes Schaub - litb
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3 Comments
Martin Ba
Technically specking it is no longer an "operator" there, or is it?
Johannes Schaub - litb
@Martin most people use the word "operator" for many things that aren't directly used for computing values. Like for "::" ("scope operator"). I don't know what the point of view of the standard is on this, exactly. In an abstract sense, one could view "->" as a functional operator mapping a sequence of types (parameters) to a return type, like the haskell operator, which is written "->" too.
musiphil
@JohannesSchaub-litb:
:: is actually an operator, like . or ->, and is called "scope resolution operator" in the standard.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."
Comments
-> 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;
Comments
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.
