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What is the difference between the dot (.) operator and -> in C++?
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Difference between Dot(.)operator is used to normally access members of a structure or union. The Arrow(->) operator exists to access the members of the structure or the unions using pointers.
The dot ( . ) operator is used to access a member of a struct, while the arrow operator ( -> ) in C is used to access a member of a struct which is referenced by the pointer in question.
if you have an actual object (or a reference to the object, declared with & in the declared type), and you use -> if you have a pointer to an object (declared with * in the declared type).
Remarks. The member access operators . and -> are used to refer to members of struct , union , and class types. Member access expressions have the value and type of the selected member.
foo->bar() is the same as (*foo).bar().
The parenthesizes above are necessary because of the binding strength of the * and . operators.
*foo.bar() wouldn't work because Dot (.) operator is evaluated first (see operator precedence)
The Dot (.) operator can't be overloaded, arrow (->) operator can be overloaded.
The Dot (.) operator can't be applied to pointers.
Also see: What is the arrow operator (->) synonym for in C++?
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For a pointer, we could just use
*pointervariable.foo But the . operator has greater precedence than the * operator, so . is evaluated first. So we need to force this with parenthesis:
(*pointervariable).foo But typing the ()'s all the time is hard, so they developed -> as a shortcut to say the same thing. If you are accessing a property of an object or object reference, use . If you are accessing a property of an object through a pointer, use ->
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