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What's the difference between . and ->

zhcnyuyang (4)
There are two operators . and -> in c++,I know when you want to get a variable or function of an object of a class,you should use .
But when I want to get a functions and variables from a pointer ,I see they both can use and seem same.So what's the difference between them?
Last edited on
MikeyBoy (5631)
a->b is the same as doing (*a).b.

So, you use -> when you have a pointer to an object. You use . when you have the object itself.

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struct MyStruct {
  int val;
}

MyStruct a;
a.val = 5;
std::cout << a.val << std::endl;  // Displays 5

MyStruct* b = &a;  // b is a pointer, pointing to a.
b->val = 10;
std::cout << b->val  << std::endl;  // Displays 10
std::cout << a.val << std::endl;  // Also displays 10
Last edited on
zhcnyuyang (4)
Hoping I can get reply quickly,thanks.
zhcnyuyang (4)
I've read this answer
a->b is the same as doing (*a).b.

So, you use -> when you have a pointer to an object. You use . when you have the object itself.

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struct MyStruct {
  int val;
}

MyStruct a;
a.val = 5;
std::cout << a.val << std::endl;  // Displays 5

MyStruct* b = &a;  // b is a pointer, pointing to a.
b->val = 10;
std::cout << b->val  << std::endl;  // Displays 10
std::cout << a.val << std::endl;  // Also displays 10



But what does b.val mean?
MikeyBoy (5631)
But what does b.val mean?

Nothing. It's illegal syntax, assuming b is a raw pointer.

Or are you talking about smart pointers?
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