dangling-pointer

问题 Swift has: Strong References Weak References Unowned References How is an unowned reference different from a weak reference? When is it safe to use an unowned reference? Are unowned references a security risk like dangling pointers in C/C++? 回答1: Both weak and unowned references do not create a strong hold on the referred object (a.k.a. they don't increase the retain count in order to prevent ARC from deallocating the referred object). But why two keywords? This distinction has to do with the

问题 I have the following code. #include <iostream> int * foo() { int a = 5; return &a; } int main() { int* p = foo(); std::cout << *p; *p = 8; std::cout << *p; } And the code is just running with no runtime exceptions! The output was 58 How can it be? Isn\'t the memory of a local variable inaccessible outside its function? 回答1: How can it be? Isn't the memory of a local variable inaccessible outside its function? You rent a hotel room. You put a book in the top drawer of the bedside table and go

问题 I know this is pretty common question, but still new for me! I don\'t understand concept of dangling pointer, was googling around, and writing test methods to find one. I just wonder is this a dangling pointer? As whatever example I found was returning something, here I\'m trying something similar! Thanks! void foo(const std::string name) { // will it be Dangling pointer?!, with comments/Answer // it could be if in new_foo, I store name into Global. // Why?! And what is safe then? new_foo