reference

问题 So, it's a bit of a simplistic case - but if I set up a circular reference like this: #!/usr/bin/perl use strict; use warnings; { my $thing; my $otherthing; $thing -> {otherthing} = \$otherthing; $otherthing -> {thing} = \$thing; } I create a memory leak - because by reference counting, the memory allocated here will never be freed, despite not having any outward 'access point'. So what I'm wondering - in this sort of scenario, is there any way I could - via debugging or similar - 'rediscover

问题 So, it's a bit of a simplistic case - but if I set up a circular reference like this: #!/usr/bin/perl use strict; use warnings; { my $thing; my $otherthing; $thing -> {otherthing} = \$otherthing; $otherthing -> {thing} = \$thing; } I create a memory leak - because by reference counting, the memory allocated here will never be freed, despite not having any outward 'access point'. So what I'm wondering - in this sort of scenario, is there any way I could - via debugging or similar - 'rediscover

问题 I've always been a bit confused about how STL containers (vector, list, map...) store values. Do they store references to the values I pass in, or do they copy/copy construct +store the values themselves? For example, int i; vector<int> vec; vec.push_back(i); // does &(vec[0]) == &i; and class abc; abc inst; vector<abc> vec; vec.push_back(inst); // does &(vec[0]) == &inst; Thanks 回答1: STL Containers copy-construct and store values that you pass in. If you want to store objects in a container

问题 When I consider the two following overloads: template <class... T> void f(const T&... x); template <class T> void f(const T& x); I have the guarantee that f(x) will always call the second function and will never lead to an ambiguity. In a sense the second version is universally prioritized compared to the first one for one argument whatever its type is. Now consider the situation where there is a universal reference and a const reference versions of a function: template <class T> void f(T&& x

问题 This is going to sound like a silly question, but I'm still learning C, so please bear with me. :) I'm working on chapter 6 of K&R (structs), and thus far through the book have seen great success. I decided to work with structs pretty heavily, and therefore did a lot of work early in the chapter with the point and rect examples. One of the things I wanted to try was changing the canonrect function (2nd Edition, p 131) work via pointers, and hence return void . I have this working, but ran

问题 I am continuing my path to deep understanding of Java Thread. Unfortunately my Java Certification didn't cover that part, so the only way of learning is to post a series of dumb questions. With so many years of Java Development, I am sometimes wondering how much I still have to learn :-) In particular my attention is now with the reference handler thread. "Reference Handler" daemon prio=10 tid=0x02da3400 nid=0xb98 in Object.wait() [0x0302f000] java.lang.Thread.State: WAITING (on object

问题 I am continuing my path to deep understanding of Java Thread. Unfortunately my Java Certification didn't cover that part, so the only way of learning is to post a series of dumb questions. With so many years of Java Development, I am sometimes wondering how much I still have to learn :-) In particular my attention is now with the reference handler thread. "Reference Handler" daemon prio=10 tid=0x02da3400 nid=0xb98 in Object.wait() [0x0302f000] java.lang.Thread.State: WAITING (on object

问题 I originally thought Python was a pure pass-by-reference language. Coming from C/C++ I can't help but think about memory management, and it's hard to put it out of my head. So I'm trying to think of it from a Java perspective and think of everything but primitives as a pass by reference. Problem: I have a list, containing a bunch of instances of a user defined class. If I use the for-each syntax, ie.: for member in my_list: print(member.str); Is member the equivalent of an actual reference to

问题 The question: Is there benefit to passing an integral type by const reference as opposed to simply by value. ie. void foo(const int& n); // case #1 vs void foo(int n); // case #2 The answer is clear for user defined types, case #1 avoids needless copying while ensuring the constness of the object. However in the above case, the reference and the integer (at least on my system) are the same size, so I can't imagine there being a whole lot of difference in terms of how long it takes for the

问题 I am confused that why following code is not able to compile int foo(const float* &a) { return 0; } int main() { float* a; foo(a); return 0; } Compiler give error as: error: invalid initialization of reference of type 'const float*&' from expression of type 'float*' but when I try to pass without by reference in foo, it is compiling fine. I think it should show same behavior whether I pass by reference or not. Thanks, 回答1: Because it isn't type-safe. Consider: const float f = 2.0; int foo