auto

Erasing item in a for(-each) auto loop

由 浪尽此生 提交于 2019-11-29 04:51:24
Is there a way to erase specific elements when using a auto variable in a for loop like this? for(auto a: m_Connections) { if(something) { //Erase this element } } I know I can either do say for(auto it=m_map.begin() ... or for(map<int,int>::iterator it=m_map.begin() ... and manually increment the iterator (and erase) but if I could do it with less lines of code I'd be happier. Thanks! No, there isn't. Range based for loop is used to access each element of a container once. Every time an element is removed from the container, iterators at or after the erased element are no longer valid (and

Is 'auto const' and 'const auto' the same?

由 十年热恋 提交于 2019-11-29 04:19:06
问题 Is there a semantic difference between auto const and const auto , or do they mean the same thing? 回答1: The const qualifier applies to the type to the immediate left unless there is nothing to the left then it applies to the type to the immediate right. So yup it's the same. 回答2: Contrived example: std::vector<char*> test; const auto a = test[0]; *a = 'c'; a = 0; // does not compile auto const b = test[1]; *b = 'c'; b = 0; // does not compile Both a and b have type char* const . Don't think

CSS3 流动边框

由 孤者浪人 提交于 2019-11-29 04:18:02
CSS3 流动边框（仿王者荣耀等待效果）的三种实现方式 原地址 https://www.jianshu.com/p/3c241aeae992 <!DOCTYPE html> <html> <head> <meta charset="utf8"> <style> :root { --border-anim-size: 10em; --border-anim-width: calc(var(--border-anim-size) / 20); --border-anim-width-double: calc(var(--border-anim-width)*2); --border-anim-duration: 5s; --border-anim-border-color: gray; --border-anim-hover-color: LightCoral; } body { display: flex; } .border-anim { width: var(--border-anim-size); height: var(--border-anim-size); position: relative; border: 1px solid var(--border-anim-border-color); } .border-anim::before, .border-anim:

How to select iterator type using auto variable?

由 偶尔善良 提交于 2019-11-29 03:55:00
I have a std::unordered_map std::unordered_map<std::string, std::string> myMap; I want to get a const iterator using find. In c++03 I would do std::unordered_map<std::string, std::string>::const_iterator = myMap.find("SomeValue"); In c++11 I would want to use auto instead to cut down on the templates auto = myMap.find("SomeValue"); Will this be a const_iterator or iterator? How does the compiler decide which to use? Is there a way I can force it to choose const? Johannes Schaub - litb It will use non-const iterators if myMap is a non-const expression. You could therefore say #include <type

auto with string literals

由 岁酱吖の 提交于 2019-11-29 01:19:46
#include <iostream> #include <typeinfo> int main() { const char a[] = "hello world"; const char * p = "hello world"; auto x = "hello world"; if (typeid(x) == typeid(a)) std::cout << "It's an array!\n"; else if (typeid(x) == typeid(p)) std::cout << "It's a pointer!\n"; // this is printed else std::cout << "It's Superman!\n"; } Why is x deduced to be a pointer when string literals are actually arrays? A narrow string literal has type "array of n const char " [2.14.5 String Literals [lex.string] §8] The feature auto is based on template argument deduction and template argument deduction behaves

css flex布局问题width:auto

由 强颜欢笑 提交于 2019-11-29 00:22:30
1.如图布局，属性和值是动态获取的，当属性值参数多的时候换行只是属性值部分换行， 又因为属性不是固定的字数，是动态获取的 2，综合以上运用flex布局特意强调width:auto;flex:1;未知属性宽度用width:auto; 来源： CSDN 作者： 远一航一 链接： https://blog.csdn.net/webZRH/article/details/82218756

Why do I need to explicitly write the 'auto' keyword?

由 妖精的绣舞 提交于 2019-11-28 23:32:29
问题 I am moving towards C++11 from C++98 and have become familiar with the auto keyword. I was wondering why we need to explicitly declare auto if the compiler is able to automatically deduce the type. I know C++ is a strongly typed language and this is a rule but was it not possible to achieve the same outcome without explicitly declaring a variable auto ? 回答1: Dropping the explicit auto would break the language: e.g. int main() { int n; { auto n = 0; // this shadows the outer n. } } where you

How to use lambda auto parameters in C++11

由 好久不见. 提交于 2019-11-28 20:20:36
I have a code in C++14. However, when I used it in C++11, it has an error at const auto . How to use it in C++11? vector<vector <int> > P; std::vector<double> f; vector< pair<double, vector<int> > > X; for (int i=0;i<N;i++) X.push_back(make_pair(f[i],P[i])); ////Sorting fitness descending order stable_sort(X.rbegin(), X.rend()); std::stable_sort(X.rbegin(), X.rend(), [](const auto&lhs, const auto& rhs) { return lhs.first < rhs.first; }); leemes C++11 doesn't support generic lambdas . That's what auto in the lambda's parameter list actually stands for: a generic parameter, comparable to

How to declare array with auto

由 瘦欲@ 提交于 2019-11-28 18:24:48
I have been playing with auto and I noticed that for most cases you can replace a variable definition with auto and then assign the type. In the following code w and x are equivalent (default initialized int , but lets not get into potential copies). Is there a way to declare z such that it has the same type as y ? int w{}; auto x = int{}; int y[5]; auto z = int[5]; bames53 TL;DR template<typename T, int N> using raw_array = T[N]; auto &&z = raw_array<int,5>{}; Your example of auto z = int[5]; isn't legal any more than auto z = int; is, simply because a type is not a valid initializer. You can

How do I get a const_iterator using auto?

由 喜你入骨 提交于 2019-11-28 17:44:45
First question: is it possible to "force" a const_iterator using auto? For example: map<int> usa; //...init usa auto city_it = usa.find("New York"); I just want to query, instead of changing anything pointed by city_it , so I'd like to have city_it to be map<int>::const_iterator . But by using auto, city_it is the same to the return type of map::find() , which is map<int>::iterator . Any suggestion? Sorry, but I just think the best suggestion is not using auto at all, since you want to perform a (implicitly valid) type conversion . auto is meant for deducing the exact type , which is not what
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