是否有一个库(例如Apache,Guava)提供了List<T>一个方法
void add(Optional<T> element)
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如果元素存在,则添加元素(如果是,则为无操作!element.isPresent())?显然很容易实现,但似乎有人可能已经做过这样一个显而易见的事情.
我正在通过将typedef转换为使用别名来将一些C ++代码更新为C + 11。鉴于以下SCCE:
#include <iostream>
#include <linux/cn_proc.h>
/**
* Legacy C structure
*/
struct sample {
enum what {
FOO,
BAR
} what;
};
void tdef( ) {
typedef enum sample::what demo;
demo a = sample::FOO;
std::cout << a << std::endl;
}
void usingdemo( ) {
using demo = enum sample::what;
demo a = sample::BAR;
std::cout << a
<< std::endl;
}
int main() {
tdef();
usingdemo();
}
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我正在使用using声明得到警告:
warning: declaration ‘enum sample::what’ does not declare anything
using demo = …Run Code Online (Sandbox Code Playgroud) 在Opensuse中运行Java Swing应用程序时,主类(com-simontuffs-onejar-Boot)的名称将显示在屏幕顶部.如何将其设置为我想要的标题?
版本信息:
/etc/SUSE-release
openSUSE 12.1 (i586)
VERSION = 12.1
CODENAME = Asparagus
uname -a
Linux pax22 3.1.0-1.2-default #1 SMP Thu Nov 3 14:45:45 UTC 2011 (187dde0) i686 i686 i386 GNU/Linux
java -version
java version "1.6.0_12"
Java(TM) SE Runtime Environment (build 1.6.0_12-b04)
Java HotSpot(TM) Server VM (build 11.2-b01, mixed mode)
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//hpp
template <typename T>
struct Demo {
template<typename U>
U convert(const T &t);
};
//cpp
template <typename T>
template <typename U>
U Demo<T>::convert(const T &t) {
return static_cast<U>(t);
}
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如何在cpp中显式实例化模板?(例如T是双,U是int)