用 g++ 编译 C++11

问题描述

我正在尝试将我的 C++ 编译器更新为 C++11.我搜索了一下，得出的结论是我必须使用标志 -std=c++0x 或 -std=gnu++0x，但是我不知道很多关于旗帜的事情.谁能帮我?(我使用的是 Ubuntu 12.04.)

当我尝试使用包含在 C++11 中的库(即数组)时，这是我从编译器中得到的错误:

#include #include int main(){std::arrayarr = {2, 3, 5};...}

<块引用>

此文件需要对即将推出的 ISO C++ 标准 C++0x 的编译器和库支持.此支持目前处于实验阶段，必须使用 -std=c++0x 或 -std=gnu++0x 编译器选项启用.

标志(或编译器选项)只不过是传递给编译器可执行文件的普通命令行参数.

假设您从命令行(终端)调用 g++:

$ g++ -std=c++11 your_file.cpp -o your_program

或

$ g++ -std=c++0x your_file.cpp -o your_program

如果上述方法不起作用.

I'm trying to update my C++ compiler to C++11. I have searched a bit and I have come to the conclusion that I have to use the flag -std=c++0x or -std=gnu++0x, but I don't know many things about flags. Can anyone help me? (I'm using Ubuntu 12.04.)

Here is the error that I get from the compiler when I attempt to use a library which is included in C++11 (i.e. array):

#include <array>
#include <iostream>

int main()
{
    std::array<int, 3> arr = {2, 3, 5};
    ...
}

This file requires compiler and library support for the upcoming ISO C++ standard, C++0x. This support is currently experimental, and must be enabled with the -std=c++0x or -std=gnu++0x compiler options.

Flags (or compiler options) are nothing but ordinary command line arguments passed to the compiler executable.

Assuming you are invoking g++ from the command line (terminal):

$ g++ -std=c++11 your_file.cpp -o your_program

or

$ g++ -std=c++0x your_file.cpp -o your_program

if the above doesn't work.

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