clang++ (3.3/Xcode) 中 std::function 的定义在哪里
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Where is definition of std::function in clang++ (3.3/Xcode)
提问by shigeya
Problem Solved => see the update at the end
问题已解决 => 看到最后的更新
I'm trying to use std::functionbut it looks like just include <functional>does not provide the definition. I have tried to compile following code:
我正在尝试使用，std::function但看起来只是包含<functional>不提供定义。我试图编译以下代码：
#include <functional>
std::function<int(int)> f = nullptr;
with c++11 as compile option:
使用 c++11 作为编译选项：
% clang++ -c -std=c++11 t.cc
cause:
原因：
t.cc:3:6: error: no type named 'function' in namespace 'std'
std::function<int(int)> f = nullptr;
~~~~~^
t.cc:3:14: error: expected unqualified-id
std::function<int(int)> f = nullptr;
^
2 errors generated.
what am I missing? I know C++ well but new to clang++/C++11 thus I lack of important knowledge, I guess.
我错过了什么？我很了解 C++，但我对 clang++/C++11 是新手，因此我想我缺乏重要的知识。
I'm using clang++ on MacOS X 10.8.
我在 MacOS X 10.8 上使用 clang++。
Update 1
更新 1
I have tried a sample at cppreference.com sitebut it won't compile too. Giving some option solve the problem?
我已经在 cppreference.com 站点尝试了一个示例，但它也无法编译。给一些选项解决问题？
Update 2
更新 2
Tried above sample from cppreference.com with clang++ -c -std=c++11 -stdlib=libc++11 x.cc, and compiler still says:
从 cppreference.com 尝试了上面的示例clang++ -c -std=c++11 -stdlib=libc++11 x.cc，编译器仍然说：
x.cc:1:10: fatal error: 'functional' file not found
#include <functional>
^
1 error generated.
Where is functional? I guess I should give -stdlib=libc++11or whatever but it does not work too:
功能在哪里？我想我应该给予-stdlib=libc++11或其他什么，但它也不起作用：
clang: error: invalid library name in argument '-stdlib=libc++11'
How I can find list of argument for -stdlib? (note: in man page, only available options are libc++and libstdc++both of them don't work)
我如何找到参数列表-stdlib？（注意：在手册页中，只有可用选项libc++并且libstdc++它们都不起作用）
Or functional just does not work?
或者功能只是不起作用？
回答by
This is not about the definition of the function. You don't have a linker error. You have a compiler error. The problem is, presumably, that the BSD/GNU/Darwin standard library installed in the real sysroot doesn't support C++11. You have to use the one that comes with Clang by specifying the -stdlib=libc++compiler flag.
这与函数的定义无关。您没有链接器错误。你有一个编译器错误。问题大概是安装在真正的 sysroot 中的 BSD/GNU/Darwin 标准库不支持 C++11。您必须通过指定-stdlib=libc++编译器标志来使用 Clang 附带的那个。
回答by Brett Hale
For C++11, it's best to always invoke clangas: clang++ -std=c++11 -stdlib=libc++
对于 C++11，最好总是这样调用clang：clang++ -std=c++11 -stdlib=libc++
I use this most of the time, so I set the environment variable $CXXto this value. That way, I'm getting the dialect and library option in both compilation and linking. -std=c++11is insufficient, as clang will still use the (old) system gcc headers in /usr/include/c++/4.2.1.
我大部分时间都使用这个，所以我将环境变量设置$CXX为这个值。这样，我在编译和链接中都获得了方言和库选项。-std=c++11是不够的，因为 clang 仍将在/usr/include/c++/4.2.1.
-stdlib=libc++will use the clang headers in /usr/lib/c++/v1such as <functional>.
-stdlib=libc++将在/usr/lib/c++/v1诸如<functional>.
There's a similar questionwith an answer by Howard Hinnant, who is (IIRC) an Apple engineer.
苹果工程师 (IIRC) 的 Howard Hinnant 回答了类似的问题。
