在静态变量初始化中使用 cout 时出现 C++ 分段错误
我有一个程序，我使用 cout 来发出调试信息.代码在静态全局变量的初始化中执行，即在程序执行的早期.当我使用自己的构建脚本构建程序时，它在第一次使用 cout 时出现段错误(只有一个字符串文字被转移到 cout，所以它不能是值).我使用 valgrind 检查早期对无效位置的写入，但没有(也没有可能生成这些写入的代码，我在输出之前没有做太多).当我将源代码复制到 Eclipse 项目并让 Eclipse 内置构建器构建它时，一切正常.我没有使用任何奇怪的构建器设置，只是用 -ggdb -std=c++0x 编译，这是仅有的两个标志.
I have a program where I use cout to emit debug information. The code is executed in the initialization of a static global variable, i.e. quite early in the program execution. When I use my own build script to build the program, it segfaults at the first use of cout (only a string literal is shifted into cout, so it cannot be the value). I used valgrind to check for earlier writes to invalid locations, but there are none (and there is also no code that would be likely to generate those writes, I dont do too much before the output). When I copy the source code to an eclipse project and let the eclipse built-in builder build it, then everything works fine. I used no weird builder settings simply compiled with -ggdb -std=c++0x, these are the only two flags.
那么，如果之前没有无效写入，带有字符串文字的 cout 段错误的原因是什么?构建配置对此有何影响?
So what can be the reason that a cout with a string literal segfaults, if there were no invalid writes before? How can the build configuration affect this?
(对不起，我不能给你一个最小的例子，因为这个例子可以简单地在你的机器上编译，就像我在使用 eclipse builder 时所做的那样)
(I am sorry I can give you no minimal example, as this example would simply compile fine at your machine, as it does for me when using the eclipse builder)
这是堆栈跟踪:
0x00007ffff7b6d7d1 in std::ostream::sentry::sentry(std::ostream&) () from /usr/lib /x86_64-linux-gnu/libstdc++.so.6
(gdb) backtrace
#0 0x00007ffff7b6d7d1 in std::ostream::sentry::sentry(std::ostream&) () from /usr/lib/x86_64-linux-gnu/libstdc++.so.6
#1 0x00007ffff7b6dee9 in std::basic_ostream<char, std::char_traits<char> >& std::__ostream_insert<char, std::char_traits<char> >(std::basic_ostream<char, std::char_traits<char> >&, char const*, long) ()
from /usr/lib/x86_64-linux-gnu/libstdc++.so.6
#2 0x00007ffff7b6e2ef in std::basic_ostream<char, std::char_traits<char> >& std::operator<< <std::char_traits<char> >(std::basic_ostream<char, std::char_traits<char> >&, char const*) ()
from /usr/lib/x86_64-linux-gnu/libstdc++.so.6
#3 0x00000000004021be inTest::fill (this=0x6120f8, funcs=...) at inTest.cpp:92
最后一帧是我的代码.第 92 行简单地写着:
The last frame is my code. Line 92 simply reads:
std::cout << "Test";
推荐答案
正如 Luchian 所指出的，你不能在第一个之前使用 std::coutios_base::Init 的实例已构建.你不必然而，定义一个实例；包括 <iostream> 应该就够了.
As Luchian has pointed out, you cannot use std::cout before the first
instance of ios_base::Init has been constructed. You don't have to
define an instance, however; including <iostream> should be enough.
初始化顺序是在单个翻译单元中定义的.如果您在所有具有静态的文件的顶部包含 <iostream>实例，你应该没问题.如果静态对象的构造函数然而，在另一个翻译单元中调用一个函数，输出是在该翻译单元中，仅包含 <iostream> 是不够的仅在执行输出的翻译单元中.你必须包括它在定义静态变量的翻译单元中.甚至如果他们不做任何输出.
Order of initialization is defined within a single translation unit.
If you include <iostream> at the top of all files which have static
instances, you should be OK. If the constructor of a static object
calls a function in another translation unit, however, and the output is
in that translation unit, it is not sufficient to include <iostream>
only in the translation unit which does the output. You must include it
in the translation unit where the static variable(s) are defined. Even
if they don't do any output.
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