[英]Why doesn't my change to clog stick?
I think I'm failing to understand some finer point of C++. 我想我无法理解C ++的一些优点。 I want to set up a log of what my program does, and discovered
std::clog
, which seems to do what I want in theory, but in practice it doesn't. 我想建立一个程序执行情况的日志,并发现
std::clog
,它在理论上似乎std::clog
我的要求,但实际上却没有。
If I do the following, clog works as expected and writes "Test 1" to the screen, and "Test 2" shows up in a file: 如果执行以下操作,则阻塞将按预期方式工作并将“ Test 1”写入屏幕,并且在文件中显示“ Test 2”:
int main ()
{
clog << "Test 1" << endl;
streambuf* original_buffer = clog.rdbuf (ofstream ("test.log").rdbuf ()));
clog << "test 2" << endl;
clog.rdbuf (original_buffer);
return 0;
}
But if I put all that into a class as such, then "Test 1" is written to the screen, test.log is created, but there's nothing inside and "Test 2" is no where to be found!: 但是,如果我将所有内容放到这样的类中,则将“ Test 1”写入屏幕,创建了test.log,但是里面什么也没有,并且“ Test 2”也找不到了!:
class IerrLog
{
std::streambuf * original_buffer;
public:
IerrLog ()
{
std::ofstream logFile ("test.log");
original_buffer = std::clog.rdbuf (logFile.rdbuf ());
}
~IerrLog ()
{
std::clog.rdbuf (original_buffer);
}
};
int main () {
clog << "Test 1" << endl;
IerrLog someLog ();
clog << "Test 2" << endl;
return 0;
}
What am I missing? 我想念什么?
EDIT: If I run the latter in valgrind, I get errors like this (the former runs clean): 编辑:如果我在valgrind中运行后者,我会得到这样的错误(前者运行干净):
Invalid read of size 8
at 0x39598993E5: std::ostream::flush() (in /usr/lib/libstdc++.so.6.0.10)
by 0x395989B5F2: std::basic_ostream<char, std::char_traits<char> >& std::endl<char, std::char_traits<char> >(std::basic_ostream<char, std::char_traits<char> >&) (in /usr/lib/libstdc++.so.6.0.10)
by 0x400F8E: main (main.cc:23)
Address 0x7ff0006c8 is just below the stack ptr. To suppress, use: --workaround-gcc296-bugs=yes
I'm not obnoxious enough to think that I (a lowly common programmer) found a compiler bug with such a simple program, but this makes me even more confused and valgrind obviously finds that the latter is somehow wrong, even though I tried to make them functionally identical. 我还不足为奇,以为我(一个不太普通的程序员)发现了一个如此简单的程序的编译器错误,但是这使我更加困惑,而valgrind显然发现后者是某种程度上的错误,即使我试图这样做也是如此。它们在功能上是相同的。
I assume you want to create a stack variable of IerrLog . 我假设您要创建IerrLog的堆栈变量。 You need to change
你需要改变
IerrLog someLog ();
to 至
IerrLog someLog;
Your original statement will be interpreted by the compiler as a declaration of function someLog() which takes no arguments and returns an IerrLog. 您的原始语句将由编译器解释为函数someLog()的声明,该声明不带任何参数并返回IerrLog。
You should also create your file as a member variable and not on the stack. 您还应该将文件创建为成员变量,而不是在堆栈上。
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