C++ 传递对 std::ifstream 的引用作为参数
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Pass a reference to std::ifstream as parameter
提问by Newbie
I'm trying to write a function with an ifstream&argument.
我正在尝试编写一个带ifstream&参数的函数。
void word_transform(ifstream & infile)
{
infile("content.txt");
//etc
}
which gave me an error:
这给了我一个错误：
Type 'ifstream' (aka 'basic_ifstream ') does not provide a call operator.
类型“ifstream”（又名“basic_ifstream”）不提供调用运算符。
Can you please me what's wrong?
你能帮我看看有什么问题吗？
回答by jpo38
call operatoris a function like operator()( params )allowing to use the syntax myObject( params ).
调用运算符是一个函数，类似于operator()( params )允许使用语法myObject( params )。
So, when you write infile(...), you are trying to us a call operator.
因此，当您编写 时infile(...)，您是在尝试向我们呼叫接线员。
What you are trying to do is to open a file, use the openmethod:
您要做的是打开一个文件，使用以下open方法：
void word_transform(ifstream & infile)
{
infile.open("content.txt",std::ios_base::in);
if ( infile.is_open() )
infile << "hello";
infile.close();
}
But, as commented, it does not really make sense to pass infile reference to such a function. You may consider:
但是，正如所评论的，将 infile 引用传递给这样的函数并没有真正意义。你可以考虑：
void word_transform(istream& infile)
{
infile << "hello";
}
int main()
{
ifstream infile;
infile.open("content.txt",std::ios_base::in);
if ( infile.is_open() )
word_transform( infile );
infile.close();
return 0;
}
Or:
或者：
void word_transform()
{
ifstream infile;
infile.open("content.txt",std::ios_base::in);
if ( infile.is_open() )
infile << "hello";
infile.close();
}
int main()
{
word_transform();
return 0;
}
回答by nvoigt
You attempt to call operator()on your parameter. That will not work. Are you trying to open a file? If you get an ifstreamas parameter, it should be open from the start because you opened it outside your function. Passing a stream and then opening it inside your function does not make sense.
您尝试调用operator()您的参数。那不管用。你想打开一个文件吗？如果你得到一个ifstreamas 参数，它应该从一开始就打开，因为你在函数之外打开了它。传递一个流然后在你的函数中打开它是没有意义的。
void word_transform(std::ifstream& infile)
{
// read something from infile
}
int main()
{
std::ifstream file("content.txt");
// error checks
word_transform(file);
return 0;
}
回答by Codor
To my understanding, the call operator is operator(), which apparently is not defined for ifstream. You have to do something different with the argument of word_transform.
据我了解，调用运算符是operator()，显然没有为ifstream. 你必须对 的参数做一些不同的事情word_transform。
回答by ravi
infile("content.txt");
Note that this would try to call operator() on already created object of type infile. As no such operator exists from ifstream , you got an error.
请注意，这将尝试在已创建的 infile 类型对象上调用 operator()。由于 ifstream 中不存在这样的运算符，因此出现错误。
Rather you should do:-
相反，您应该这样做：-
infile.open("content.txt");
回答by Walter
Typically, (references to) streams are passed to functions for I/O. This means the function should take
std::istreamor std::ostreamargument, but not std::ifstreamor std::ofstream. This way your function can be used with any stream object derived from std::istream, including std::cin, std::istrstream, and std::ifstream.
通常，（对）流的引用被传递给 I/O 函数。这意味着函数应该采用
std::istreamorstd::ostream参数，而不是std::ifstreamor std::ofstream。这样，你的函数可以用任何流对象使用源自std::istream包括std::cin，std::istrstream，和std::ifstream。
As nvoigt said, passing an std::ifstreamto a function for opening makes little sense. It is definitely not clear to the caller that its stream may be closed and opened to another file.
正如 nvoigt 所说，将 an 传递std::ifstream给打开的函数毫无意义。调用者肯定不清楚它的流可能被关闭并打开到另一个文件。
