将cout分配给变量名 | 将cout分配给变量名

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问题描述

在ANSI C ++中，如何将cout流分配给变量名？我想做的是，如果用户指定了输出文件名，我在那里发送输出，否则，将其发送到屏幕。例如：

In ANSI C++, how can I assign the cout stream to a variable name? What I want to do is, if the user has specified an output file name, I send output there, otherwise, send it to the screen. So something like:

ofstream outFile;
if (outFileRequested)
    outFile.open("foo.txt", ios::out);
else
    outFile = cout;  // Will not compile because outFile does not have an
                     // assignment operator

outFile << "whatever" << endl;

我试过这样做一个宏函数：

I tried doing this as a Macro function as well:

#define OUTPUT outFileRequested?outFile:cout

OUTPUT << "whatever" << endl;

但是这给我一个编译器错误。

But that gave me a compiler error as well.

我想我可以为每个输出使用一个IF-THEN块，但我想避免，如果我可以。有任何想法吗？

I supposed I could either use an IF-THEN block for every output, but I'd like to avoid that if I could. Any ideas?

推荐答案

使用参考。请注意，引用必须是 std :: ostream 类型，而不是 std :: ofstream ，因为 std :: cout 是 std :: ostream ，因此必须使用最小公分母。

Use a reference. Note that the reference must be of type std::ostream, not std::ofstream, since std::cout is an std::ostream, so you must use the least common denominator.

std::ofstream realOutFile;

if(outFileRequested)
    realOutFile.open("foo.txt", std::ios::out);

std::ostream & outFile = (outFileRequested ? realOutFile : std::cout);

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