[英]Print a #define macro using std::cout
I am trying to do this我正在尝试这样做
#define _TEST_ test
#include <iostream>
int main()
{
std::cout << "_TEST_" << std::endl;
}
As far as my understanding, I expect this output.据我了解,我希望这个 output。
test
However, the output I get is但是,我得到的 output 是
_TEST_
Why am I doing wrong here?为什么我在这里做错了?
Macro expansion in the C/C++ preprocessor only happens to tokens. C/C++ 预处理器中的宏扩展只发生在标记上。 Variables names, for instance, are tokens.
例如,变量名称是标记。 But the inside of a string is not a token;
但是字符串内部不是标记; it's a part of a larger token (namely, the string literal itself).
它是更大标记的一部分(即字符串文字本身)。
If you want the macro to expand to something within quotation marks, you need to use stringification .如果你想让宏扩展到引号内的东西,你需要使用stringification 。
#define xstr(x) str(x)
#define str(x) #x
#define _TEST_ test
#include <iostream>
int main()
{
std::cout << xstr(_TEST_) << std::endl;
}
You can read the above link for why we need those extra two layers of indirection ( xstr
and str
), but the basic idea is that #
itself doesn't do macro expansion, so by calling xstr
, we force a macro expansion of the argument ( _TEST_
into test
, namely), and then separately we call str
to stringify that.您可以阅读上面的链接了解为什么我们需要那些额外的两层间接(
xstr
和str
),但基本思想是#
本身不进行宏扩展,因此通过调用xstr
,我们强制对参数进行宏扩展( _TEST_
进入test
,即),然后我们分别调用str
对其进行字符串化。 If we had just called str
directly, it would see #_TEST_
and not perform macro expansion.如果我们只是直接调用
str
,它会看到#_TEST_
而不会执行宏扩展。
"_TEST_"
is a string literal and not a macro. "_TEST_"
是字符串文字而不是宏。 So no macro replacement will be done due to "_TEST_"
.因此,由于
"_TEST_"
,将不会进行任何宏替换。 To achieve your expected output you need to remove the surrounding double quotes and also change the macro to as shown below要实现预期的 output,您需要删除周围的双引号并将宏更改为如下所示
//-------------vvvvvv--->double quotes added here
#define _TEST_ "test"
#include <iostream>
int main()
{
//-------------------vvvvvv---------------> not withing quotes
std::cout << _TEST_ << std::endl;
}
The output of the above modified program is:上面修改后的程序的output为:
test
In the modified program, the macro _TEST_
stands for the string literal "test"
.在修改后的程序中,宏
_TEST_
代表字符串文字"test"
。 And thus when we use that macro in the statement std::cout << _TEST_ << std::endl;
因此,当我们在语句
std::cout << _TEST_ << std::endl;
中使用该宏时, it will be replaced by the string literal, producing the expected output. ,它将被字符串文字替换,产生预期的 output。
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