不使用宏确定呼叫线路

发布于 2024-09-29 09:59:07 字数 547 浏览 3 评论 0原文

是否可以在不借助宏的情况下确定调用函数的行号？

考虑这段代码：

#include <iostream>

#define PrintLineWithMacro() \
  std::cout << "Line: " << __LINE__ << std::endl;   // Line 4

void PrintLine()
{
  std::cout << "Line: " << __LINE__ << std::endl;   // Line 8
}

int main(int argc, char **argv)
{
  PrintLine();           // Line 13
  PrintLineWithMacro();  // Line 14
  return 0;
}

它输出以下内容：

Line: 8
Line: 14

我理解为什么每个人都会打印他们所做的事情。我更感兴趣是否可以在不使用宏的情况下模拟宏功能。

原文

Is it possible to determine the line number that calls a function without the aid of a macro?

Consider this code:

#include <iostream>

#define PrintLineWithMacro() \
  std::cout << "Line: " << __LINE__ << std::endl;   // Line 4

void PrintLine()
{
  std::cout << "Line: " << __LINE__ << std::endl;   // Line 8
}

int main(int argc, char **argv)
{
  PrintLine();           // Line 13
  PrintLineWithMacro();  // Line 14
  return 0;
}

which outputs the following:

Line: 8
Line: 14

I understand why each prints what they do. I am more interested if it's possible to mimic the macro function without using a macro.

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霊感 2024-10-06 09:59:07

我会执行以下操作：

#define PrintLine() PrintLine_(__LINE__)

void PrintLine_(int line) {
    std::cout << "Line: " << line << std::endl;
}

我知道这并没有完全删除预处理器，但它确实将大部分逻辑移至实际函数中。

I would do the following:

#define PrintLine() PrintLine_(__LINE__)

void PrintLine_(int line) {
    std::cout << "Line: " << line << std::endl;
}

I know that this doesn't completely remove the preprocessor, but it does move most of the logic into an actual function.

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故人的歌 2024-10-06 09:59:07

C++在c++20标准中已经赶上了这个问题。

在 c++20 中，您现在可以这样做：

#include <iostream>
#include <source_location>

void PrintLine(std::source_location location = std::source_location::current())
{
    std::cout << "Line: " << location.line() << std::endl;   // Line 8
}

int main(int argc, char **argv)
{
    PrintLine();           // Line: 11
    PrintLine();           // Line: 12
    return 0;
}

另一个问题是您感兴趣的编译器是否已经赶上了标准。
请注意，该构造在 c++20 之前就已存在，您可以尝试：

#include <experimental/source_location>

C++ has caught up with this question in the c++20 standard.

In c++20 you may now do:

#include <iostream>
#include <source_location>

void PrintLine(std::source_location location = std::source_location::current())
{
    std::cout << "Line: " << location.line() << std::endl;   // Line 8
}

int main(int argc, char **argv)
{
    PrintLine();           // Line: 11
    PrintLine();           // Line: 12
    return 0;
}

Another question is whether the compiler of your interest has caught up with the standard.
Note that the construct was present pre c++20 and you could try: