C ++ 20是否要求将源代码存储在文件中？ [英] Does C++20 mandate source code being stored in files?
问题描述
但是，如果我没记错的话，一个稍微奇怪的问题是C ++源代码不需要文件系统来存储其文件。
A slightly strange question, however, if I remember correctly, C++ source code doesn't require a file system to store its files.
具有可扫描的编译器通过相机手写的纸将是一个符合要求的实现。
Having a compiler that scans handwritten papers via a camera would be a conforming implementation. Although practically not making that much sense.
但是C ++ 20现在使用 文件名 。
However C++20 now adds source location with file_name. Does this now imply that source code should always be stored in a file?
推荐答案
不，源代码不一定要存储在文件中吗？
您可以在管道内完全编译（并链接）C ++，例如将编译器置于中间。
No, source code doesn't have to come from a file (nor go to a file).
You can compile (and link) C++ completely within a pipe, putting your compiler in the middle, e.g.
generate_source | g++ -o- -xc++ - | do_something_with_the_binary
几十年来一直如此。另请参见：
and it's been like that for decades. See also:
- Is it possible to get GCC to read from a pipe?
- How to make GCC output to stdout?
std的引入： C ++ 20中的：source_location 不会更改此状态。只是有些代码没有明确定义的源位置（或者可能定义明确，但意义不大）。实际上，我想说的是坚持使用文件定义 std :: source_location 有点近视……尽管公平地说，它只不过是 __ FILE __ 和 __ LINE __ 在C ++（和C）中已经存在。
The introduction of std::source_location in C++20 doesn't change this state of affairs. It's just that some code will not have a well-defined source location (or it may be well-defined, but not very meaningful). Actually, I'd say that the insistence on defining std::source_location using files is a bit myopic... although in fairness, it's just a macro-less equivalent of __FILE__ and __LINE__ which already exist in C++ (and C).
@ HBv6指出，如果从标准输入流使用GCC进行编译时，如果打印 __ FILE __ 的值，则：
@HBv6 notes that if you print the value of __FILE__ when compiling using GCC from the standard input stream:
echo -e '#include <iostream>\n int main(){std::cout << __FILE__ ;}' | g++ -xc++ -
运行生成的可执行文件< stdin> 。
running the resulting executable prints <stdin>.
@Morwenn指出代码：
@Morwenn notes that this code:
#include <https://raw.githubusercontent.com/Morwenn/poplar-heap/master/poplar.h>
// Type your code here, or load an example.
void poplar_sort(int* data, size_t size) {
poplar::make_heap(data, data + size);
poplar::sort_heap(data, data + size);
}
可在GodBolt上运行（但无法在您的计算机上运行-没有流行的编译器支持此功能。）
works on GodBolt (but won't work on your machine - no popular compiler supports this.)
在语言标准中，对于C ++程序源是否需要来自文件的问题并没有明确回答。查看C ++ 17标准（n4713）的草稿，第5.1节[lex.separate]读取：
The question of whether C++ program sources need to come from files is not answered clearly in the language standard. Looking at a draft of the C++17 standard (n4713), section 5.1 [lex.separate] reads:
- 程序的文本以本文档中称为源文件的单元形式保存。通过预处理指令#include将源文件连同所有标头（20.5.1.2）和包含的源文件（19.2）一起，减去通过任何条件包含（19.1）预处理指令跳过的任何源行，称为转换单元。
因此，源代码本身不一定保存在文件中，而是保存在单元称为源文件。但是，包含项从何而来？有人会假设它们来自文件系统上的命名文件...但是也没有强制要求。
So, the source code is not necessarily kept in a file per se, but in a "unit called a source file". But then, where do the includes come from? One would assume they come from named files on the filesystem... but that too is not mandated.
无论如何， std :: source_location 似乎并没有改变C ++ 20中的措辞或影响其解释（AFAICT）。
At any rate, std::source_location does not seem to change this wording in C++20 or to affect its interpretation (AFAICT).
这篇关于C ++ 20是否要求将源代码存储在文件中？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
