英文:
How to Return a Writer and a Reader from a function in zig?
问题
我正在创建一个简单的程序,它从文件中读取数据然后处理该文件。
我看到了一个读取文件的模式,使用了 std.io.bufferedReader
和 std.io.fixedBufferStream().writer()
所以我正在尝试创建一个函数来返回Writer和Reader两者。
为了做到这一点,我试图创建一个结构体,其中字段的类型是 std.io.Writer
和 std.io.Reader
,但它们是返回类型而不是实际类型。
那么,我该如何创建它?
我的当前代码如下:
const std = @import("std");
const BufferedWriterAndReader = struct {
writer: std.io.Writer,
reader: std.io.Reader,
};
fn getBufferedWriterAndReader(path: []u8) BufferedWriterAndReader {
var file = try std.fs.cwd().openFile(path, .{});
defer file.close();
var buf_reader = std.io.bufferedReader(file.reader());
var file_reader = buf_reader.reader();
var lines: [256]u8 = undefined;
var lines_buf = std.io.fixedBufferStream(&lines);
var lines_writer = lines_buf.writer();
return BufferedWriterAndReader{
.writer = lines_writer,
.reader = file_reader,
};
}
pub fn main() !void {
getBufferedWriterAndReader("strategy.txt");
}
(Note: I've removed HTML escape codes for special characters to provide a cleaner translation.)
英文:
I'm creating a simple program that reads from a file and then process that file.
I saw a pattern for reading using a std.io.bufferedReader
and std.io.fixedBufferStream().writer()
So I'm trying to create a function to return both Writer and Reader.
For doing such thing I'm trying to create a struct which the fields are from type std.io.Writer
and std.io.Reader
but they are functions that return the type and not an actual type.
So, how can I create it?
My current looks like:
const std = @import("std");
const BufferedWriterAndReader = struct {
writer: std.io.Writer,
reader: std.io.Reader,
};
fn getBufferedWriterAndReader(path: []u8) BufferedWriterAndReader {
var file = try std.fs.cwd().openFile(path, .{});
defer file.close();
var buf_reader = std.io.bufferedReader(file.reader());
var file_reader = buf_reader.reader();
var lines: [256]u8 = undefined;
var lines_buf = std.io.fixedBufferStream(&lines);
var lines_writer = lines_buf.writer();
return BufferedWriterAndReader{
.writer = lines_writer,
.reader = file_reader,
};
}
pub fn main() !void {
getBufferedWriterAndReader("strategy.txt");
}
答案1
得分: 1
你在这里尝试的做法由于不同的原因而行不通:
var file = try std.fs.cwd().openFile(path, .{});
defer file.close();
你在函数结束时关闭了文件,因此读取/写入文件的读者/写者将无法继续操作文件。
buf_reader
和 lines_buf
也位于栈上或使用栈内存,因此它们在函数调用结束时会无效。
所以我建议你退一步,重新思考你实际想要在这里做什么。例如,如果你只想要一个简单的解决方案,不关心性能,可以将整个文件读取到堆上(file.readToEndAlloc()
)。
现在来实际回答你的问题:
你可以在编译时检查值的类型。因此,你可以这样查询确切的类型:
@compileLog(@TypeOf(lines_writer));
@compileLog(@TypeOf(file_reader));
(现在这些的结果看起来很复杂。鉴于 Zig 没有内置继承,没有简化这些类型的简单方法。)
英文:
What you are trying to do here won't work for different reasons:
var file = try std.fs.cwd().openFile(path, .{});
defer file.close();
You are closing the file at the end of the function. Therefor the reader/writer won't be able to read/write to the file anymore.
The buf_reader
, lines_buf
also sit on the stack/use stack memory, so they get invalidated at the end of the function call.
So I would suggest that you take a step back and rethink what you actually want to do here. For example if you just want a simple solution and don't care about performance, just read the entire file onto the heap (file.readToEndAlloc()
)
Now to actually answer your question:
You can check out the type of a value at compile-time. So you could just do this to query the exact type:
@compileLog(@TypeOf(lines_writer));
@compileLog(@TypeOf(file_reader));
(Now the results of that will look complicated. And there is no easy way to simplify the types given that zig has no builtin inheritance)
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