有人可以解释这是如何工作的：const print = @import("std").debug.print;

Question

How can this const print variable behave like a function?

const print = @import("std").debug.print;
print("hello, world", .{});

I understand you can assign expressions to variables. But this seems to behave like a precompiler macro in c/c++, I wouldn't have guessed that. Is it because "all variables declared in a comptime expression are implicitly comptime variables" and @ makes it a comptime expression, so it is evaluated before compilation, much like a macro would in c? Could someone elaborate a bit? This seems a very powerful feature.

Answer 1

@ does not indicate a comptime expression; rather @ prefixes built-in functions in Zig. @import returns a struct that provides access to the declarations made public by the imported file.

The expression @import("std").debug.print evaluates to the print function defined in the standard library file std/debug.zig ; it is not the expression that is assigned to print in the posted code, but the function . That is, the posted code works because print in OP code is actually a function. This can be seen by running the code below:

const print = @import("std").debug.print;

pub fn main() void {
    print("@TypeOf print: {}\n", .{ @TypeOf(print) });
}

Results:

$ zig run print_type.zig 
@TypeOf print: fn(comptime []const u8, anytype) void

Function Assignment

OP has asked for another example of assigning a function that is not imported to an identifier:

const print = @import("std").debug.print;

fn my_function(msg: []const u8) void {
    print("{s}\n", .{ msg });
}

const my_function_alias = my_function;

pub fn main() void {
    const another_function_alias = my_function;
    const yet_another_function_alias = my_function_alias;

    my_function("mf");
    my_function_alias("mfa");
    another_function_alias("afa");
    yet_another_function_alias("yafa");
}

Program output:

$ zig run function_assignment.zig 
mf
mfa
afa
yafa