结合函数参数包和默认参数 [英] Combining function parameter pack and default arguments

问题描述

我有一个带参数包的函数:

I have a function with a parameter pack:

template<typename... Targs>
void tprintf(const char* format, Targs... args) {}

(实现无关紧要，只是签名).我想使用 GCC/Clang 内置函数将源位置添加为默认参数.类似的东西

(the implementation shouldn't matter, just the signature). I want to add source position as default argument, using the GCC/Clang builtins. Something like

template<typename... Targs>
void tprintf(const char* format, Targs... args,
             const char* file = __builtin_FILE(),
             unsigned line = __builtin_LINE()) {}

这可以编译，但调用它并没有像我希望的那样将参数传递给 args；例如

This compiles, but calls to it are not passing parameters to args as I hoped; e.g.

tprintf("%d%s", 0, "a");

给出(在 Clang 10 上)

gives (on Clang 10)

<source>:7:5: error: no matching function for call to 'tprintf'
    tprintf("%d%s", 0, "a");    
    ^~~~~~~
<source>:2:6: note: candidate function template not viable: no known conversion from 'const char [2]' to 'unsigned int' for 3rd argument
void tprintf(const char* format, Targs... args,    
     ^

这似乎表明 args 是空的，0 是 file，而 "a" 是 行.

which seems to indicate args is empty, 0 is file, and "a" is line.

实际上，在编写问题时，我发现显式传递 Targs 有效:

Actually, while writing the question I've found that explicitly passing Targs works:

tprintf<int, char*>("%d%s", 0, "a");

是否可以避免这种情况?

Is it possible to avoid this?

推荐答案

解决方案是给我们 C++20 的 std::source_location:

The solution would be to us C++20's std::source_location:

#include <iostream>
#include <source_location>

template<typename... Targs, auto location = source_location::current()>
auto tprintf(char const* format, Targs const&... args) -> void {
    std::cout
        << std::format("{}:{}: ", location.file_name(), location.line())
        << std::format(format, args...);
}

auto main() -> int {
    tprintf("hello {}", "world"); // prints "example.cpp:12: hello world"
}

如果 C++20 不是一个选项(尤其是在当前，编译器不支持源位置)，那么有一种方法可以不用宏来实现.您可以简单地对编译器内置程序进行一些间接操作.

If C++20 is not an option (especially at this current time, compilers don't support source location) then there is a way to do it without macros. You can simply do a little indirection over the compiler built-ins.

不能将默认参数放在可变参数之后，但可以将它们放在第一个参数的构造函数中以达到相同的效果:

You cannot put defaulted arguments after the variadic arguments, but you can put them in the constructor of the first parameter to have the same effect:

#include <cstdio>

struct location_and_format {
    constexpr location_and_format(
        char const* _format,
        char const* _file_name = __builtin_FILE(),
        unsigned _line = __builtin_LINE()
    ) noexcept : format{_format}, file_name{_file_name}, line{_line} {}

    char const* format;
    char const* file_name;
    unsigned line;
};

template<typename... Targs>
void tprintf(location_and_format format, Targs... args) {
    printf("%s:%u: ", format.file_name, format.line);
    printf(format.format, args...);
}

int main() {
    tprintf("hello %s", "world"); // prints example.cpp:22: hello world
}

实例

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