c++ - 将带有捕获的 lambda 传递给遗留回调

Question

我在 C++11 项目中使用 C 库，这个 C 库提供了一个需要函数指针的函数。我想将一个 C++11 lambda 传递给它，它可以正常工作，除非我捕获变量。这是一个简短的例子:

#include <cstdio>
#include <functional>

typedef int (*Callback)();

void legacy(Callback callback) {
    printf("%i\n", callback());
}

int stdCallback() {    
    return 1; 
}

int main(int argc, char* argv[]) {
    int number = 3;    

    // Standard C callback works
    legacy(stdCallback);

    // Lambda without capturing works
    legacy([]() { return 2; });    

    // Lambda with capturing doesn't work
    legacy([&]() { return number; });

    return 0;
}

GNU C++ 编译器在第三次调用 legacy 函数时给出以下错误消息:

test.cpp: In function ‘int main(int, char**)’:
test.cpp:24:36: error: cannot convert ‘main(int, char**)::<lambda()>’ to ‘Callback {aka int (*)()}’ for argument ‘1’ to ‘void legacy(Callback)’
 legacy([&]() { return number; });

我该如何解决这个问题？或者在技术上不可能将捕获的 lambda 用作 C 函数指针？

Answer 1

不，如果 lambda 捕获任何内容，则不能将其转换为函数指针。

C++ 标准，第 5.1.2/6 节:[expr.prim.lambda]，强调我的:

The closure type for a non-generic lambda-expression with no lambda-capture has a public non-virtual non-explicit const conversion function to pointer to function with C ++ language linkage (7.5) having the same parameter and return types as the closure type’s function call operator. The value returned by this conversion function shall be the address of a function that, when invoked, has the same effect as invoking the closure type’s function call operator