模板函数内的 C++ 静态变量初始化

Question

我注意到函数模板中静态变量初始化的一个非常奇怪的行为。考虑以下示例:

MyFile * createFile()
{
    std::cout << "createFile" << std::endl;
    return nullptr;
}

template <typename T>
void test(const T& t)
//void test(T t)
{
    static MyFile *f = createFile();
}

void main()
{
    test("one");
    //test("two");
    test("three");
}

只要 test 中的 f 是静态的，我就希望 createFile 只被调用一次。但是，它被调用了两次。

在花了一些时间解决这个问题后，我注意到从 test 中的参数中删除 const 引用可以修复它。另一个有趣的是，传递给函数的字符串的长度也会影响初始化:当参数长度相等时，静态变量只初始化一次，否则，将进行新的初始化。

谁能解释一下？非常欢迎除上述之外的解决方案/解决方法。

Answer 1

文字“一”是一个const char [4] .

这段代码:

test("one")

理想情况下想调用test(const char (&)[4])

这适用于 test(const T&) (因为 const char (&) [4] 可以绑定(bind)到 const char (const&) [4] )。

但它不适用于 test(T t)因为您不能按值传递字符串文字。它们通过引用传递。

然而，const char[4]可以衰减到const char* , 可以匹配 template<class T> void func(T t) .

证据就在布丁里:

#include <cstdint>
#include <iostream>
#include <typeinfo>

template <typename T, std::size_t N>
void test_const(const T(&t)[N])
{
    std::cout << __func__ << " for literal " << t << " T is a " << typeid(T).name() << " and N is " << N << std::endl;
}

template <typename T>
void test_mutable(T &t)
{
    std::cout << __func__ << " for literal " << t << " T is a " << typeid(T).name() << std::endl;
}

template <typename T>
void test_const_ref(const T &t)
{
    std::cout << __func__ << " for literal " << t << " T is a " << typeid(T).name() << std::endl;
}

template <typename T>
void test_copy(T t)
{
    std::cout << __func__ << " for literal " << t << " T is a " << typeid(T).name() << std::endl;
}

int main()
{
    test_const("one");
    test_const("three");
    test_mutable("one");
    test_mutable("three");
    test_const_ref("one");
    test_const_ref("three");
    test_copy("one");
    test_copy("three");
}

示例结果(clang):

test_const for literal one T is a c and N is 4
test_const for literal three T is a c and N is 6
test_mutable for literal one T is a A4_c
test_mutable for literal three T is a A6_c
test_const_ref for literal one T is a A4_c
test_const_ref for literal three T is a A6_c
test_copy for literal one T is a PKc
test_copy for literal three T is a PKc

这里是一个带有 demangled 名称的版本(将在 clang 和 gcc 上编译):

#include <cstdint>
#include <iostream>
#include <typeinfo>
#include <cstdlib>
#include <cxxabi.h>

std::string demangle(const char* name)
{
    int status = -1;
    // enable c++11 by passing the flag -std=c++11 to g++
    std::unique_ptr<char, void(*)(void*)> res {
        abi::__cxa_demangle(name, NULL, NULL, &status),
        std::free
    };

    return (status==0) ? res.get() : name ;
}

template <typename T, std::size_t N>
void test_const(const T(&t)[N])
{
    std::cout << __func__ << " for literal " << t << " T is a " << demangle(typeid(T).name()) << " and N is " << N << std::endl;
}

template <typename T>
void test_mutable(T &t)
{
    std::cout << __func__ << " for literal " << t << " T is a " << demangle(typeid(T).name()) << std::endl;
}

template <typename T>
void test_const_ref(const T &t)
{
    std::cout << __func__ << " for literal " << t << " T is a " << demangle(typeid(T).name()) << std::endl;
}

template <typename T>
void test_copy(T t)
{
    std::cout << __func__ << " for literal " << t << " T is a " << demangle(typeid(T).name()) << std::endl;
}

int main()
{
    test_const("one");
    test_const("three");
    test_mutable("one");
    test_mutable("three");
    test_const_ref("one");
    test_const_ref("three");
    test_copy("one");
    test_copy("three");
}

预期输出:

test_const for literal one T is a char and N is 4
test_const for literal three T is a char and N is 6
test_mutable for literal one T is a char [4]
test_mutable for literal three T is a char [6]
test_const_ref for literal one T is a char [4]
test_const_ref for literal three T is a char [6]
test_copy for literal one T is a char const*
test_copy for literal three T is a char const*