C模板函数内的静态变量初始化

我注意到函数模板中静态变量初始化的一个奇怪的行为.请考虑以下示例：

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");
}

只要测试中的f是静态的,我希望createFile只被调用一次.但是,它被调用两次.

花了一些时间来解决问题,我注意到从测试中的参数中删除const引用修复了它.另一个有趣的事情是传递给函数的字符串的长度也会影响初始化：当参数的长度相等时,静态变量只初始化一次,否则会发生新的初始化.

有人可以解释一下吗？除了上面提到的解决方案/解决方案之外,我们非常欢迎.

解决方法

文字“一”是一个const char [4].

这段代码：

test("one")

理想情况下会调用test(const char(&)[4])

这适用于测试(const T&)(因为const char(&)[4]可以绑定到const char(const&)[4]).

但它无法用于测试(T t),因为您无法按值传递字符串文字.它们通过引用传递.

但是,const char [4]可以衰减为const char *,它可以匹配模板< class T> void func(T t).

证据在于布丁：

#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,&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");
}

预期产量：

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*

C模板函数内的静态变量初始化

解决方法

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