Question

template<typename T>
void print_size(const T& x)
{
    std::cout << sizeof(x) << '\n';
}

int main()
{
    print_size("If you timidly approach C++ as just a better C or as an object-oriented language, you are going to miss the point.");
    // prints 115
}

这在最近的g ++编译器上打印115。显然，T被推断为数组（而不是指针）。标准是否保证了这种行为？我有点惊讶，因为以下代码打印了指针的大小，我认为auto的行为与模板参数推导完全相同？

int main()
{
    auto x = "If you timidly approach C++ as just a better C or as an object-oriented language, you are going to miss the point.";
    print_size(x);
    // prints 4
}

Answer 1

auto完全符合¹之类的模板参数推导。完全像T！

比较一下：

template<typename T>
void print_size(T x)
{
    std::cout << sizeof(x) << '\n';
}

int main()
{
    print_size("If you timidly approach C++ as just a better C or as an object-oriented language, you are going to miss the point.");
    // prints 4
    auto x = "If you timidly approach C++ as just a better C or as an object-oriented language, you are going to miss the point.";
    print_size(x);
    // prints 4
}

有了这个：

template<typename T>
void print_size(const T& x)
{
    std::cout << sizeof(x) << '\n';
}

int main()
{
    print_size("If you timidly approach C++ as just a better C or as an object-oriented language, you are going to miss the point.");
    // prints 115
    const auto& x = "If you timidly approach C++ as just a better C or as an object-oriented language, you are going to miss the point.";
    print_size(x);
    // prints 115
}

¹不完全，但这不是一个极端的案例。

字符串文字的模板参数推导

1 个答案: