a puzzle: "decltype definition"

原创于 2013-12-19 12:27:02 发布 · 446 阅读

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本文通过两个示例对比了C++中使用decltype初始化不同类型的变量时的区别。重点介绍了常量整型与引用类型在内存地址上的差异，并强调了在使用引用类型时必须进行初始化的重要性。

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Today, I>#include <iostream> int main() { const int ci = 0, &cj = ci; decltype(ci) x = 0; decltype(ci) y = x; std::cout << &x << "," << &y << std::endl; return 0; }

#include <iostream>
int main()
{
const int ci = 0, &cj = ci;
decltype(cj) x = 0;
decltype(cj) y = x;
std::cout << &x << "," << &y << std::endl;
return 0;
}

The results are different. Why? You should made your own conclusion? Thinking it right now..

Conclusion:

the first program: The type of "x" is int-type, the same as the "y".. their addresses are different.

The second one: the type of "x" is a reference-type... the initialization value is 0.. the second is the same type as "x", but it point to the address of "x". so they have the same address.. Remeber the reference-type needs the initialization before using. It's necessary to know that... Don't forget that..