检查向量是否为空 [英] Checking whether a vector is empty

问题描述

假设我有一个 std::vector 说 Vector

现在对向量执行一些操作(插入或删除)后，我想检查向量是否为空，并在此基础上我想执行一些操作.

Now after performing some operations on the vector(either insertion or deletion) I want to check if the vector is empty and on the basis of that I want to perform some operations.

哪种方法更好

方法一

if (Vector.size() == 0){ /* operations */ }

方法二

if (Vector.empty()) { /* operations */ }

1 或 2 哪个更好?

推荐答案

v.size() == 0 说我正在比较大小"，但这样做是为了检查容器是否空的.在你知道它做什么之前，有一个小算法需要消化(非常小，因为它只包含一个比较).
OTOH，v.empty() 完全符合它所说的:它检查 v 是否为空.
由于这个原因，我显然更喜欢#2，因为它按照它所说的去做.毕竟，这就是发明 empty() 的原因.

v.size() == 0 says "I'm comparing the size", but does so to check whether the container empty. There's a small algorithm to digest (very small, as it only consists of a comparison) before you know what it does.
OTOH, v.empty() does exactly what it says: it checks whether v is empty.
Due to this, I clearly prefer #2, as it does what it says. That's why empty() was invented, after all.

但也有一个算法原因更喜欢 empty():如果有人后来将 std::vector 更改为 std::list, v.size() 可能有 O(n).(在 C++ 03 中，std::vector 保证为 O(1)，但对于 std::list 则不保证.根据 James 对 Prasoon 的答案所有em> C++1x 中的容器.)

But there's also an algorithmic reason to prefer empty(): If someone later changes std::vector into a std::list, v.size() might have O(n). (In C++ 03 it's guaranteed to be O(1) for std::vector, but not for std::list. According to James' comment to Prasoon's answer it will be O(1) for all containers in C++1x.)

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