Question

can swap function be used to free memory as in case of vectors as mentioned below?

     std::vector<int> v1;

     // somehow increase capacity

    std::vector<int>().swap(v1);

Answer 1

我不敢肯定我会理解你的问题,但你问,能否以同样的方式 empty一 set?

如果是,那么:

Answer 2

关于病媒的<代码>swap的定义是交换两种病媒的内容和。

set doesn t have anything like capacity in its public interface, so it s not defined to swap capacity, just contents. There s no particular reason why a set implementation should over-allocate beyond what it needs. But if it does then there s no standard way to ensure that the "spare" memory is freed, since it s permitted to leave the spare memory where it is on a swap.

Answer 3

是的,这一骗局使你能够把病媒的能力降至零:

std::vector<int> intVec;
intVec.reserve(100);
std::vector<int>().swap(intVec);

or

确切计算要素

std::vector<int> intVec;    
intVec.reserve(100);
intVec.push_back(1);
std::cout << intVec.capacity() << std::endl;  // prints 100
std::vector<int>(intVec).swap(intVec);
std::cout << intVec.capacity() << std::endl;  // prints 1

for more on that look here: http://www.gotw.ca/gotw/054.htm

Answer 4

采用病媒的这种方式的理由是,如果你简单使用病媒代码< clear()或resize()方法,那么你的大小在逻辑上可能会被降到0,但所分配的记忆将留在那里重新使用。

如果你想要的是实际去除多余的记忆,那么你就会用空洞的病媒冲动,而这种媒介从来就从未分配过(尽管它可能分配一些)。

Doing the same with set may work too if set has allocated some "nodes" and decides to hold on to them for re-use although a set does not have a reserve() function so really it is up to the implementor of STL as to whether to release the memory or hold onto it.

Answer 5

由于<条码><>>/代码>不大可能保持超负荷能力,因此,互换成切碎。

此外,C++11在集装箱中添加了一个<代码>shrink_to_fit()的成员,如果可能有用的话。因此,仅仅利用这一点,如果无法得到,那就不担心。

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