[英]Is it possible to apply std::sort on std::unique<T[ ]>?
Suppose I have a dynamic array that I want to sort, I could do 假设我有一个我要排序的动态数组,我可以做
std::vector<int> v(100);
for (int i = 0; i < 100; i++) v[i] = rand();
std::sort(v.begin(), v.end());
but for performance critical code, the initialization overhead is unacceptable, more details at https://stackoverflow.com/a/7269088/3667089 但对于性能关键代码,初始化开销是不可接受的,更多细节请访问https://stackoverflow.com/a/7269088/3667089
I could also do 我也可以
int *v = new int[100];
for (int i = 0; i < 100; i++) v[i] = rand();
std::sort(v, v + 100);
but having to manage memory ourselves is bound to memory leak in large codebases. 但是必须自己管理内存必然会在大型代码库中发生内存泄漏。
So it seems that the most feasible approach is 所以似乎最可行的方法是
std::unique_ptr<int[]> v(new int[100]);
for (int i = 0; i < 100; i++) v[i] = rand();
std::sort(v, v + 100);
No initialization overhead nor need to worry about memory management, but this returns a long compilation error. 没有初始化开销也不需要担心内存管理,但这会返回一个很长的编译错误。 Could someone let me know what I am doing wrong?
有人能让我知道我做错了什么吗?
I am on Ubuntu 14.04, GCC as compiler. 我在Ubuntu 14.04,GCC作为编译器。
EDIT: Change the code so the data is not already sorted 编辑:更改代码,以便数据尚未排序
std::sort
still needs iterators, and unique_ptr
is not an iterator. std::sort
仍然需要迭代器,而unique_ptr
不是迭代器。 However, it does hold onto something that can be used as one: its pointer: 但是,它确实保留了可以用作一个的东西:它的指针:
std::sort(v.get(), v.get() + 100);
or 要么
std::sort(&*v, &*v + 100);
or 要么
std::sort(&v[0], &v[0] + 100); // N.B. &v[100] invokes undefined behavior
But what you really want is a vector
allocator that default-initializes instead of value-initializes. 但你真正想要的是一个
vector
分配器,它默认初始化而不是值初始化。 That's where the performance difference is coming from - using std::vector
with the default allocator will zero-initialize all your int
s first and then assign them some value, whereas your other options do not have this extra zero-initialization step. 这就是性能差异的来源 - 使用带有默认分配器的
std::vector
将首先对所有int
零初始化,然后为它们分配一些值,而其他选项则没有这个额外的零初始化步骤。
Check out Casey's implementation of such a thing and then just do: 查看Casey实现这样的事情,然后做:
std::vector<int, default_init_allocator<int>> v(100); // not zero-initialized
for (int i = 0; i < 100; i++) v[i] = i;
std::sort(v.begin(), v.end());
A different approach which is simpler in the sense that you don't have to deal with allocators (though more annoying on the code-wise) is to introduce a wrapper for int
for which value-initialization does not do anything: 一种不同的方法,在你不必处理分配器的意义上更简单(虽然在代码方面更烦人)是为
int
引入一个包装器,值初始化不会做任何事情:
template <class T>
struct Wrapper {
Wrapper() { }
T value;
};
std::vector<Wrapper<int>> v(100); // not zero-initialized
for (int i = 0; i < 100; i++) v[i].value = i; // but have to do this...
Note that simply using reserve()
and push_back()
is insufficient - that's still quite a bit more work that needs to be done than simply assigning by index after default-initialization, and if you're latency sensitive enough to have asked this question, that can be significant. 注意,简单地使用
reserve()
和push_back()
是不够的 - 除了在默认初始化之后简单地通过索引分配之外,还需要做更多的工作,并且如果你对延迟敏感到足以提出这个问题,这可能很重要。
Reading the link from the question, it seems you'd be happy using a vector
if it didn't call an unnecessary constructor for every element. 从问题中读取链接,如果它没有为每个元素调用不必要的构造函数,那么使用
vector
似乎很高兴。 There are techniques to eliminate this overhead. 有一些技术可以消除这种开销。
std::vector<int> v;
v.reserve(100);
for (int i = 0; i < 100; i++) v.emplace_back(rand());
std::sort(v.begin(), v.end());
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