提升STL使用效率的50个具体方法

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"Effective STL: 50 Specific Ways to Improve Your Use of Standard Template Library" 这篇资源是计算机科学的经典著作，专注于提升使用STL（Standard Template Library，标准模板库）的效率和质量。STL是C++编程语言中的一个核心组件，包含了一系列容器、迭代器、算法和函数对象，为程序员提供了高效的数据结构和操作。 1. 选择合适的容器：不同的容器有不同的性能特性和用途，如vector适合动态数组，list适合链表操作，map和set则用于关联容器。选择时需考虑插入、删除、查找等操作的需求。 2. 避免对容器独立代码的幻觉：虽然STL设计目标是可移植，但容器的实现可能因编译器和库而异，编写依赖特定容器特性的代码可能导致意外行为。 3. 优化容器内对象的复制：确保复制构造函数和赋值运算符有效且高效，特别是在容器中复制大量对象时。 4. 使用empty而非size()检查空容器：empty方法更直接地表明容器是否为空，避免了与零值比较的潜在错误。 5. 优先使用范围成员函数：例如，用`for_each`代替单独的元素操作，这样可以提高效率并减少出错的可能性。 6. 注意C++的最烦人解析（Most Vexing Parse）：某些语句可能被解析为函数声明而非对象初始化，需小心处理。 7. 在使用动态分配指针的容器时，记得在容器销毁前删除指针：避免内存泄漏，确保正确管理资源。 8. 避免使用auto_ptr在容器中：auto_ptr不适用于容器，因为其所有权转移特性可能导致意外行为，推荐使用智能指针如unique_ptr或shared_ptr。 9. 慎重选择删除选项：在容器中删除元素时，了解不同方法（如erase，clear）的差异和影响。 10. 了解allocator（分配器）的约定和限制：分配器控制对象的内存分配，不同的容器可能有不同的默认分配器，自定义分配器需谨慎使用。 11. 对定制分配器有合理的期望：它们可以优化内存管理，但不是所有场景下都适用，可能引入复杂性。 12. 关于STL容器的线程安全，保持现实的期待：大部分STL容器不是线程安全的，需要自己添加同步机制。 13. 优先使用vector和string：它们提供了高效的动态数组功能，优于手动管理的动态数组。 14. 使用reserve避免不必要的重新分配：在预知容量需求时，通过reserve预留空间，减少因插入导致的内部数据移动。 15. 注意字符串实现的差异：不同的编译器和库可能有不同的字符串实现，可能影响性能和行为。 16. 知道如何将vector和string数据传递给旧版API：旧API可能需要原始指针，需要转换或复制数据。 17. 利用“swap技巧”修剪多余容量：交换两个容器或对象的容量，可以使一个容器释放其不再需要的内存。这些知识点提供了对STL深入理解和有效使用的指南，帮助程序员写出更高效、更可靠的C++代码。

Compared to arrays. STL containers are much more civilized. They create (by

copying) only as many objects as you ask for, they do it only when you direct them to,

and they use a default constructor only when you say they should. Yes, STL containers

make copies, and yes, you need to understand that, but don't lose sight of the fact that

they're still a big step up from arrays.

Item 4. Call empty instead of checking size() against zero.

For any container c, writing

if (c.size() == 0)...

is essentially equivalent to writing

if (c.empty())...

That being the case, you might wonder why one construct should be preferred to the

other, especially in view of the fact that empty is typically implemented as an inline

function that simply returns whether size returns 0.

You should prefer the construct using empty, and the reason is simple: empty is a

constant-time operation for all standard containers, but for some list implementations,

size takes linear time.

But what makes list so troublesome? Why can't it, too. offer a constant-time size? The

answer has much to do with list's unique splicing functions. Consider this code:

list<int> list1;

list<int> Iist2;

...

list1.splice( // move ail nodes in Iist2

list1.end(), Iist2, // from the first occurrence

find(list2.begin(), Iist2.end(), 5), // of 5 through the last

find(list2.rbegin(), Iist2.rend(), 10).base() // occurrence of 10 to the

); //end of list1. See Item 28

// for info on the "base()" call

This code won't work unless Iist2 contains a 10 somewhere beyond a 5, but let's

assume that's not a problem. Instead, let's focus on this question: how many elements

are in list1 after the splice? Clearly, list1 after the splice has as many elements as it did

before the splice plus however many elements were spliced into it. But how many

elements were spliced into it? As many as were in the range defined by

find(list2.begin(), Iist2.end(), 5) and find(list2.rbegin(), Iist2.rend(), 10).base(). Okay,

how many is that? Without traversing the range and counting them, there's no way to

know. And therein lies the problem.

Suppose you're responsible for implementing list, list isn't just any container, it's a

standard container, so you know your class will be widely used. You naturally want

your implementation to be as efficient as possible. You figure that clients will

commonly want to find out how many elements are in a list, so you'd like to make size

a constant- time operation. You'd thus like to design list so it always knows how many

elements it contains.

At the same time, you know that of all the standard containers, only list offers the

ability to splice elements from one place to another without copying any data. You

reason that many list clients will choose list specifically because it offers high-

efficiency splicing. They know that splicing a range from one list to another can be

accomplished in constant time, and you know that they know it, so you certainly want

to meet their expectation that splice is a constant-time member function.

This puts you in a quandary. If size is to be a constant-time operation, each list

member function must update the sizes of the lists on which it operates. That includes

splice. But the only way for splice to update the sizes of the lists it modifies is for it to

count the number of elements being spliced, and doing that would prevent splice from

achieving the constant-time performance you want for it. If you eliminate the

requirement that splice update the sizes of the lists it's modifying, splice can be made

constant-time, but then size becomes a linear-time operation. In general, it will have to

traverse its entire data structure to see how many elements it contains. No matter how

you look at it, something — size or splice — has to give. One or the other can be a

constant-time operation, but not both.

Different list implementations resolve this conflict in different ways, depending on

whether their authors choose to maximize the efficiency of size or splice. If you

happen to be using a list implementation where a constant-time splice was given

higher priority than a constant-time size, you'll be better off calling empty than size,

because empty is always a constant-time operation. Even if you're not using such an

implementation, you might find yourself using such an implementation in the future.

For example, you might port your code to a different platform where a different

implementation of the STL is available, or you might just decide to switch to a

different STL implementation for your current platform.

No matter what happens, you can't go wrong if you call empty instead of checking to

see if size() == 0. So call empty whenever you need to know whether a container has

zero elements.

Item 5. Prefer range member functions to their single-element

counterparts.

Quick! Given two vectors, v1 and v2, what's the easiest way to make v1’s contents be

the same as the second half of v2's? Don't agonize over the definition of "half when v2

has an odd number of elements, just do something reasonable.

Time's up! If your answer was

v1.assign(v2.begin() + v2.size() /2, v2.end());

or something quite similar, you get full credit and a gold star. If your answer involved

more than one function call, but didn't use any kind of loop, you get nearly full credit,

but no gold star. If your answer involved a loop, you've got some room for

improvement, and if your answer involved multiple loops, well, let's just say that you

really need this book.

By the way, if your response to the answer to the question included "Huh?", pay close

attention, because you're going to learn something really useful.

This quiz is designed to do two things. First, it affords me an opportunity to remind

you of the existence of the assign member function, a convenient beast that too many

programmers overlook. It's available for all the standard sequence containers (vector,

string, deque, and list). Whenever you have to completely replace the contents of a

container, you should think of assignment. If you're just copying one container to

another of the same type, operator= is the assignment function of choice, but as this

example demonstrates, assign is available for the times when you want to give a

container a completely new set of values, but operator= won't do what you want.

The second reason for the quiz is to demonstrate why range member functions are

superior to their single-element alternatives. A range member function is a member

function that, like STL algorithms, uses two iterator parameters to specify a range of

elements over which something should be done. Without using a range member

function to solve this Item's opening problem, you'd have to write an explicit loop,

probably something like this:

vector<Widget> v1, v2; // assume v1 and v2 are vectors

//of Widgets

v1.clear();

for ( vector<Widget>::const_iterator ci = v2.begin() + v2.size() / 2;

ci != v2.end();

++ci)

v1.push_back(*ci);

Item 43 examines in detail why you should try to avoid writing explicit loops, but you

don't need to read that Item to recognize that writing this code is a lot more work than

is writing the call to assign. As we'll see shortly, the loop also happens to impose an

efficiency penalty, but we'll deal with that in a moment.

One way to avoid the loop is to follow the advice of Item 43 and employ an algorithm

instead:

v1.clear();

copy(v2.begin() + v2.size() / 2, v2.end(), back_inserter(v1 ));

Writing this is still more work than writing the call to assign. Furthermore, though no

loop is present in this code, one certainly exists inside copy (see Item 43). As a result,

the efficiency penalty remains. Again. I'll discuss that below. At this point, I want to

digress briefly to observe that almost all uses of copy where the destination range is

specified using an insert iterator (i.e.. via inserter, back_inserter. or front_inserter) can

be — should be — replaced with calls to range member functions. Here, for example,

the call to copy can be replaced with a range version of insert:

v1 .insert(v1 .end(), v2.begin() + v2.size() / 2, v2.end());

This involves slightly less typing than the call to copy, but it also says more directly

what is happening: data is being inserted into v1. The call to copy expresses that, too,

but less directly. It puts the emphasis in the wrong place. The interesting aspect of

what is happening is not that elements are being copied, it's that v1 is having new data

added to it. The insert member function makes that clear. The use of copy obscures it.

There's nothing interesting about the fact that things are being copied, because the STL

is built on the assumption that things will be copied. Copying is so fundamental to the

STL. it's the topic of Item 3 in this book!

Too many STL programmers overuse copy, so the advice I just gave bears repeating:

Almost all uses of copy where the destination range is specified using an insert iterator

should be replaced with calls to range member functions.

•

Returning to our assign example, we've already identified two reasons to prefer

range member functions to their single-element counterparts:

•

It's generally less work to write the code using the range member functions.

•

Range member functions tend to lead to code that is clearer and more

straightforward.

In short, range member functions yield code that is easier to write and easier to

understand. What's not to like'.'

Alas, some will dismiss these arguments as matters of programming style, and

developers enjoy arguing about style issues almost as much as they enjoy arguing

about which is the One True Editor. (As if there's any doubt. It's Emacs.) It would be

helpful to have a more universally agreed-upon criterion for establishing the

superiority of range member functions to their single-element counterparts. For the

standard sequence containers, we have one: efficiency. When dealing with the

standard sequence containers, application of single-element member functions makes

more demands on memory allocators, copies objects more frequently, and/or performs

redundant operations compared to range member functions that achieve the same end.

For example, suppose you'd like to copy an array of ints into the front of a vector. (The

data might be in an array instead of a vector in the first place, because the data came

from a legacy C API. For a discussion of the issues that arise when mixing STL

containers and C APIs, see Item 16.) Using the vector range insert function, it's

honestly trivial:

int data[numValues]; // assume numValues is

// defined elsewhere

vector<int> v;

...

v.insert(v.begin(), data, data + numValues); // insert the ints in data

//into v at the front

Using iterative calls to insert in an explicit loop, it would probably look more or less

like this:

vector<int>::iterator insertLoc(v.begin());

for (int i = 0; i < numValues; ++i) {

insertLoc = v.insert(insertLoc, data[i]);

}

Notice how we have to be careful to save the return value of insert for the next loop

iteration. If we didn't update insertLoc after each insertion, we'd have two problems.

First, all loop iterations after the first would yield undefined behavior, because each

insert call would invalidate insertLoc. Second, even if insertLoc remained valid, we'd

always insert at the front of the vector (i.e., at v.begin()), and the result would be that

the ints copied into v would end up in reverse order.

If we follow the lead of Item 43 and replace the loop with a call to copy, we get

something like this:

copy(data. data + numValues, inserter(v, v.begin()));

By the time the copy template has been instantiated, the code based on copy and the

code using the explicit loop will be almost identical, so for purposes of an efficiency

analysis, we'll focus on the explicit loop, keeping in mind that the analysis is equally

valid for the code employing copy. Looking at the explicit loop just makes it easier to

understand where the efficiency hits come from. Yes, that's "hits." plural, because the

code using the single-element version of insert levies up to three different performance

taxes on you, none of which you pay if you use the range version of insert.

The first tax consists of unnecessary function calls. Inserting numValues elements into

v one at a time naturally costs you numValues calls to insert. Using the range form of

insert, you pay for only one function call, a savings of numValues-1 calls. Of course,

it's possible that inlining will save you from this tax, but then again, it's possible that it

won't. Only one thing is sure. With the range form of insert, you definitely won't pay

it.

Inlining won't save you from the second tax, which is the cost of inefficiently moving

the existing elements in v to their final post-insertion positions. Each time insert is

called to add a new value to v. every element above the insertion point must be moved

up one position to make room for the new element. So the element at position p must

be moved up to position p+1, etc. In our example, we're inserting numValues elements

at the front of v. That means that each element in v prior to the insertions will have to

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提升STL使用效率的50个具体方法

Effective STL

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