好吧，因为vector::resize()比普通内存分配(由malloc)做更多的处理。

尝试在复制构造函数中设置断点(定义它以便可以设置断点!)，就会增加处理时间。

这似乎取决于编译器标志。下面是一个基准代码:

#include <chrono>
#include <cmath>
#include <ctime>
#include <iostream>
#include <vector>


int main(){

    int size = 1000000; // reduce this number in case your program crashes
    int L = 10;

    std::cout << "size=" << size << " L=" << L << std::endl;
    {
        srand( time(0) );
        double * data = new double[size];
        double result = 0.;
        std::chrono::steady_clock::time_point start = std::chrono::steady_clock::now();
        for( int l = 0; l < L; l++ ) {
            for( int i = 0; i < size; i++ ) data[i] = rand() % 100;
            for( int i = 0; i < size; i++ ) result += data[i] * data[i];
        }
        std::chrono::steady_clock::time_point end   = std::chrono::steady_clock::now();
        auto duration = std::chrono::duration_cast<std::chrono::microseconds>(end - start).count();
        std::cout << "Calculation result is " << sqrt(result) << "\n";
        std::cout << "Duration of C style heap array:    " << duration << "ms\n";
        delete data;
    }

    {
        srand( 1 + time(0) );
        double data[size]; // technically, non-compliant with C++ standard.
        double result = 0.;
        std::chrono::steady_clock::time_point start = std::chrono::steady_clock::now();
        for( int l = 0; l < L; l++ ) {
            for( int i = 0; i < size; i++ ) data[i] = rand() % 100;
            for( int i = 0; i < size; i++ ) result += data[i] * data[i];
        }
        std::chrono::steady_clock::time_point end   = std::chrono::steady_clock::now();
        auto duration = std::chrono::duration_cast<std::chrono::microseconds>(end - start).count();
        std::cout << "Calculation result is " << sqrt(result) << "\n";
        std::cout << "Duration of C99 style stack array: " << duration << "ms\n";
    }

    {
        srand( 2 + time(0) );
        std::vector<double> data( size );
        double result = 0.;
        std::chrono::steady_clock::time_point start = std::chrono::steady_clock::now();
        for( int l = 0; l < L; l++ ) {
            for( int i = 0; i < size; i++ ) data[i] = rand() % 100;
            for( int i = 0; i < size; i++ ) result += data[i] * data[i];
        }
        std::chrono::steady_clock::time_point end   = std::chrono::steady_clock::now();
        auto duration = std::chrono::duration_cast<std::chrono::microseconds>(end - start).count();
        std::cout << "Calculation result is " << sqrt(result) << "\n";
        std::cout << "Duration of std::vector array:     " << duration << "ms\n";
    }

    return 0;
}

不同的优化标志给出不同的答案:

$ g++ -O0 benchmark.cpp 
$ ./a.out 
size=1000000 L=10
Calculation result is 181182
Duration of C style heap array:    118441ms
Calculation result is 181240
Duration of C99 style stack array: 104920ms
Calculation result is 181210
Duration of std::vector array:     124477ms
$g++ -O3 benchmark.cpp
$ ./a.out 
size=1000000 L=10
Calculation result is 181213
Duration of C style heap array:    107803ms
Calculation result is 181198
Duration of C99 style stack array: 87247ms
Calculation result is 181204
Duration of std::vector array:     89083ms
$ g++ -Ofast benchmark.cpp 
$ ./a.out 
size=1000000 L=10
Calculation result is 181164
Duration of C style heap array:    93530ms
Calculation result is 181179
Duration of C99 style stack array: 80620ms
Calculation result is 181191
Duration of std::vector array:     78830ms

您的确切结果会有所不同，但这在我的机器上是非常典型的。

好吧，因为vector::resize()比普通内存分配(由malloc)做更多的处理。

尝试在复制构造函数中设置断点(定义它以便可以设置断点!)，就会增加处理时间。

Martin York的回答让我很困扰，因为他似乎试图掩盖初始化问题。但他将冗余的默认构造确定为性能问题的根源是正确的。

[编辑:Martin的回答不再建议更改默认构造函数。]

对于眼前的问题，你当然可以调用2参数版本的向量<Pixel> ctor:

std::vector<Pixel> pixels(dimension * dimension, Pixel(255, 0, 0));

如果你想用一个常数值初始化，这是一种常见的情况。但更普遍的问题是:如何有效地初始化比常数值更复杂的东西?

为此，您可以使用back_insert_iterator，这是一个迭代器适配器。这里有一个int类型的向量的例子，尽管一般的思想也适用于像素:

#include <iterator>
// Simple functor return a list of squares: 1, 4, 9, 16...
struct squares {
    squares() { i = 0; }
    int operator()() const { ++i; return i * i; }

private:
    int i;
};

...

std::vector<int> v;
v.reserve(someSize);     // To make insertions efficient
std::generate_n(std::back_inserter(v), someSize, squares());

或者，您可以使用copy()或transform()来代替generate_n()。

缺点是，构造初始值的逻辑需要移动到一个单独的类中，这比将其放在原位更不方便(尽管c++ 1x中的lambdas使这更好)。此外，我希望这仍然不会像基于malloc()的非stl版本那样快，但我希望它会接近，因为它只对每个元素进行一次构造。

GNU's STL (and others), given vector<T>(n), default constructs a prototypal object T() - the compiler will optimise away the empty constructor - but then a copy of whatever garbage happened to be in the memory addresses now reserved for the object is taken by the STL's __uninitialized_fill_n_aux, which loops populating copies of that object as the default values in the vector. So, "my" STL is not looping constructing, but constructing then loop/copying. It's counter intuitive, but I should have remembered as I commented on a recent stackoverflow question about this very point: the construct/copy can be more efficient for reference counted objects etc..

So:

vector<T> x(n);

or

vector<T> x;
x.resize(n);

是-在许多STL实现中-类似于:

T temp;
for (int i = 0; i < n; ++i)
    x[i] = temp;

The issue being that the current generation of compiler optimisers don't seem to work from the insight that temp is uninitialised garbage, and fail to optimise out the loop and default copy constructor invocations. You could credibly argue that compilers absolutely shouldn't optimise this away, as a programmer writing the above has a reasonable expectation that all the objects will be identical after the loop, even if garbage (usual caveats about 'identical'/operator== vs memcmp/operator= etc apply). The compiler can't be expected to have any extra insight into the larger context of std::vector<> or the later usage of the data that would suggest this optimisation safe.

这可以与更明显的直接实现形成对比:

for (int i = 0; i < n; ++i)
    x[i] = T();

我们可以期待一个编译器优化。

为了更明确地解释vector行为的这一方面，可以考虑:

std::vector<big_reference_counted_object> x(10000);

显然，如果我们创建10000个独立对象，而不是创建10000个引用相同数据的对象，这是一个很大的区别。有一种合理的观点认为，保护普通c++用户不意外地做一些如此昂贵的事情的好处超过了现实世界中难以优化的拷贝构造的非常小的成本。

原始答案(供参考/理解评论): 没有机会。Vector和数组一样快，至少如果你合理地保留空间. ...

一个更好的基准测试(我认为…)，编译器由于优化可以改变代码，因为分配的向量/数组的结果不会在任何地方使用。 结果:

$ g++ test.cpp -o test -O3 -march=native
$ ./test 
UseArray inner completed in 0.652 seconds
UseArray completed in 0.773 seconds
UseVector inner completed in 0.638 seconds
UseVector completed in 0.757 seconds
UseVectorPushBack inner completed in 6.732 seconds
UseVectorPush completed in 6.856 seconds
The whole thing completed in 8.387 seconds

编译器:

gcc version 6.2.0 20161019 (Debian 6.2.0-9)

CPU:

model name  : Intel(R) Core(TM) i7-3630QM CPU @ 2.40GHz

代码是:

#include <cstdlib>
#include <vector>

#include <iostream>
#include <string>

#include <boost/date_time/posix_time/ptime.hpp>
#include <boost/date_time/microsec_time_clock.hpp>

class TestTimer
{
    public:
        TestTimer(const std::string & name) : name(name),
            start(boost::date_time::microsec_clock<boost::posix_time::ptime>::local_time())
        {
        }

        ~TestTimer()
        {
            using namespace std;
            using namespace boost;

            posix_time::ptime now(date_time::microsec_clock<posix_time::ptime>::local_time());
            posix_time::time_duration d = now - start;

            cout << name << " completed in " << d.total_milliseconds() / 1000.0 <<
                " seconds" << endl;
        }

    private:
        std::string name;
        boost::posix_time::ptime start;
};

struct Pixel
{
    Pixel()
    {
    }

    Pixel(unsigned char r, unsigned char g, unsigned char b) : r(r), g(g), b(b)
    {
    }

    unsigned char r, g, b;
};

void UseVector(std::vector<std::vector<Pixel> >& results)
{
    TestTimer t("UseVector inner");

    for(int i = 0; i < 1000; ++i)
    {
        int dimension = 999;

        std::vector<Pixel>& pixels = results.at(i);
        pixels.resize(dimension * dimension);

        for(int i = 0; i < dimension * dimension; ++i)
        {
            pixels[i].r = 255;
            pixels[i].g = 0;
            pixels[i].b = 0;
        }
    }
}

void UseVectorPushBack(std::vector<std::vector<Pixel> >& results)
{
    TestTimer t("UseVectorPushBack inner");

    for(int i = 0; i < 1000; ++i)
    {
        int dimension = 999;

        std::vector<Pixel>& pixels = results.at(i);
            pixels.reserve(dimension * dimension);

        for(int i = 0; i < dimension * dimension; ++i)
            pixels.push_back(Pixel(255, 0, 0));
    }
}

void UseArray(Pixel** results)
{
    TestTimer t("UseArray inner");

    for(int i = 0; i < 1000; ++i)
    {
        int dimension = 999;

        Pixel * pixels = (Pixel *)malloc(sizeof(Pixel) * dimension * dimension);

        results[i] = pixels;

        for(int i = 0 ; i < dimension * dimension; ++i)
        {
            pixels[i].r = 255;
            pixels[i].g = 0;
            pixels[i].b = 0;
        }

        // free(pixels);
    }
}

void UseArray()
{
    TestTimer t("UseArray");
    Pixel** array = (Pixel**)malloc(sizeof(Pixel*)* 1000);
    UseArray(array);
    for(int i=0;i<1000;++i)
        free(array[i]);
    free(array);
}

void UseVector()
{
    TestTimer t("UseVector");
    {
        std::vector<std::vector<Pixel> > vector(1000, std::vector<Pixel>());
        UseVector(vector);
    }
}

void UseVectorPushBack()
{
    TestTimer t("UseVectorPush");
    {
        std::vector<std::vector<Pixel> > vector(1000, std::vector<Pixel>());
        UseVectorPushBack(vector);
    }
}


int main()
{
    TestTimer t1("The whole thing");

    UseArray();
    UseVector();
    UseVectorPushBack();

    return 0;
}