Arc*_*hie 26 c++ performance lambda c++11
我只是试图比较C++ 11中lambda表达式的性能,所以我做了测试 - 计算double值向量中元素的总和.这是实施:
#include <vector>
#include <algorithm>
#include <iostream>
#include <ctime>
#define LOG(x) { std::cout << #x << " = " << (x) << "\n"; }
#define TIME(t) { std::cout << ((double)(clock() - (t)) / CLOCKS_PER_SEC) << " s\n"; }
double sum(const std::vector<double>& v)
{
double s = 0.0;
for (auto i = v.cbegin(); i != v.cend(); ++i)
s += *i;
return s;
}
int main()
{
const size_t MAX = 1; // number of tests
const size_t SIZE = 100000000; // length of the vector
std::vector<double> v(SIZE, 1.0);
double out;
clock_t clk;
std::cout << "iterator\n";
clk = clock();
out = 0.0;
for (size_t i = 0; i < MAX; ++i)
out += sum(v);
TIME(clk)
LOG(out)
std::cout << "\nlambda\n";
clk = clock();
out = 0.0;
for (size_t i = 0; i < MAX; ++i)
std::for_each(v.cbegin(), v.cend(), [&](double d) { out += d; });
TIME(clk)
LOG(out)
return 0;
}
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以下是此程序的结果(在VS2010 SP1中,在发布模式下编译):
iterator 0.32 s out = 1e+008 lambda 0.326 s out = 1e+008
可以看出,性能几乎没有差异.但是,如果我将10作为值MAX(它表示求和将执行10次而不是1次),结果会有所不同:
iterator 0.287 s out = 1e+009 lambda 2.84 s out = 1e+009
对λ表达的测试花费大约10倍的时间.为什么?我认为这可能是由于事实造成的,即每次迭代都会创建新的lambda,但是我试过这个:
out = 0.0;
auto f = [&](double d) { out += d; };
for (size_t i = 0; i < MAX; ++i)
std::for_each(v.cbegin(), v.cend(), f);
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结果没有改变.有人可以向我解释这种行为吗?
Arc*_*hie 49
原来,这不是lambda表达式的任何问题,只是编译器在第一种情况下通过缓存sum()函数的结果来优化外部循环.将第一个案例更改为此表单后:
out = 0.0;
for (size_t i = 0; i < MAX; ++i)
{
out += sum(v);
v[i] = 1.0; // this adds O(1) time and prevents caching
}
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两个案例的时间大致相等,lambda是最喜欢的.