我对矢量查找与地图查找感到好奇并为它编写了一个小测试程序..它看起来像矢量总是比我使用它的方式更快..我还应该考虑其他什么吗?测试是否有偏见?运行的结果在底部..它以纳秒为单位,但gcc似乎不支持我的平台。
使用字符串进行查找当然会改变很多事情。
我正在使用的编译行是这样的:g ++ -O3 --std = c ++ 0x -o lookup lookup.cpp
#include <iostream>
#include <vector>
#include <map>
#include <unordered_map>
#include <chrono>
#include <algorithm>
unsigned dummy = 0;
class A
{
public:
A(unsigned id) : m_id(id){}
unsigned id(){ return m_id; }
void func()
{
//making sure its not optimized away
dummy++;
}
private:
unsigned m_id;
};
class B
{
public:
void func()
{
//making sure its not optimized away
dummy++;
}
};
int main()
{
std::vector<A> v;
std::unordered_map<unsigned, B> u;
std::map<unsigned, B> m;
unsigned elementCount = 1;
struct Times
{
unsigned long long v;
unsigned long long u;
unsigned long long m;
};
std::map<unsigned, Times> timesMap;
while(elementCount != 10000000)
{
elementCount *= 10;
for(unsigned i = 0; i < elementCount; ++i)
{
v.emplace_back(A(i));
u.insert(std::make_pair(i, B()));
m.insert(std::make_pair(i, B()));
}
std::chrono::time_point<std::chrono::steady_clock> start = std::chrono::high_resolution_clock::now();
for(unsigned i = 0; i < elementCount; ++i)
{
auto findItr = std::find_if(std::begin(v), std::end(v),
[&i](A & a){ return a.id() == i; });
findItr->func();
}
auto tp0 = std::chrono::high_resolution_clock::now()- start;
unsigned long long vTime = std::chrono::duration_cast<std::chrono::nanoseconds>(tp0).count();
start = std::chrono::high_resolution_clock::now();
for(unsigned i = 0; i < elementCount; ++i)
{
u[i].func();
}
auto tp1 = std::chrono::high_resolution_clock::now()- start;
unsigned long long uTime = std::chrono::duration_cast<std::chrono::nanoseconds>(tp1).count();
start = std::chrono::high_resolution_clock::now();
for(unsigned i = 0; i < elementCount; ++i)
{
m[i].func();
}
auto tp2 = std::chrono::high_resolution_clock::now()- start;
unsigned long long mTime = std::chrono::duration_cast<std::chrono::nanoseconds>(tp2).count();
timesMap.insert(std::make_pair(elementCount ,Times{vTime, uTime, mTime}));
}
for(auto & itr : timesMap)
{
std::cout << "Element count: " << itr.first << std::endl;
std::cout << "std::vector time: " << itr.second.v << std::endl;
std::cout << "std::unordered_map time: " << itr.second.u << std::endl;
std::cout << "std::map time: " << itr.second.m << std::endl;
std::cout << "-----------------------------------" << std::endl;
}
std::cout << dummy;
}
./lookup
Element count: 10
std::vector time: 0
std::unordered_map time: 0
std::map time: 1000
-----------------------------------
Element count: 100
std::vector time: 0
std::unordered_map time: 3000
std::map time: 13000
-----------------------------------
Element count: 1000
std::vector time: 2000
std::unordered_map time: 29000
std::map time: 138000
-----------------------------------
Element count: 10000
std::vector time: 22000
std::unordered_map time: 287000
std::map time: 1610000
-----------------------------------
Element count: 100000
std::vector time: 72000
std::unordered_map time: 1539000
std::map time: 8994000
-----------------------------------
Element count: 1000000
std::vector time: 746000
std::unordered_map time: 12654000
std::map time: 154060000
-----------------------------------
Element count: 10000000
std::vector time: 8001000
std::unordered_map time: 123608000
std::map time: 2279362000
-----------------------------------
33333330
答案 0 :(得分:5)
我一点也不震惊测试的矢量比其他任何东西都好。它的asm代码(实际反汇编)分解为此(在我的Apple LLVM 4.2上完全选择):
0x100001205: callq 0x100002696 ; symbol stub for: std::__1::chrono::steady_clock::now()
0x10000120a: testl %r13d, %r13d
0x10000120d: leaq -272(%rbp), %rbx
0x100001214: je 0x100001224 ; main + 328 at main.cpp:78
0x100001216: imull $10, %r14d, %ecx
0x10000121a: incl 7896(%rip) ; dummy
0x100001220: decl %ecx
0x100001222: jne 0x10000121a ; main + 318 [inlined] A::func() at main.cpp:83
main + 318 at main.cpp:83
0x100001224: movq %rax, -280(%rbp)
0x10000122b: callq 0x100002696 ; symbol stub for: std::__1::chrono::
注意'循环'(jne 0x10000121a
)。 “find_if”已经完全优化,结果实际上是通过递减寄存器扫描数组,以计算递增全局的次数。 这就是所有正在做的事情;在这里没有任何类型的搜索。
所以,是的,你是如何使用它的。
答案 1 :(得分:4)
首先,您似乎没有在测试之间清除容器。所以他们不包含你认为他们做的事情。
其次,根据您的时间,您的矢量呈现线性时间,这是不可能的,因为算法中的复杂度为O(N * N)。可能它已经优化了。我建议不要试图打击优化,而不是试图对抗优化。
第三,您的值对于矢量来说太可预测了。这可能会大大影响它。尝试随机值(或random_shuffle())