我编写了以下基准:
#include <iostream> // cout
#include <math.h> // pow
#include <chrono> // high_resolution_clock
using namespace std;
using namespace std::chrono;
int64_t calculate(int);
int main()
{
high_resolution_clock::time_point t1, t2;
// Test 1
t1 = high_resolution_clock::now();
calculate(200);
t2 = high_resolution_clock::now();
cout << "RUNTIME = " << duration_cast<nanoseconds>(t2 - t1).count() << " nano seconds" << endl;
// Test 2
t1 = high_resolution_clock::now();
calculate(200000);
t2 = high_resolution_clock::now();
cout << "RUNTIME = " << duration_cast<nanoseconds>(t2 - t1).count() << " nano seconds" << endl;
}
int64_t calculate(const int max_exponent)
{
int64_t num = 0;
for(int i = 0; i < max_exponent; i++)
{
num += pow(2, i);
}
return num;
}
在Odroid XU3上运行此基准测试时,将产生以下输出(运行8次):
RUNTIME TEST 1 = 1250 nano seconds
RUNTIME TEST 2 = 1041 nano seconds
RUNTIME TEST 1 = 1292 nano seconds
RUNTIME TEST 2 = 1042 nano seconds
RUNTIME TEST 1 = 1250 nano seconds
RUNTIME TEST 2 = 1083 nano seconds
RUNTIME TEST 1 = 1292 nano seconds
RUNTIME TEST 2 = 1083 nano seconds
RUNTIME TEST 1 = 1209 nano seconds
RUNTIME TEST 2 = 1084 nano seconds
RUNTIME TEST 1 = 1166 nano seconds
RUNTIME TEST 2 = 1083 nano seconds
RUNTIME TEST 1 = 1292 nano seconds
RUNTIME TEST 2 = 1042 nano seconds
RUNTIME TEST 1 = 1166 nano seconds
RUNTIME TEST 2 = 1250 nano seconds
RUNTIME TEST 1 = 1250 nano seconds
RUNTIME TEST 2 = 1250 nano seconds
第二个指数是第一个指数的1000倍。为什么第二通电话有时会更快结束?
我将GCC(4.8)用作带有-Ofast
标志的编译器。
更新:我可以在我的i7 4770k上重现类似的行为。
简短的答案是“消除死代码”。编译器会发现您永远不会使用调用该函数的结果(并且该函数没有副作用),因此它仅消除了调用该函数。
打印出函数的结果,情况会有所变化。例如:
Ignore: -9223372036854775808 RUNTIME = 0 nano seconds
Ignore: -9223372036854775808 RUNTIME = 23001300 nano seconds
修改后的代码,以防万一:
#include <iostream> // cout
#include <math.h> // pow
#include <chrono> // high_resolution_clock
using namespace std;
using namespace std::chrono;
int64_t calculate(int);
int main() {
high_resolution_clock::time_point t1, t2;
// Test 1
t1 = high_resolution_clock::now();
auto a = calculate(200);
t2 = high_resolution_clock::now();
std::cout << "Ignore: " << a << "\t";
cout << "RUNTIME = " << duration_cast<nanoseconds>(t2 - t1).count() << " nano seconds" << endl;
// Test 2
t1 = high_resolution_clock::now();
auto b = calculate(200000);
t2 = high_resolution_clock::now();
std::cout << "Ignore: " << b << "\t";
cout << "RUNTIME = " << duration_cast<nanoseconds>(t2 - t1).count() << " nano seconds" << endl;
}
int64_t calculate(const int max_exponent) {
int64_t num = 0;
for (int i = 0; i < max_exponent; i++) {
num += pow(2, i);
}
return num;
}
从那里开始,您有一个很小的细节,即您正在溢出int64_t
(多次)给出不确定行为的范围-但是至少有了这个,我们有合理的希望打印出来的时间反映执行指定计算的时间。
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