Ant*_*emi 15 c++ performance gcc exception visual-c++
在C++中测量异常处理开销/性能的最佳方法是什么?
请提供独立的代码示例.
我的目标是Microsoft Visual C++ 2008和gcc.
我需要从以下案例中获得结果:
这是我想出的测量代码.你觉得它有什么问题吗?
到目前为止适用于Linux和Windows,编译:
g++ exception_handling.cpp -o exception_handling [ -O2 ]
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或者例如Visual C++ Express.
要获得基本案例("完全从语言中删除异常支持"),请使用:
g++ exception_handling.cpp -o exception_handling [ -O2 ] -fno-exceptions -DNO_EXCEPTIONS
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或MSVC中的类似设置.
这里有一些初步结果.由于机器负载的不同,它们可能都很糟糕,但它们确实对相关的异常处理开销有所了解.(执行摘要:当没有异常被抛出时,没有或很少,当它们实际被抛出时是巨大的.)
#include <stdio.h>
// Timer code
#if defined(__linux__)
#include <sys/time.h>
#include <time.h>
double time()
{
timeval tv;
gettimeofday(&tv, 0);
return 1.0 * tv.tv_sec + 0.000001 * tv.tv_usec;
}
#elif defined(_WIN32)
#include <windows.h>
double get_performance_frequency()
{
unsigned _int64 frequency;
QueryPerformanceFrequency((LARGE_INTEGER*) &frequency); // just assume it works
return double(frequency);
}
double performance_frequency = get_performance_frequency();
double time()
{
unsigned _int64 counter;
QueryPerformanceCounter((LARGE_INTEGER*) &counter);
return double(counter) / performance_frequency;
}
#else
# error time() not implemented for your platform
#endif
// How many times to repeat the whole test
const int repeats = 10;
// How many times to iterate one case
const int times = 1000000;
// Trick optimizer to not remove code
int result = 0;
// Case 1. No exception thrown nor handled.
void do_something()
{
++result;
}
void case1()
{
do_something();
}
// Case 2. No exception thrown, but handler installed
#ifndef NO_EXCEPTIONS
void do_something_else()
{
--result;
}
void case2()
{
try
{
do_something();
}
catch (int exception)
{
do_something_else();
}
}
// Case 3. Exception thrown and caught
void do_something_and_throw()
{
throw ++result;
}
void case3()
{
try
{
do_something_and_throw();
}
catch (int exception)
{
result = exception;
}
}
#endif // !NO_EXCEPTIONS
void (*tests[])() =
{
case1,
#ifndef NO_EXCEPTIONS
case2,
case3
#endif // !NO_EXCEPTIONS
};
int main()
{
#ifdef NO_EXCEPTIONS
printf("case0\n");
#else
printf("case1\tcase2\tcase3\n");
#endif
for (int repeat = 0; repeat < repeats; ++repeat)
{
for (int test = 0; test < sizeof(tests)/sizeof(tests[0]); ++test)
{
double start = time();
for (int i = 0; i < times; ++i)
tests[test]();
double end = time();
printf("%f\t", (end - start) * 1000000.0 / times);
}
printf("\n");
}
return result; // optimizer is happy - we produce a result
}
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