raii可以实现,通过new delete控制调用前后,很懒,只想加几个字符,不想换行
snippet mlt
#include <sys/time.h>
#define MLLOG printf
#ifndef MLLOG
#error "please define MLLOG"
#endif
struct MLT{
MLT(struct timeval start_time, const char* label_tag)
:t_start(start_time), label(label_tag)
{
}
template <typename T>
MLT& operator = (T)
{
return *this;
}
~MLT()
{
gettimeofday(&t_end, 0);
long time_diff = (t_end.tv_sec - t_start.tv_sec)*1000 + (t_end.tv_usec - t_start.tv_usec)/1000;
MLLOG("%s cost time is %ld ms\n", label, time_diff);
}
private:
struct timeval t_start;
struct timeval t_end;
const char* label;
};
MLT mlt_helper(struct timeval start_time, const char* label_tag)
{
return MLT(start_time, label_tag);
}
// caNNot use for void return functions
// and seems tedious
#define mlt(f) struct timeval t_start; \
gettimeofday(&t_start, 0); \
mlt_helper(t_start, #f) = \
// cannot use for void return functions
#define mt struct timeval t_start; \
gettimeofday(&t_start, 0); \
mlt_helper(t_start, "") = \
测试:
int f(void)
{
float k = 0.0;
for(auto i = 0; i < 10010; i++)
for(auto m = 0; m < 10090; m++)
{
float j = ((float)i*4.19)/4.342;
k += j;
}
}
int main(int argc, char* argv[])
{
mlt(xxx)
f();
cout << "after " << endl;
return 0;
}
有两个问题,void不能作为operator=的参数,所以benchmark的函数必须有返回值,mlt不能直接取f 的名称,导致要写个(),比较烦。懒。