linux c 时间问题
错误：‘starttime’的存储大小未知
‘stoptime’的存储大小未知

#include <stdio.h>
#include <time.h>
#include <math.h>

void func()
{
unsigned int i,j;
double y;
for( i=0; i<1000 ; ++i )
{
for( j=0; j<1000 ; ++j )
{
y=sin((double)i);
}

}

}

int main(int argc, char *argv[])
{
struct timeval starttime,stoptime;
double usetime;
gettimeofday(starttime,NULL);
func();
gettimeofday(stoptime,NULL);
usetime=(starttime.tv_sec - stoptime.tv_sec)*
1000000+
starttime.tv_usec - stoptime.tv_usec;
usetime /= 1000000;
printf("the func run use time %lf\n",usetime);
return 0;
}


[解决办法]

C/C++ code

#include <stdio.h>#include <time.h>#include <math.h>#include <sys/time.h>void func(){  unsigned int i,j;  double y;  for( i=0; i<1000 ; ++i )  {  for( j=0; j<1000 ; ++j )  {  y=sin((double)i);  }  }}int main(int argc, char *argv[]){  struct timeval starttime;  struct timeval stoptime;  double usetime;  gettimeofday(&starttime,NULL);  func();  gettimeofday(&stoptime,NULL);  usetime=(starttime.tv_sec - stoptime.tv_sec)*   1000000+  starttime.tv_usec - stoptime.tv_usec;  usetime /= 1000000;  printf("the func run use time %lf\n",usetime);  return 0;}
[解决办法]
gettimeofday(starttime,NULL);
调用函数后，传入的参数是不会改变的，尽管在函数里操作了；而如果是传入参数的地址，则调用函数操作，可以改变参数值。而且gettimeofday（&starttime, NULL）是这样用的。