#include <stdio.h>
int main()
{
double n=21.195;
printf("%.2f\n",n);
return 0;
}
结果:21.20
#include <stdio.h>
int main()
{
double n=21.195;
n*=100;//2119.5
n=(int)n;//2119
n=n/100;//21.19
printf("%.2f\n",n);
return 0;
}
结果:21.19
简化一下代码:
#include <stdio.h>
int main()
{
double n=21.195;
n=(double)(int)(100*n)/100;
printf("%.2f\n",n);
return 0;
}