(int)x*0.5, how to understand

(int)x*0.5;

In this case, only the x value is rounded, and the type is not converted. The rounded number is then multiplied by 0.5.

Let’s look at the title first:

double x=5.5,(int)X*0.5
请问这时的x=?

int main()
{
    double x=5.5;
    x=(int)x*0.5;//对x的初值进行取整，不对类型进行转换
    printf("x=%0.1lf",x);//0.1只获取到小数点后一位数
    return 0;
}

The result is:

x=2.5