Question

b的输出，从阅读代码预期输出为0而不是1。任何人都可以解释如何达到这个输出？

  int a=5, b=6, c=1;
  double x=0.5, y=1.0, z=1.5;
  c = fcn1(a, b);
  y = fcn2(y, a);
  b = fcn3(x, y);
  z = fcn3(c, b);
  System.out.println("a="+a+", b="+b+", c="+c);
  System.out.println("x="+x+", y="+y+", z="+z);
  }
  static int fcn1(int i, int j){
  int k = i-j;
  return (++k);
  }
  static double fcn2(double t, int n){
  return (t*n);

  }

  static int fcn3(double u, double v){
  return fcn1((int)(u*v), 2);
  }
  static double fcn3(int r, int s){
  return fcn2(r,s);

Answer 1

<强>输出：
a=5, b=1, c=0
x=0.5, y=5.0, z=0.0

<强>代码

import java.io.*;
import java.util.*;
public class Main
{
    public static int fcn1(int i, int j){
        int k = i-j;
        return (++k);
    }
    public static double fcn2(double t, int n){
        return (t*n);
    }
    public static int fcn3(double u, double v){
        return fcn1((int)(u*v), 2);
    }
    public static double fcn3(int r, int s){
        return fcn2(r,s);
    }
    public static void main(String[] args)
    {
        int a=5, b=6, c=1;
        double x=0.5, y=1.0, z=1.5;
        c = fcn1(a, b);  //c=0
        y = fcn2(y, a);  //y=5.0
        b = fcn3(x, y);  //b=fcn1((int)2.5, 2)  //b=1
        z = fcn3(c, b);  //z=c*b                //z=0.0
        System.out.println("a="+a+", b="+b+", c="+c);
        System.out.println("x="+x+", y="+y+", z="+z);
    }  
}

Answer 2

我会给你一个提示，你会自己理解它：

当您将double投射到int时，您将只获得自然部分，例如：

double d = 2.5;
int i = (int) d;
//i in this case equal to 2 and not 2.5

这种情况发生在这个方法中：

static int fcn3(double u, double v) {
    return fcn1((int) (u * v), 2);//u = 0.5 v = 5.0 ## 0.5 * 5.0 = 2.5 ## (int) 2.5 = 2 
    //-----------------------------------^^------^^-----------------^^----------------^
}

此后的所有计算都很简单：

static int fcn1(int i, int j) {
    int k = i - j; // i = 2 ## j = 2 ## 2 - 2 = 0
    return (++k);//++0 = 1 <--------------------b
}

确定简单程序的输出

2 个答案: