確定一個簡單的程序的輸出

Question

輸出爲b，從閱讀代碼預期產量0代替我1。 任何人都可以解釋如何到達這個輸出？

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

輸出：
a=5, b=1, c=0
x=0.5, y=5.0, z=0.0

代碼

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

我會給你一個提示，你將與你的自我理解：

當你施放一個double到int將只得到本質的部分，例如：

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

發生在這裏的這個方法：

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 
    //-----------------------------------^^------^^-----------------^^----------------^ 
} 

所有此之後，計算容易：

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