1过程是怎么样的呀public static void main(String[] args) {\x05int[] x = {1,2,3,4,5};\x05increase(x);\x05int[] y = {1,2,3,4,5};\x05increase(y[0]);\x05System.out.println(x[0] + " " + y[0]);\x05}\x05public static void increase(int[] x) {\x05for (
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/28 23:19:28
![1过程是怎么样的呀public static void main(String[] args) {\x05int[] x = {1,2,3,4,5};\x05increase(x);\x05int[] y = {1,2,3,4,5};\x05increase(y[0]);\x05System.out.println(x[0] +](/uploads/image/z/6762677-5-7.jpg?t=1%E8%BF%87%E7%A8%8B%E6%98%AF%E6%80%8E%E4%B9%88%E6%A0%B7%E7%9A%84%E5%91%80public+static+void+main%28String%5B%5D+args%29+%7B%5Cx05int%5B%5D+x+%3D+%7B1%2C2%2C3%2C4%2C5%7D%3B%5Cx05increase%28x%29%3B%5Cx05int%5B%5D+y+%3D+%7B1%2C2%2C3%2C4%2C5%7D%3B%5Cx05increase%28y%5B0%5D%29%3B%5Cx05System.out.println%28x%5B0%5D+%2B+%22+%22+%2B+y%5B0%5D%29%3B%5Cx05%7D%5Cx05public+static+void+increase%28int%5B%5D+x%29+%7B%5Cx05for+%28)
1过程是怎么样的呀public static void main(String[] args) {\x05int[] x = {1,2,3,4,5};\x05increase(x);\x05int[] y = {1,2,3,4,5};\x05increase(y[0]);\x05System.out.println(x[0] + " " + y[0]);\x05}\x05public static void increase(int[] x) {\x05for (
1过程是怎么样的呀
public static void main(String[] args) {
\x05int[] x = {1,2,3,4,5};
\x05increase(x);
\x05int[] y = {1,2,3,4,5};
\x05increase(y[0]);
\x05System.out.println(x[0] + " " + y[0]);
\x05}
\x05public static void increase(int[] x) {
\x05for (int i = 0; i < x.length; i++)
\x05x[i]++;
\x05}
\x05public static void increase(int y) {
\x05y++;
\x05}}
1过程是怎么样的呀public static void main(String[] args) {\x05int[] x = {1,2,3,4,5};\x05increase(x);\x05int[] y = {1,2,3,4,5};\x05increase(y[0]);\x05System.out.println(x[0] + " " + y[0]);\x05}\x05public static void increase(int[] x) {\x05for (
我的理解是这样:
不知道你懂不懂点基础:
定义数组X,之后调用的方法increase,会把数组里面的每一项+1,则在输出的时候取X数组第一项X[0]的时候是1+1=2;
而数组Y为什么没有改变,是因为,在调用ingcrease方法的时候的参数是int的数值,它值的改变不会影响到数组y的改变,所以依旧是1.
好吧,我也是菜逼,这是我的理解希望能帮到你.