如果这样的类生成，该怎么办？

看第四个就可以，前三个都挺简单

a2.m1() 你看 a2是一个c的实例, 但是c里面没有m1方法. 所以他调用的是B的m1.
a2.m2()同上, 调用的是A中的m2
a2.m3()是C自己的

readman 发表于 2014-7-6 17:12
a2.m1() 你看 a2是一个c的实例, 但是c里面没有m1方法. 所以他调用的是B的m1.
a2.m2()同上, 调用的是A中的m ...

答案是B'm1 C'm2 A'm3....

TonyJang 发表于 2014-7-6 20:17
答案是B'm1 C'm2 A'm3....

c 没有自己的m2啊

/System/Library/Java/JavaVirtualMachines/1.6.0.jdk/Contents/Home/bin/java -Didea.launcher.port=7532 "-Didea.launcher.bin.path=/Applications/IntelliJ IDEA 13.app/bin" -Dfile.encoding=UTF-8 -classpath "/System/Library/Java/JavaVirtualMachines/1.6.0.jdk/Contents/Home/lib/deploy.jar:/System/Library/Java/JavaVirtualMachines/1.6.0.jdk/Contents/Home/lib/dt.jar:/System/Library/Java/JavaVirtualMachines/1.6.0.jdk/Contents/Home/lib/javaws.jar:/System/Library/Java/JavaVirtualMachines/1.6.0.jdk/Contents/Home/lib/jce.jar:/System/Library/Java/JavaVirtualMachines/1.6.0.jdk/Contents/Home/lib/jconsole.jar:/System/Library/Java/JavaVirtualMachines/1.6.0.jdk/Contents/Home/lib/management-agent.jar:/System/Library/Java/JavaVirtualMachines/1.6.0.jdk/Contents/Home/lib/plugin.jar:/System/Library/Java/JavaVirtualMachines/1.6.0.jdk/Contents/Home/lib/sa-jdi.jar:/System/Library/Java/JavaVirtualMachines/1.6.0.jdk/Contents/Classes/alt-rt.jar:/System/Library/Java/JavaVirtualMachines/1.6.0.jdk/Contents/Classes/alt-string.jar:/System/Library/Java/JavaVirtualMachines/1.6.0.jdk/Contents/Classes/charsets.jar:/System/Library/Java/JavaVirtualMachines/1.6.0.jdk/Contents/Classes/classes.jar:/System/Library/Java/JavaVirtualMachines/1.6.0.jdk/Contents/Classes/jsse.jar:/System/Library/Java/JavaVirtualMachines/1.6.0.jdk/Contents/Classes/ui.jar:/System/Library/Java/JavaVirtualMachines/1.6.0.jdk/Contents/Home/lib/ext/apple_provider.jar:/System/Library/Java/JavaVirtualMachines/1.6.0.jdk/Contents/Home/lib/ext/dnsns.jar:/System/Library/Java/JavaVirtualMachines/1.6.0.jdk/Contents/Home/lib/ext/localedata.jar:/System/Library/Java/JavaVirtualMachines/1.6.0.jdk/Contents/Home/lib/ext/sunjce_provider.jar:/System/Library/Java/JavaVirtualMachines/1.6.0.jdk/Contents/Home/lib/ext/sunpkcs11.jar:/Users/Readman/Dropbox/美国工作/leetcode/out/production/leetcode:/Applications/IntelliJ IDEA 13.app/lib/idea_rt.jar" com.intellij.rt.execution.application.AppMain C
m1 in B
m2 in A
m3 in C

Process finished with exit code 0

C继承了B，B继承了A。同时B覆盖了A中的m1()方法，所以B的m1是自己的方法，m2，m3和A一样；C继承时覆盖了B的m3，所以C的方法是B的m1，A的m2，C自己的m3。  而a2是类C的对象，所以我觉得答案应该是： B‘s m1   A’s m2   C‘s m3 13.

p.s. 我记得我当时自己做的时候就觉得是答案错了，跟小伙伴讨论了一下也是这样觉得。若是理解有误请各位大侠拔刀相助！

TonyJang 发表于 2014-7-6 20:17
答案是B'm1 C'm2 A'm3....

答案肯定错了，c哪有m2

实测，结果和第二组一样
B‘s m1   A’s m2   C‘s m3 13.

我记得我在head first java 里面看过这个~答案应该是有问题的~
如果就调用方法来说的话~我记得这本书里面有一句话非常有用
“The Lowest Win”
就是说对于继承关系来说~如果本类中没有相应的被调用方法的话~就要再继承关系当中找~越矮的类中被overriden的方法才会被调用~

所以只要记住这句话就行了~“The Lowest Win”
所以正确答案应该是大家说的~
B‘s m1   A’s m2   C‘s m3  13