char *a[]={"diamonds","clubs","spades","hearts"};
char **p[]={a+3,a+2,a+1,a};
char ***ptr=p;
cout<<*ptr[2][2];
为什么会显示,请解释2个tr执行阵列及其内容。
请注意,x[y]对条码的紧接于<条码>*x,因此,<条码>*ptr (2) 被解释为<条码>*(ptr (2)]。
还注意到x[y]=*(x+y)。
因此,
*(ptr[2][2]) == *(p[2][2]) // ptr = p
== *((a+1)[2]) // p[2] == a+1
== *(*(a+1+2)) // x[y] == *(x+y)
== *(*(a+3)) // 1+2 == 3
== *(a[3]) // *(x+y) == x[y]
== *("hearts") // a[3] == "hearts"
== "hearts"[0] // *x == *(x+0) == x[0]
== h
见KennyTM对解释的出色回答......但我认为,这是教授利用你的夸张来“视像”记忆的完美例子。
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