C++ 指针和数组内存

我对以下代码的输出感到困惑：

char *s[]={"knowledge","is","power"};
char t[]={"Hello"};
char **p,*q;
p=s;
q=t;
cout<<++*p;
cout<<*p++;
cout<<++*p;
cout<<++*q;

现在，输出为Knowledge Nowlede s I

请解释内存分配是如何在两个阵列中进行的

char *s[] = {"knowledge", "is", "power"};

       ===========================================================
addr1: | 'k' | 'n' | 'o' | 'w' | 'l' | 'e' | 'd' | 'g' | 'e' | 0 |
       ===========================================================

       =================
addr2: | 'i' | 's' | 0 |
       =================

       ===================================
addr3: | 'p' | 'o' | 'w' | 'e' | 'r' | 0 |
       ===================================

       =========================
addr4: | addr1 | addr2 | addr3 |
       =========================

    s: addr4

---

看这张照片。然后，简化语句以理解它们。通过使用无效的指针值访问数组，也会有未定义的行为。指针算术是需要小心处理的。请尝试指针算术下的此链接。这应该会澄清你的疑问，非常类似的问题：为什么不打印e@Max Vollmer，但它取H+1=73+1=74.chart[]={Hello}；与char*t[]={Hello}；非常不同；。您没有指向字符数组指针数组的指针，而是指向字符数组的指针。因此，当您取消引用该指针时，您不会得到指向char数组的指针，而是得到char。因此，您增加的是字符，而不是指针。

char *t[] = {"Hello"};

       ===================================
addr5: | 'H' | 'e' | 'l' | 'l' | 'o' | 0 |
       ===================================

       =========
addr6: | addr5 |
       =========

    t: addr6

char **p,**q;
p=s;
q=t;

    p: addr4
    q: addr6

++*p;

     p: addr4
    *p: addr1
  ++*p: addr1 + 1
(!) *p: addr1 + 1

*p++;

     p: addr4
    *p: addr1 + 1
  *p++: addr1 + 1
(!)  p: addr4 + 1

++*p;

     p: addr4 + 1
    *p: addr2
  ++*p: addr2 + 1
(!) *p: addr2 + 1

++*q;

     q: addr6
    *q: addr5
  ++*q: addr5 + 1
(!) *q: addr5 + 1