所以,我一直在玩 C 指针和指针算法,因为我对它们并不完全满意。我想出了这段代码。
char* a[5] = { "Hi", "My", "Name", "Is" , "Dennis"};
char** aPtr = a; // This is acceptable because 'a' is double pointer
char*** aPtr2 = &aPtr; // This is also acceptable because they are triple pointers
//char ***aPtr2 = &a // This is not acceptable according to gcc 4.8.3, why ?
//This is the rest of the code, the side notes are only for checking
printf("%s\n",a[0]); //Prints Hi
printf("%s\n",a[1]); //Prints My
printf("%s\n",a[2]); //Prints Name
printf("%s\n",a[3]); //Prints Is
printf("%s\n",a[4]); //Prints Dennis
printf("%s\n",*(a+0)); //Prints Hi
printf("%s\n",*(a+1)); //Prints My
printf("%s\n",*(a+2)); //Prints Name
printf("%s\n",*(a+3)); //Prints Is
printf("%s\n",*(a+4)); //Prints Dennis
printf("%s\n",*(*(aPtr2) +0)); //Prints Hi
printf("%s\n",*(*(aPtr2) +1)); //Prints My // ap = a, *ap = *a, *(ap)+1 = *a+1 ?
printf("%s\n",*(*(aPtr2) +2)); //Prints Name
printf("%s\n",*(*(aPtr2) +3)); //Prints Is
printf("%s\n",*(*(aPtr2) +4)); //Prints Dennis
char*** aPtr2 = &a
根据 gcc 4.8.3 是 Not Acceptable ,为什么?
抱歉忘记添加编译器警告:
warning: initialization from incompatible pointer type [enabled by default]
可能不清楚我想说什么,所以我不得不添加这个链接:
- 这是有效的代码:http://ideone.com/4ePj4h . (第 7 行。注释掉)
- 这是无效的代码:http://ideone.com/KMG7OS . (第 6 行。注释掉)
注意注释掉的行。
最佳答案
a
是 5 个指针的缓冲区的地址并且是不可变的(即它是一个固定的指针)。如果你允许
char ***aPtr2 = &a;
然后
*aPtr2 = &a[3];
实际上会修改地址a
(这是禁止的)。
关于c - 关于双指针和三指针/双维数组,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30579784/