问题描述
根据我的书，创建多维数组char a[10][10]时，它表示必须使用类似于char a[][10]的参数才能将数组传递给函数.
When you create the multi-dimensional array char a[10][10], according to my book it says you must use a parameter similar to char a[][10] to pass the array to a function.
为什么必须这样指定长度?您不只是传递一个双指针指向with，并且那个双指针是否已经指向已分配的内存?那么为什么参数不能为char **a?您是否要通过提供第二个10重新分配任何新的内存?
Why must you specify the length as such? Aren't you just passing a double pointer to being with, and doesn't that double pointer already point to allocated memory? So why couldn't the parameter be char **a? Are you reallocating any new memory by supplying the second 10.
推荐答案
指针不是数组
被取消引用的char **是类型为char *的对象.
A dereferenced char ** is an object of type char *.
被取消引用的char (*)[10]是类型为char [10]的对象.
A dereferenced char (*)[10] is an object of type char [10].
数组不是指针
有关此主题的信息，请参见 c-faq条目.
See the c-faq entry about this very subject.
假设你有
char **pp;
char (*pa)[10];
并且出于争论的目的，它们都指向同一位置:0x420000.
and, for the sake of argument, both point to the same place: 0x420000.
pp == 0x420000; /* true */
(pp + 1) == 0x420000 + sizeof(char*); /* true */
pa == 0x420000; /* true */
(pa + 1) == 0x420000 + sizeof(char[10]); /* true */
(pp + 1) != (pa + 1) /* true (very very likely true) */
，这就是为什么自变量不能为char**类型的原因.另外，char**和char (*)[10]是不兼容的类型，因此参数的类型(衰变数组)必须与参数(函数原型中的类型)相匹配
and this is why the argument cannot be of type char**. Also char** and char (*)[10] are not compatible types, so the types of arguments (the decayed array) must match the parameters (the type in the function prototype)
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