Question

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.

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.

Answer 1

Pointers are not arrays

A dereferenced char ** is an object of type char *.

A dereferenced char (*)[10] is an object of type char [10].

Arrays are not pointers

See the c-faq entry about this very subject.

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Assume you have

char **pp;
char (*pa)[10];

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) */

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)