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.
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.
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)