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)
C language standard, draft n1256:
6.3.2.1 Lvalues, arrays, and function designators
...
3 Except when it is the operand of thesizeofoperator or the unary&operator, or is a string literal used to initialize an array, an expression that has type ‘‘array of type’’ is converted to an expression with type ‘‘pointer to type’’ that points to the initial element of the array object and is not an lvalue. If the array object has register storage class, the behavior is undefined.
Given a declaration of
char a[10][10];
the type of the array expression a is "10-element array of 10-element array of char". Per the rule above, that gets coverted to type "pointer to 10-element array of char", or char (*)[10].
Remember that in the context of a function parameter declaration, T a[N] and T a[] are identical to T *a; thus, T a[][10] is identical to T (*a)[10].
