I think that it is because the former is an array of pointers to char and the latter is a pointer to an array of chars, and we need to properly specify the size of the object being pointed to for our function definition. In the former;
function(char * p_array[])
the size of the object being pointed to is already included (its a pointer to char), but the latter
function(char (*p_array)[])
needs the size of the array p_array points to as part of p_array's definition? I'm at the stage where I've been thinking about this for too long and have just confused myself, someone please let me know if my reasoning is correct.
char* is a pointer to a character, which can be the beginning of a C-string. char* and char[] are used for C-string and a string object is used for C++ springs. char[] is an array of characters that can be used to store a C-string.
Difference between char s[] and char *s in CThe s[] is an array, but *s is a pointer. For an example, if two declarations are like char s[20], and char *s respectively, then by using sizeof() we will get 20, and 4.
The fundamental difference is that in one char* you are assigning it to a pointer, which is a variable. In char[] you are assigning it to an array which is not a variable.
C programming does not allow to return an entire array as an argument to a function. However, you can return a pointer to an array by specifying the array's name without an index.
Both are valid in C but not C++. You would ordinarily be correct:
char *x[]; // array of pointers to char
char (*y)[]; // pointer to array of char
However, the arrays decay to pointers if they appear as function parameters. So they become:
char **x; // Changes to pointer to array of pointer to char
char (*y)[]; // No decay, since it's NOT an array, it's a pointer to an array
In an array type in C, one of the sizes is permitted to be unspecified. This must be the leftmost one (whoops, I said rightmost at first). So,
int valid_array[][5]; // Ok
int invalid_array[5][]; // Wrong
(You can chain them... but we seldom have reason to do so...)
int (*convoluted_array[][5])[][10];
There is a catch, and the catch is that an array type with []
in it is an incomplete type. You can pass around a pointer to an incomplete type but certain operations will not work, as they need a complete type. For example, this will not work:
void func(int (*x)[])
{
x[2][5] = 900; // Error
}
This is an error because in order to find the address of x[2]
, the compiler needs to know how big x[0]
and x[1]
are. But x[0]
and x[1]
have type int []
-- an incomplete type with no information about how big it is. This becomes clearer if you imagine what the "un-decayed" version of the type would be, which is int x[][]
-- obviously invalid C. If you want to pass a two-dimensional array around in C, you have a few options:
Pass a one-dimensional array with a size parameter.
void func(int n, int x[])
{
x[2*n + 5] = 900;
}
Use an array of pointers to rows. This is somewhat clunky if you have genuine 2D data.
void func(int *x[])
{
x[2][5] = 900;
}
Use a fixed size.
void func(int x[][5])
{
x[2][5] = 900;
}
Use a variable length array (C99 only, so it probably doesn't work with Microsoft compilers).
// There's some funny syntax if you want 'x' before 'width'
void func(int n, int x[][n])
{
x[2][5] = 900;
}
This is a frequent problem area even for C veterans. Many languages lack intrinsic "out-of-the-box" support for real, variable size, multidimensional arrays (C++, Java, Python) although a few languages do have it (Common Lisp, Haskell, Fortran). You'll see a lot of code that uses arrays of arrays or that calculates array offsets manually.
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