Does char *fld[NUMFLD] mean I have a pointer to a array of strings? So NUMFLD will contain strings located at NUMFLD 0,1,2 and so on... I then reference them with my pointer like *fld[1], fld[2]... this has got me all confused after I thought I was making progress. Can someone tell me what this means and give me a example in how it works. Thanks.
-Carl
No -- there's really no such thing as a "pointer to an array." The declaration means an array of pointers (to strings, presumably). fld[0] points to the first string, fld[1] to the second, etc.Originally Posted by cjohnman
There is such a thing, but it's rarely used, in my experience. C sneaks them in behind your back when you're passing multidimensional arrays to functions:
Here x is actually a pointer to an array of 5 int. This could also be written:Code:void f(int x[][5]);
You can, of course, also explicitly use pointers to arrays in your code, though I don't ever recall having to do this:Code:void f(int (*x)[5]);
Not especially useful, but there you are. If you mentioned that there's really no such thing as pointers to arrays in order to not confuse those new to C, sorry for blowing the operation.Code:#include <stdio.h> int main(void) { int a[1] = { 5 }; int (*p)[1] = &a; printf("%d\n", (*p)[0]); return 0; }
No, there's no such thing as a pointer to an array. Even your example is a plain pointer. You're just using a different syntax to tell the compiler the dimensions of the array.
You can see this for more info:
http://cpwiki.sourceforge.net/A_pointer_on_pointers
Last edited by Elysia; 04-10-2008 at 11:52 PM.
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
Or, of course, there's this:
Of course, I didn't think of allocating an array on the heap which would technically make it a pointer to an array.
Well, whatever goes. I don't know if I want to call cas's example as a pointer to an array because that's technically wrong. It's defined as a pointer to an array, yet it's a multi-dimensional array - types do not agree.
What are you talking about?
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
Looks to my like a pointer a array of 1 dimension with 5 elements (though it may not be, deceiving syntax).Code:void f(int (*x)[5]);
And in the end, it's just a simple pointer that's passed to the function because the address of the start of the array is passed, so technically it's just a pointer.
Type is important for things such as ++x.
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
If you read the C standard, it's quite clear that pointers to arrays exist, and that what I posted is an example of one. However, you must first acknowledge that pointers and arrays are separate things, I suppose, and if you do not do that, it would be hard to convince you that a pointer to an array can exist.
I see things from a lower perspective. An array is just a block of memory. And to address that block of memory, a pointer is required. Although if it's placed on the stack, then the compiler just needs to move memory from a stack pointer offset, wheras if the array is passed to another function, the address to the start of the array is passed, thus effectively making it a pointer that must be dereferenced first.
