hi ..
the question may sound bit absurd..but answer to it would be of gr8 help to me !!
how can i pass a 2D array to a function whose argument needs a 2D pointer...
e.g.
i want to pass a
to a functionCode:char buff[4][4];
Code:void fun(char **inpbuff, int row, int col) { int i,j; /*do some operation here on all elements of the array passed. e.g.*/ for(i = 0; i < row; i++) for (j = 0; j <= col; j++ ) inpbuff[i][j] = 'a' + i + j; }
No, doesn't work (we've had a discussion on that before, how an array is basically a pointer). I believe this does:
int func(int two_d[][4])
Note: just like array syntax, but specify outer dimension (columns).
Originally Posted by Adak
It is because array is not a pointer. It is casted to pointer when needed. The automatic cast in opposite direction is not available
All problems in computer science can be solved by another level of indirection,
except for the problem of too many layers of indirection.
– David J. Wheeler
I disagree. An array is a pointer. Do you want to argue?
casting it should work....
edit: nevermind it doesn't =(
Last edited by dra; 02-12-2008 at 01:16 AM.
You could read again the thread... 2 types having different size could not be the same.
All problems in computer science can be solved by another level of indirection,
except for the problem of too many layers of indirection.
– David J. Wheeler
An array requires a special pointer syntax that specifies the size of the memory to be allocated, thus the compiler can also see the actual size of the array inside the function it's defined. When passed to another function, its true nature shows as a pointer and you can't get the size.
I could well argue about arrays beings pointers, but this is not the place.
You are confused, you need help.
An array and a pointer are DIFFERENT. You have a serious misconception.
An array is a collection of homogeneous objects stored in contiguous memory.
A pointer contains a reference to an object or function.
THERE IS NO SIMILARITY THERE!!
Your confusion, which is common, comes from all the cases where arrays decay to a pointer to their first element. There are only a handful of cases where they do not (sizeof operator, ++, &, that's all I can think of off the top of my head).
An array of arrays decays to a pointer to its first element, which is another array, so the type is "pointer to array" NOT A POINTER TO POINTER!!
That is also why casts DO NOT WORK!
Stop posting in threads like this until you understand what you are doing with pointers, arrays, and indirect addressing. There is already enough confusion and misunderstanding.
What's the big deal? I know fully what pointers and arrays are. No misconception.
However, believe what you want, I don't think there is anything "right."
Believe what makes it easier for you.
Last edited by Elysia; 02-12-2008 at 01:47 AM.
You know they do not - look at the first post
array is r-value, pointer is l-value. You see the differencies, but still try to argue...
All problems in computer science can be solved by another level of indirection,
except for the problem of too many layers of indirection.
– David J. Wheeler
The whole l-value, r-value stuff is confusing, I'll give you that.
No matter, this isn't going anywhere, so let's all just think what we want and get along.
That's the spirit! Heaven forbid that you actually learn something. Hard-headed ignorance is bliss. /sarcasm
technosavvy, if you don't want to change your function (especially if you need to use it on arrays of different sizes), you can also use an intermediate array of pointers:
Don't mind my blustering, Elysia's tip about changing the parameter type is also correct, even though the explanation was wrong. Choose what works best for you, but consider using a vector instead of an array altogether.Code:char buff[4][4]; char *pb[4]; for ( int i = 0; i < 4; ++i ) pb[i] = buff[i]; // use the automatic array-to-pointer conversion fun(pb, 4, 4); ... void fun(char **inpbuff, int row, int col) { int i,j; /*do some operation here on all elements of the array passed. e.g.*/ for(i = 0; i < row; i++) for (j = 0; j <= col; j++ ) inpbuff[i][j] = 'a' + i + j; }
There's no need to learn anything here. I know what arrays and pointers are and the whole ordeal that arrays decay into pointers is just absurd, so I'm sticking with my way of thinking.
There's a reason for everything.
An array is not a pointer, and the C++ Standard says so.An array is a pointer.
You are arguing against the C++ Standard. There is nothing to argue, since what you state is factually incorrect.Do you want to argue?
Look up a C++ Reference and learn How To Ask Questions The Smart Way
Even if it may sound rude or arrogant, I don't care what the standard says. I believe in assembly and assembly says it's a pointer.
This silly discussion will get nowhere, so why bother?
