how send the size of an array as parameters through a function in c++?

Question

I work with Xcode, but now I want to learn and use Visual Studio C++ and my first challenge is to send and array through a function and its size as parameters, how can I accomplish this?

void Llena2(int R, int C, int (*XY)[C]); //in xcode

void Llena2(int R, int C, int (*XY)[C]); //error C2057: expected constant expression
                              //error C2466: cannot allocate an array of constant size 0

Is it possible to do some similar to xcode? Thanks in advance

@Claudio I have a real problem 'cause I've been reading about how to send the content of an array of characters (pointers type char) to another array of characters and until now I have no idea how to accomplish this in C++. Do you have any idea? — Hypatia
– Hypatia, Commented Jun 21, 2013 at 12:06

Community · Accepted Answer · 2020-06-20 09:12:55Z

4

You should use std::vector in this case.

Edit:

According to Incompatibilities Between ISO C and ISO C++, this feature: void test(int R, int C, int (*XY)[R][C]) (VLA) is only valid in C99 but not in C++.

C99 also provides new declaration syntax for function parameters of VLA types, allowing a variable identifier or a '*' to occur within the brackets of an array function parameter declaration in place of a constant integer size expression.

...

C++ does not support VLAs.

Xcode defaults to use C99, so it is valid in Xcode.

edited Jun 20, 2020 at 9:12

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answered Jun 5, 2013 at 10:56

Naruil

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2 Comments

Hypatia Over a year ago

@Naruil, I need to sorting the array of characters *Name[] = {"Al","Fe","Zr"}; I mean, at the end this array of characters must change into this {"Fe", "Zr","Al"}, how can I do that in C++? Thanks in advance

Naruil Over a year ago

The order in the result seems werid. It is not in ascending order or descending order. Can you explain more about your requirement?

Janman · Accepted Answer · 2013-06-05 12:34:33Z

1

You are initializing a new array in your function. An Array is (almost) a pointer. Therefore you can write:

void Llena2(int R, int C, int *XY);

edited Jun 5, 2013 at 12:34

answered Jun 5, 2013 at 11:01

Janman

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4 Comments

Hypatia Over a year ago

Thank you so much but I forgot to mention that is an 2D array, that's why I send in xcode R and C

Rome_Leader Over a year ago

This is correct. Your error is in this part: int(*XY)[C] You cannot pass the array like this, because C++ passes arrays by reference and doesn't take the array as an argument. You can, as pointed out, basically treat the array like a pointer, where *XY is a pointer that points to the first element of the array. If the size of the array is known to be C, you could also pass int C But if the size is unknown at compile time, I would suggest, as the above user has, to use a vector instead.

Janman Over a year ago

@Hypatia You can do that by passing an array of pointers. It is described here pretty well: stackoverflow.com/questions/8767166/… Anyways, vectors have improved a lot in C++11, and they are a lot easier to use. Try to use them instead.

Janman Over a year ago

@ComicSansMS Yea, that was before I knew it was a multi-dimensional array. I'm deleting it now.

Rick Yorgason · Accepted Answer · 2013-06-05 11:35:29Z

0

It's surprising that XCode allows this.

Is C constant? If so, you can turn the C paramter into a non-type template parameter:

#include <cstddef>

template <std::size_t C>
void Llena2(int R, int XY[][C]);

The compiler will automatically infer C at compile-time this way, but this only works with the outermost dimension.

answered Jun 5, 2013 at 11:35

Rick Yorgason

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2 Comments

Hypatia Over a year ago

C is not a constant, is the result of an equation and I take it as the size of the Column of the array

Hypatia Over a year ago

My real problem is that I translated an equations program (in MathLab) to C and now my tutor suggested me to translate the same program (in ANSI C) to a Visual Studio C++ because will be necessary to create a graphic interface, so now I really need to learn the best I can C++. Thank you so much for your help, means a lot to me

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