Home > Gcc Error > Gcc Error Variable-size Type Declared Outside Of Any Function

Gcc Error Variable-size Type Declared Outside Of Any Function

Powered by vBulletin Version 4.2.3 Copyright © 2016 vBulletin Solutions, Inc. Browse more C / C++ Questions on Bytes Question stats viewed: 1801 replies: 12 date asked: Nov 14 '05 Follow this discussion BYTES.COM © 2016 Formerly "TheScripts.com" from 2005-2008 About Bytes Thus, for this particular case, you need to switch to macros.When you need an array size to be determined by a variable ( const or not ), you can do it BTW. this contact form

This is probably the problem you're encountering. Alex Nov 15 '05 #2 P: n/a Emmanuel Delahaye Bi*************@gmail.com wrote on 01/09/05 : Hi, The following code: #include // const int const_asize = 10; #define define_asize = 10; int Anyideas? #include #define arrsize(a) (sizeof(a) / sizeof(*a)) int array1[3]={3,2,1}; int array2[arrsize(array1)]; /*should be equiv. Frounze wrote in message news:11**********************@o13g2000cwo.googlegr oups.com... .... // const int const_asize = 10; #define define_asize = 10; int array[define_asize] = {1,2,3,4,5,6,7,8,9,0}; .... http://stackoverflow.com/questions/1435004/variable-size-type-declared-outside-of-any-function

I must never have been bitten by that one (yet) ;-) -leor -- Leor Zolman --- BD Software --- www.bdsoft.com On-Site Training in C/C++, Java, Perl and Unix C++ users: download Obviously, VLA are not permitted on static arrays. I'm wondering however what happens if a const is used to declare the size of an array. rcgldr, Sep 23, 2011 Sep 23, 2011 #5 uart Science Advisor Here's an interesting little twist to this.

The lack of a good way to create genuine constants in C is arguably a flaw in the language, but we're stuck with it. Comment 2 Jakub Jelinek 2001-02-05 16:08:30 EST Fixed in gcc-2.96-74. I know I'm doing something wrong, and that its something small. You can take rid of that restriction by templating the function on the array sizes.

A templated example might look like the following aberration: template void populate(int (&(&random)[Width])[Height], int x, int y); This is ugly. For ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ I don't think that's what the OP wanted (see underlined text). `array2'shouldn't have 3 elements because `array1' has three elements, but becausethe zeroth element of `array1' is 3. The keyword "const" doesn't create a constant. http://cboard.cprogramming.com/cplusplus-programming/13701-variable-size-type-declared-outside-any-function.html All rights reserved.

Save your draft before refreshing this page.Submit any pending changes before refreshing this page. As I said before, you can't have an unspecified array size upon declaration. The stuff in square brackets is the array dimensions, and those have to be set whenever you declare the array. For example:int array[3]={3,2,1};int x=array[0];main() { ... " Actually, the issues don't change much depending on /what/ you want to use that value for...the bottom line, as you've acknowledged below, is that

It is more likely that (2) will not have a symbol, leaving you wondering what it is. - (1) cannot be used as a dimension for arrays at global scope; both int extract_value( int (&a)[10][10], int row, int col ) { return a[row][col]; } int main() { int a[10][10] = {}; a[5][5] = 1; std::cout << extract_value( a, 5, 5 ) << int dim = 4; int ar[dim]; If dim were const, it would work because the compiler would be able to tell exactly how wide ar should be (because the value of By Drahcir in forum C Programming Replies: 8 Last Post: 02-04-2006, 01:53 AM Please Help - Problem with Compilers By toonlover in forum C++ Programming Replies: 5 Last Post: 07-23-2005, 10:03

Andrew->, Это чтобы не заниматься ерундой и объявить константу N для задания размерности @Andrew-> 0 / 0 / 0 Регистрация: 12.01.2010 Сообщений: 20 12.01.2010, 19:32 [ТС] ошибка variable-size type The more time we spend trying to distinguish between a possible typo and whatever is actually causing the problem you're asking about, the less time we can spend answering your question. The zeroed and unitialized data just have size information in a typcial program image (I think a MSDOS type .com program contains everything in one large chunk). For example: int array1[3]={3,2,1}; int array2[array1[0]]; //should be equiv.

What is so gosh-darn important about having to "extract it out of the array", anyway? But in the following piece of code: #include // const int const_asize = 10; #define define_asize = 10; int array[define_asize] = {1,2,3,4,5,6,7,8,9,0}; int main(int argc, char * argv[]) { return C++1 void fcn(int arr[100]); Тут размерность массива сразу указана, а тут C++1 void vvod(int N,int A[N],int I); Задается переменная, и эта переменная сразу выступает в кач-ве размерности массива. navigate here My guess is in the case of the C++ compiler, in the compile time symbol table, it's treats the const int ListSize also as a define, while the C compiler doesn't

If you must, then make "array2" a pointer to int and malloc the second array. Hurkyl, Sep 23, 2011 (Want to reply to this thread? What actually are virtual particles?

Try making it static.

e.g. Note You need to log in before you can comment on or make changes to this bug. In terms of allocation at compile time versus run time, it depends on what the operating system's loader is capable of when loading a program. In the following: int main(void) { #define TEN 10 const int ten = 10; int arr0[TEN]; int arr1[ten]; ... } arr0 is an ordinary array, but arr1 is a VLA.

Even if you can clearly read in the signature the dimensions of the array, those dimensions are ignored by the compiler. For most operating systems, a typical program includes these sections: code, data, stack, with data separated into initialized data, zeroed data (statics in c), and unitialized data. Also, are you compiling to the C99 standard, or the older one? his comment is here share|improve this answer answered Sep 16 '09 at 21:40 David Rodríguez - dribeas 151k12180385 add a comment| up vote 1 down vote I shall illustrate with an eg: //globals const int

Note also that VLAs are a new feature in C99, and some compilers may not yet implement them. One possibility is to define a macro which expands to `3', and use it to both initialize the zeroth element of `array1' and define the size of `array2'. The compiler must know, at compile time, the size of the array. My opinion is that it is not the best behaviour to differentiate between a literal and a const int when they're declared global, but make no distinction when they're local.

They seem to have come to the same conclusion as here. C++1 void fcn(int arr[100]); Тут размерность массива сразу указана, а тут C++1 void vvod(int N,int A[N],int I); Задается переменная, и эта переменная сразу выступает в кач-ве размерности массива. Yup; a case of overloaded 3's ;-) Thanks, -leor The answer is: it cannot be done directly. It's quick & easy.

De kio “saluton” estas la rekta objekto? Similar topics struct losing it's value when in a global array BYTE array size const as array size Variables for array size global array defined by parameters passed. Yes I know that the memory doesn't get allocated at compile time, what I meant is whether it would be literally using "ListSize" as a constant (80 in this case) at The difference is that variable length arrays (VLAs) are legal in a local context.

In C, the array sizer must be a constant expression (C99 allows VLA under certain conditions. One possibility is to define a macro which expands to `3', and use it to both initialize the zeroth element of `array1' and define the size of `array2'. If the compiler doesn't know how big the array is at the time the array is declared, it will tell you so and refuse to compile. I'm using Dev-C++ and the version of "gcc" in its bin directory is 3.4.2 This is not terribly important, I really don't have a problem using #define here.