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Gcc Error Variable-size Type Declared Outside Of Any Function

For example: int array1[3]={3,2,1}; int array2[array1[0]]; //should be equiv. Why was the identity of the Half-Blood Prince important to the story? Note You need to log in before you can comment on or make changes to this bug. To the question Which one is better to use among the below statements in c: static const int var=5; or #define var 5 I thought this response was good and very his comment is here

this is on gcc compiling with the following: gcc -o consttest main.c adding a -std=c99 does not make any difference. For example consider the two possible methods below. If the array has extern scope, the array size had also better have extern scope. As far as sizing an array, a "static const int" is not the way to go in C++ unless the array itself has static scope. http://stackoverflow.com/questions/1435004/variable-size-type-declared-outside-of-any-function

So the compiler differentiates between a literal and a const int when they're declared global, but make no distinction when they're local. i.e. However if you make that as an hash define #define ARRAY_SIZE 16 struct ArrayType_t arrayType[ARRAY_SIZE] ===> this works because ARRAY_SIZE is defined at compile time and compiler can know the size Has it been defined as part of C to refuse to compile above code with a const int specifying array size, when the array is declared global?

Meaning of "oh freak" reading through the definition of `\cfrac` in AMSMath Why (in universe) are blade runners called blade runners? rcgldr, Sep 23, 2011 Sep 23, 2011 #5 uart Science Advisor Here's an interesting little twist to this. Part 4: Cosmic Acoustics Interview with a Physicist: David J. Eh?

I know I'm doing something wrong, and that its something small. Martin -- ,--. Yup; a case of overloaded 3's ;-) Thanks, -leor The answer is: it cannot be done directly. http://cboard.cprogramming.com/cplusplus-programming/13701-variable-size-type-declared-outside-any-function.html The error message "variable-sized type declared outside of any function" seems to imply that it wouldn't be a problem if it was inside a function, but as you know error messages

Miki ustuzou% 03-23-2002 #8 SilentStrike View Profile View Forum Posts Visit Homepage geek Join Date Aug 2001 Location NJ Posts 1,141 Yeah.. Unfortunately I don't know if this is C99 or older. Anyideas? #include #define arrsize(a) (sizeof(a) / sizeof(*a)) int array1[3]={3,2,1}; int array2[arrsize(array1)]; /*should be equiv. Has it been defined as part of C to refuse to compile above code with a const int specifying array size, when the array is declared global?

to: int array2[3]; */ int main() { printf("sizeof(array1) = %d\n", sizeof(array1)); printf("sizeof(array2) = %d\n", sizeof(array2)); return 0; } Output: sizeof(array1) = 12 sizeof(array2) = 12 -leor -- Leor Zolman --- BD https://gcc.gnu.org/ml/gcc-help/2005-09/msg00001.html Bug39605 - "error: variable-size type declared outside of any function" is issued twice Summary: "error: variable-size type declared outside of any function" is issued twice Status: RESOLVED FIXED Alias: None Product: I've always gotten the feeling there was more to this problem than you've shown, but that if you were to fully explain the problem you're trying to solve, ultimately we'd all You can take rid of that restriction by templating the function on the array sizes.

Note that today's implementations of VLA are broken, even with gcc 4.x). this content Nov 15 '05 #4 P: n/a Bilgehan.Balban Keith Thompson wrote: It does distinguish between a literal and a const int in either context. This is something. more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed

Comment 1 Jakub Jelinek 2001-01-21 13:31:15 EST Verified both in 2.96-71 and current FSF CVS head. 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 De kio “saluton” estas la rekta objekto? http://imagextension.com/gcc-error/gcc-error-xitwas-not-declared-in-this-scope.php 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.

My recommendation would be to follow this last path: encapsulate the N-dimensional array into a class that provides conversions into a 1D vector (the C++FAQ lite uses a raw pointer, I phinds, Sep 23, 2011 Sep 23, 2011 #7 Hurkyl Staff Emeritus Science Advisor Gold Member uart said: ↑ The latter doesn't compile in standard C as it doesn't seem to recognize Index Nav: [DateIndex] [SubjectIndex] [AuthorIndex] [ThreadIndex] Message Nav: [DatePrev][DateNext] [ThreadPrev][ThreadNext] Other format: [Raw text] Re: "variable-size type declared" error for const sized array From: Bahadir Balban

Technically, using (1) would imply the use of a VLA (variable-length array), though the dimension referenced by 'var' would of course be fixed at size 5. - (1) cannot be used

Stay logged in Physics Forums - The Fusion of Science and Community Forums > Other Sciences > Programming and Computer Science > Menu Forums Featured Threads Recent Posts Unanswered Threads Videos Also, are you compiling to the C99 standard, or the older one? C++1 void vvod(int N,int A[],int I);C++1 void vvod(int N,int* A,int I); MoreAnswers 37091 / 29110 / 5898 : 17.06.2006 : 43,301 12.01.2010, 22:05 variable-size type declared outside The compiler can count the initializers and size the array for you : int array[] = {1,2,3,4,5,6,7,8,9,0}; You can retreive the size by applying the definition of an array (sequence of

template deduction of array size Browse more C / C++ Questions on Bytes Question stats viewed: 17683 replies: 4 date asked: Nov 15 '05 Follow this discussion BYTES.COM 2016 Formerly If you declare these variables global, then there available to all functions just by using their names. I tried making it static but it didn't change the error. check over here It's quick & easy.

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'. No, create an account now. These are called Variable-sized arrays.105 Views · Answer requested by Ankur GoyalView More AnswersRelated QuestionsObjective-C (programming language): How are member functions and variables declared as public or private?Why, in C, are These were all easily fixed but the one with the const just bothered me a bit as to why it had that behavior.

Are you familiar with neural networks? the following would be just fine, even preferred: Code (Text): // myheader.hh static const size_t my_array_size = 32; extern int my_array[my_array_size]; // mysource.cc #include "myheader.hh" int my_array[my_array_size]; Hurkyl, Sep

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