I'm trying to define a stack c-style array whose size is taken from const array and is known in compile-time.
const int size[2]={100, 100};
int ar[size[0]]; //error: expression must have a constant value
It fails. How it can be fixed?
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1It works with g++4.4. It works also on ideonecpp– cpp2013-09-24 14:59:39 +00:00Commented Sep 24, 2013 at 14:59
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if you know the size at compile time you can always stick that in as the array sizemsam– msam2013-09-24 15:06:14 +00:00Commented Sep 24, 2013 at 15:06
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1@cpp: That's because GCC supports C-style variable-length arrays as a non-standard extension.Mike Seymour– Mike Seymour2013-09-24 15:20:37 +00:00Commented Sep 24, 2013 at 15:20
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3 Answers
"array whose size is taken from const array and is known in compile-time"
With C++11 you can have :
constexpr int size[2]={100, 100}; // size[0] is Compile-time constant
Use -std=c++11 or -std=c++0x to compile
1 Comment
Mike Seymour
@cpp: No it doesn't. It works with a hybrid of C++98 and C99 that supports variable-length arrays, but in standard C++ you need
constexpr to use array elements as compile-time constants - and that wasn't available before 2011.Some options (with varying degrees of popularity):
- Use constexpr (not supported in Visual Studio)
- Use a Vector
- Use dynamic allocation
- Use a
const int(C99 or newer or C++) - Use an
enum - Use a MACRO to define the size (since it's known at compile time)
2 Comments
Mark Kahn
1. constexpr could be a the best solution but it is not supported in Visual Studio 2010 2.3. Use a Vector/Use dynamic allocation -- this is not what I need 4. Use a MACRO / define - is not better than just using consts
msam
Since you're using C++ you can do that, this is also possible if using a modern C compiler (C99 or newer - this will technically be a variable length array) but will not work on MSVC if you compile the code as C code.
C++ array sizes must be constant expressions, not just constant data. Array data, even though const, is not a constant expression.
