gcc: static allocation of array with variable size

[enri]

Hi everyone.
Is anybody able to explain to me why gcc will compile without the slightest warning the following code?

#include <iostream>
 
int main()
{
   int dim;
 
   cin>> dim;
   double array[dim];
 
   // operations on the array
 
   return 0;
}

[sicarie]

Hi everyone.
Is anybody able to explain to me why gcc will compile without the slightest warning the following code?

#include <iostream>
 
int main()
{
   int dim;
 
   cin>> dim;
   double array[dim];
 
   // operations on the array
 
   return 0;
}

Because nothing is illegal. You declare an int, read it in from the user, and then declare an array of doubles that size. If you tried to use it as the array name, I'm pretty sure gcc would get upset.

[enri]

Because nothing is illegal. You declare an int, read it in from the user, and then declare an array of doubles that size. If you tried to use it as the array name, I'm pretty sure gcc would get upset.

For what I knew, static allocation, as "double vect[dim];", requires the compiler to know the dimension of the required memory at compile time. In my piece of
code this is not the case.
Am I right?

To make things worse, the same code is not compiled by Visual C++, since the value of "dim" is not known...

[horace1]

as sicarie says the array defininition should be OK - however, you should have using namespace std;

#include <iostream>
using namespace std;
 

what exactly was the error from the VC compiler?

[Ganon11]

VC++ and DevC++ both require that a regular array (no pointers) have a constant value of elements. Thus,

const int MAX_SIZE = 10;
int arr1[10];
double arr2[MAX_SIZE];

are both legal declarations - at compile time, the compiler knows that both arrays will have 10 elements. But you cannot create an array of variable size - so

int dim;
cin >> dim;
int arr3[dim];

is illegal, as the compiler cannot know what the user will enter. In order to implement this, a pointer must be used:

int dim;
cin >> dim;
int *arr4 = new int[dim];

Apparantly, the OPs compiler does not support this standard and allows a variable sized array to be created.

[horace1]

C99 compilers allow one to define a array whose size is determined at runtime. For example, my version of DEV-C++ which uses gcc accepts

   int i, dim;
 
   cin>> dim;
   double array[dim];
   cout << " array size " << sizeof(array) <<  endl;
   for(i=0;i<10;i++) array[i]=i;
   for(i=0;i<10;i++) cout << array[i] << " " ;
 

have a look at
http://www.informit.com/guides/content.asp?g=cplusplus&seqNum=215&rl=1

one tends to forget about older pre C99 compilers - Borland C++ V5.6 gives an error message saying it requires a constant expression.

[enri]

C99 compilers allow one to define a array whose size is determined at runtime. For example, my version of DEV-C++ which uses gcc accepts

   int i, dim;
 
   cin>> dim;
   double array[dim];
   cout << " array size " << sizeof(array) <<  endl;
   for(i=0;i<10;i++) array[i]=i;
   for(i=0;i<10;i++) cout << array[i] << " " ;
 

have a look at
http://www.informit.com/guides/content.asp?g=cplusplus&seqNum=215&rl=1

one tends to forget about older pre C99 compilers - Borland C++ V5.6 gives an error message saying it requires a constant expression.

Thank you all, very much,
that shed some light.

So, for what I understand, gcc uses C99 standard even when compiling plain C++, even if standard C++ does not support some of its features.

[AdrianH]

Thank you all, very much,
that shed some light.

So, for what I understand, gcc uses C99 standard even when compiling plain C++, even if standard C++ does not support some of its features.

Yeah, gcc/g+ has a lot of extensions. If you want it to be ANSI C++ compliant, use switch -ansi. To specify a particular standard gcc uses, use switch -std=. For more information do a 'info g++' on the command line and search for -ansi.


Adrian

[nmadct]

Actually that array isn't static, it's allocated on the stack. Stack memory is allocated at runtime so there's no problem in determining the size. As already mentioned, GCC allows this as a GNU extension. If you read the GCC documentation it clearly describes what features are GNU extensions. There are command-line switches to disable the extensions, and the -Wall switch makes GCC give better warnings. (Generally speaking, you should ALWAYS compile with -Wall and fix all warnings, as these warnings usually indicate a bug in your code, or a future bug waiting to happen.)

[nmadct]

So, for what I understand, gcc uses C99 standard even when compiling plain C++, even if standard C++ does not support some of its features.

GCC does not use C99 to compile C++, this would be incoherent because C and C++ are two different languages. It uses its own version of C++, which is basically the ISO C++ standard with GNU extensions. Some of the GNU extensions happen to involve adding features that also exist in the C99 standard.