can an array be declared by a non-constant value?

[nand]

Hello,

There is much documentation on the internet about always declaring arrays by constant values, eg. int array[10]; because

NOTE: The elements field within brackets [] which represents the number of elements the array is going to hold, must be a constant value, since arrays are blocks of non-dynamic memory whose size must be determined before execution. In order to create arrays with a variable length dynamic memory is needed, which is explained later in these tutorials.

http://www.cplusplus.com/doc/tutorial/arrays/

Please consider the code below:

#include <iostream>
#include<cstdlib>
using namespace std;
int main(int argc, char** argv)
{
	int n;
	cout<<"\n Please enter array lenghth";
        cin>>n;
	int sir[n];
	return 0;
}

This seems valid C/C++ code to me and it compiles on my machine. I am using GCC 4.3.2
Does declaring an array by a non-constant value seem correct to you?
Please try if this compiles on your compiler.
Thank you,
Andrei

[Athar]

It's not correct, the size must be a compile-time constant in C++.
However, GCC supports variable length arrays as an extension (note that C supports them since C99).

[shahj]

Yes, You can do that but before you as a number from a user, just define int n = 1 ( or anything)
Therefore, when user enter their value size will change.

#include <iostream>
#include<cstdlib>
using namespace std;
int main(int argc, char** argv)
{
	int n = 1;
	cout<<"\n Please enter array lenghth";
        cin>>n;
	int sir[n];
	return 0;
}

This should work...

[Athar]

This should work...

...
Whether you do some superfluous initialization of n or not has no effect on it being (not) a compile-time constant.

That's how you do it correctly in C++:

#include <iostream>
#include <vector>
using namespace std;

int main()
{
	int n;
	cout << "Please enter array length: ";
        cin >> n;
	vector<int> sir(n);
}

[ne555]

man gcc wrote:

-pedantic
Issue all the warnings demanded by strict ISO C and ISO C++; reject all programs that use forbidden extensions, and some other programs that do not follow ISO C and ISO C++.

[...]it finds some non-ISO practices, but not all---only those for which ISO C requires a diagnostic, and some others for which diagnostics have been added.

A feature to report any failure to conform to ISO C might be useful in some instances, but would require considerable additional work and would be quite different from -pedantic. We don't have plans to support such a feature in the near future.

-pedantic-errors
Like -pedantic, except that errors are produced rather than warnings.

[Mathhead200]

Athar: http://cplusplus.com/forum/general/45467/#msg246746

There is no guarantee that a vector is stored consecutively in memory. If you want dynamic memory allocation (for arrays) you should use new T[N], where T is the data type and N is the size.
http://cplusplus.com/doc/tutorial/dynamic

[LB]

If there is no guarantee that it is stored consecutively in memory, then why do vectors have a capacity and why must they reallocate when max capacity is reached? Wouldn't this suggest that they use consecutive memory locations to store information?

[Disch]

There is no guarantee that a vector is stored consecutively in memory.

AFAIK, there is. That's the whole point of vector.

Though, for other containers like deque, string, etc, you'd be right.

[jsmith]

Yes, vector nowadays guarantees contiguous allocation.

[shahj]

vector will work but the code that i have pasted earlier is working ( i checked it.) I am not sure why its not working in yours.

[firedraco]

vector will work but the code that i have pasted earlier is working ( i checked it.) I am not sure why its not working in yours.

It doesn't matter whether it works for you. The point is that it doesn't have to.