Need help in my array

[sdlt85]

This is part of my program but for some reason it is not taken the
value of c in the array P[n]. Can some one help me please.

int main()
{
int x, c, p;

cout<<"Enter the value for x: ";
cin>>x;
cout<<"How many coefficients? ";
cin>>c;

const int n = c-1;
int P[n]; // here are the errors: expected constant
expression,
//cannot allocate an
array of constant size 0, 'P': unknown size.
}

Thanks

[Jon Slaughter]

<sd****@gmail.comwrote in message
news:11*********************@q75g2000hsh.googlegro ups.com...

This is part of my program but for some reason it is not taken the
value of c in the array P[n]. Can some one help me please.

int main()
{
int x, c, p;

cout<<"Enter the value for x: ";
cin>>x;
cout<<"How many coefficients? ";
cin>>c;

const int n = c-1;
int P[n]; // here are the errors: expected constant
expression,
//cannot allocate an
array of constant size 0, 'P': unknown size.
}

You cannot allocate a static array with a dynamic size.
The compiler doesn't know what n is until after you run the program... so it
cannot allocate the space before hand... it doesn't know thatyou want a
dynamic array.

you need something more special:

int P[] = new int[c-1];

Then it should work.

Better yet you might want to look at STL vector or list... they are similar
but implemented in classes and have some other functionality

you can do that by

#include <vector>

vector<intP;

and you don't have to worry about the size as it will deal with it
automatically.

to add elements though you need to use something like P.push_back(342);

Look up details on it for more info.

[Andre Kostur]

sd****@gmail.com wrote in
news:11*********************@q75g2000hsh.googlegro ups.com:

This is part of my program but for some reason it is not taken the
value of c in the array P[n]. Can some one help me please.

int main()
{
int x, c, p;

cout<<"Enter the value for x: ";
cin>>x;
cout<<"How many coefficients? ";
cin>>c;

const int n = c-1;
int P[n]; // here are the errors: expected
constant
expression,

Yep. Arrays must have a compile-time constant size. Yours does not. c
isn't known at compile time, thus c-1 isn't known, thus n isn't known.

You may wish to consider using std::vector<int>:

std::vector<intP(n);

//cannot allocate an
array of constant size 0, 'P': unknown size.
}

Thanks

[sumedh.....]

yes,
Array is nothing but a pointer.. that means: int p[4]; can be
represented as *(p+i)
so another way of declaring it is
int *p=new int[n]; //and n doesnot need to be a constant..

thanks,
Sumedh

[Andre Kostur]

"sumedh....." <su**********@gmail.comwrote in
news:11*********************@i13g2000prf.googlegro ups.com:

yes,
Array is nothing but a pointer.. that means: int p[4]; can be
represented as *(p+i)
so another way of declaring it is
int *p=new int[n]; //and n doesnot need to be a constant..

Not the same thing. The two allocations aren't necessarily from the same
location. In some implementations "int p[4]" only requires a stack pointer
to be moved. new int[4] requires a visit to a dynamic memory allocator
(generally slower).

[=?ISO-8859-1?Q?Erik_Wikstr=F6m?=]

On 2007-06-06 09:11, Andre Kostur wrote:

"sumedh....." <su**********@gmail.comwrote in
news:11*********************@i13g2000prf.googlegro ups.com:

>yes,
Array is nothing but a pointer.. that means: int p[4]; can be
represented as *(p+i)
so another way of declaring it is
int *p=new int[n]; //and n doesnot need to be a constant..

Not the same thing. The two allocations aren't necessarily from the same
location. In some implementations "int p[4]" only requires a stack pointer
to be moved. new int[4] requires a visit to a dynamic memory allocator
(generally slower).

And it requires a manual deallocation, which makes it a bit more error
prone.

--
Erik Wikström

[James Kanze]

On Jun 6, 8:07 am, "sumedh....." <[e-mail]:

Array is nothing but a pointer..

Nonsense. Arrays and pointers are two completely different
things.

that means: int p[4]; can be
represented as *(p+i)

I think you mean that the expression p[i] is the same as *(p+i).
This is a misfeature of C, and doesn't hold if you use C++ style
arrays (i.e. std::vector). And it only holds in expressions; it
has nothing to do with the type of any objects.

so another way of declaring it is
int *p=new int[n]; //and n doesnot need to be a constant..

There is never any reason to use array new in C++. The correct
way to declare an array is:
std::vector< int array( n ) ;
About the only exception is when you need static initialization,
to avoid order of initialization issues.

--
James Kanze (GABI Software) email:ja*********@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, [phone]

[sdlt85]

On Jun 6, 5:07 am, James Kanze <[e-mail]:

On Jun 6, 8:07 am, "sumedh....." <[e-mail]:

Array is nothing but a pointer..

Nonsense. Arrays and pointers are two completely different
things.

that means: int p[4]; can be
represented as *(p+i)

I think you mean that the expression p[i] is the same as *(p+i).
This is a misfeature of C, and doesn't hold if you use C++ style
arrays (i.e. std::vector). And it only holds in expressions; it
has nothing to do with the type of any objects.

so another way of declaring it is
int *p=new int[n]; //and n doesnot need to be a constant..

There is never any reason to use array new in C++. The correct
way to declare an array is:
std::vector< int array( n ) ;
About the only exception is when you need static initialization,
to avoid order of initialization issues.

--
James Kanze (GABI Software) email:[e-mail]
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, [phone]

Alright thanks ^_^