"error C2057: expected constant expression", "error C2466: cannot allocate an array of constant size 0". Why doesn't my simple program work???

[hn.ft.pris]

Hi:
I have the following simple program:

#include<iostre am>
using namespace std;
int main(int argc, char* argv[]){

const double L = 1.234;
const int T = static_cast<con st int>(L);
int arr[T];
return 0;
}

But I get the error message shown in title. Why doesn't my program
work? Thanks for help!

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

On 2007-01-20 17:08, hn********@gmai l.com wrote:

Hi:
I have the following simple program:

#include<iostre am>
using namespace std;
int main(int argc, char* argv[]){

const double L = 1.234;
const int T = static_cast<con st int>(L);
int arr[T];
return 0;
}

But I get the error message shown in title. Why doesn't my program
work? Thanks for help!

Because T is not constant at compile-time, which it needs to be for the
compiler to know how much space to allocate on the stack. That, of
course, is not the correct explanation according to the standard but it
quite nicely describes what is going on. Consider the following code in
which T is also constant but can wary with every run of the application:

#include<iostre am>

int main()
{
double L;
std::cin >L;
const int T = static_cast<con st int>(L);
return 0;
}

--
Erik Wikström

[Daniel T.]

hn********@gmai l.com wrote:

Hi:
I have the following simple program:

#include<iostre am>
using namespace std;
int main(int argc, char* argv[]) {
const double L = 1.234;
const int T = static_cast<con st int>(L);
int arr[T];
return 0;
}

Why doesn't my program work?

Good question. The code compiled fine for my g++ compiler, but not for
http://www.comeaucomputing.com/tryitout I have some questions about
comeau's output:

"ComeauTest .c", line 6: warning: type qualifier is meaningless on cast type
const int T = static_cast<con st int>(L);
^

What is this? Isn't the 'const' required in the above context?

"ComeauTest .c", line 7: error: constant value is not known
int arr[T];
^

Why is that?

[Kai-Uwe Bux]

Erik Wikström wrote:

On 2007-01-20 17:08, hn********@gmai l.com wrote:

>Hi:
I have the following simple program:

#include<iostre am>
using namespace std;
int main(int argc, char* argv[]){

const double L = 1.234;
const int T = static_cast<con st int>(L);
int arr[T];
return 0;
}

But I get the error message shown in title. Why doesn't my program
work? Thanks for help!

Because T is not constant at compile-time,

Huh? How is that?

which it needs to be for the
compiler to know how much space to allocate on the stack. That, of
course, is not the correct explanation according to the standard but it
quite nicely describes what is going on. Consider the following code in
which T is also constant but can wary with every run of the application:

#include<iostre am>

int main()
{
double L;

You are cheating in the line above. The OP had:

const double L = 1.234;
^^^^^

std::cin >L;

With the line of the OP, this would be undefined behavior.

const int T = static_cast<con st int>(L);
return 0;
}

Best

Kai-Uwe Bux

[Pete Becker]

Daniel T. wrote:

hn********@gmai l.com wrote:

>Hi:
I have the following simple program:

#include<iostre am>
using namespace std;
int main(int argc, char* argv[]) {
const double L = 1.234;
const int T = static_cast<con st int>(L);
int arr[T];
return 0;
}

Why doesn't my program work?

Good question. The code compiled fine for my g++ compiler, but not for
http://www.comeaucomputing.com/tryitout I have some questions about
comeau's output:

>"ComeauTest.c" , line 6: warning: type qualifier is meaningless on cast type
const int T = static_cast<con st int>(L);
^

What is this? Isn't the 'const' required in the above context?

>"ComeauTest.c" , line 7: error: constant value is not known
int arr[T];
^

Why is that?

In order for T to be a compile-time constant, it has to have an
initializer that's a compile-time constant. Floating-point values aren't.

--

-- Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com)
Author of "The Standard C++ Library Extensions: a Tutorial and
Reference." (www.petebecker.com/tr1book)

[Daniel T.]

Pete Becker <pe**@versatile coding.comwrote :

Daniel T. wrote:

hn********@gmai l.com wrote:

Hi:
I have the following simple program:
>
#include<iostre am>
using namespace std;
int main(int argc, char* argv[]) {
const double L = 1.234;
const int T = static_cast<con st int>(L);
int arr[T];
return 0;
}
>
Why doesn't my program work?

Good question. The code compiled fine for my g++ compiler, but not for
http://www.comeaucomputing.com/tryitout I have some questions about
comeau's output:

"ComeauTest .c", line 6: warning: type qualifier is meaningless on cast type
const int T = static_cast<con st int>(L);
^

What is this? Isn't the 'const' required in the above context?

What about this warning?

"ComeauTest .c", line 7: error: constant value is not known
int arr[T];
^

Why is that?

In order for T to be a compile-time constant, it has to have an
initializer that's a compile-time constant. Floating-point values aren't.

I didn't know that. Thanks.

[hn.ft.pris]

"Erik Wikström 写�：
"

On 2007-01-20 17:08, hn********@gmai l.com wrote:

Hi:
I have the following simple program:

#include<iostre am>
using namespace std;
int main(int argc, char* argv[]){

const double L = 1.234;
const int T = static_cast<con st int>(L);
int arr[T];
return 0;
}

But I get the error message shown in title. Why doesn't my program
work? Thanks for help!

Because T is not constant at compile-time, which it needs to be for the
compiler to know how much space to allocate on the stack. That, of
course, is not the correct explanation according to the standard but it
quite nicely describes what is going on. Consider the following code in
which T is also constant but can wary with every run of the application:

#include<iostre am>

int main()
{
double L;
std::cin >L;
const int T = static_cast<con st int>(L);
return 0;
}

--
Erik Wikström

Thanks for help, and here I've made some modification to the program.

#include<iostre am>

int main()
{
#define L 1.234
#define T static_cast<con st int>(L)
int arr[T];
return 0;
}

Does it mean that now T is a "compile-time determined" variable?
If it does, then could I suppose that "#define" is expanded and
calculated at complie-time?
Does all "#define" perform in a same way?
Thanks a lot for helping me.

[Clark S. Cox III]

Daniel T. wrote:

Pete Becker <pe**@versatile coding.comwrote :

>Daniel T. wrote:

>>hn********@gmai l.com wrote:

Hi:
I have the following simple program:

#include<iostre am>
using namespace std;
int main(int argc, char* argv[]) {
const double L = 1.234;
const int T = static_cast<con st int>(L);
int arr[T];
return 0;
}

Why doesn't my program work?
Good question. The code compiled fine for my g++ compiler, but not for
http://www.comeaucomputing.com/tryitout I have some questions about
comeau's output:

"ComeauTest. c", line 6: warning: type qualifier is meaningless on cast type
const int T = static_cast<con st int>(L);
^
What is this? Isn't the 'const' required in the above context?

What about this warning?

For built-in types, there is no difference between a const rvalue and a
non-const rvalue. That means that there is no difference between the
following casts (they both produce an rvalue of type int):

static_cast<int >(1.234)
static_cast<con st int>(1.234)

So the const is meaningless in this context.
--
Clark S. Cox III
cl*******@gmail .com

[=?iso-8859-1?q?Erik_Wikstr=F6m?=]

On Jan 20, 8:10 pm, Kai-Uwe Bux <jkherci...@gmx .netwrote:

Erik Wikström wrote:

On 2007-01-20 17:08, hn.ft.p...@gmai l.com wrote:

Hi:
I have the following simple program:

#include<iostre am>
using namespace std;
int main(int argc, char* argv[]){

const double L = 1.234;
const int T = static_cast<con st int>(L);
int arr[T];
return 0;
}

But I get the error message shown in title. Why doesn't my program
work? Thanks for help!

Because T is not constant at compile-time,

Huh? How is that?

Because the program have to execute the static_cast before the value of
T can be determined. And to declare an array on the stack you need to
know the size of the array at compile-time. Perhaps I was a bit
unclear, what I meant was that T was not a constant value at
compile-time.

which it needs to be for the
compiler to know how much space to allocate on the stack. That, of
course, is not the correct explanation according to the standard but it
quite nicely describes what is going on. Consider the following code in
which T is also constant but can wary with every run of the application:

#include<iostre am>

int main()
{
double L;

You are cheating in the line above. The OP had:

const double L = 1.234;
^^^^^

Yes, but it's the value (and constness) of T that is of interest in
this problem.

--
Erik Wikström

[=?iso-8859-1?q?Erik_Wikstr=F6m?=]

On Jan 21, 10:51 am, hn.ft.p...@gmai l.com wrote:

Thanks for help, and here I've made some modification to the program.

#include<iostre am>

int main()
{
#define L 1.234
#define T static_cast<con st int>(L)
int arr[T];
return 0;

}

Does it mean that now T is a "compile-time determined" variable?

No, what you have done is to use macros to restructure your code a bit,
after this has been through the pre-processor this is what you'll get:

#include<iostre am>

int main()
{
int arr[static_cast<con st int>(1.234)];
return 0;
}

If it does, then could I suppose that "#define" is expanded and
calculated at compile-time?

They are expanded, yes, but not calculated. Or rather the static_cast
is not evaluated until runtime, which is to late since the size of the
array needs to be known at compile-time.

Does all "#define" perform in a same way?

All defines are expanded by the pre-processor, but that is all, if you
want code evaluated at compile-time you can do some things with
templates, but I don't think that's what you want.

What you can do is to allocate the array on the heap:

int main()
{
int* arr = new int[static_cast<con st int>(1.234)];
delete[] arr; // don't forget this when you are done
return 0;
}

--
Erik Wikström