"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<iostream>
using namespace std;
int main(int argc, char* argv[]){

const double L = 1.234;
const int T = static_cast<const 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********@gmail.com wrote:

Hi:
I have the following simple program:

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

const double L = 1.234;
const int T = static_cast<const 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<iostream>

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

--
Erik Wikström

[Daniel T.]

hn********@gmail.com wrote:

Hi:
I have the following simple program:

#include<iostream>
using namespace std;
int main(int argc, char* argv[]) {
const double L = 1.234;
const int T = static_cast<const 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<const 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********@gmail.com wrote:

>Hi:
I have the following simple program:

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

const double L = 1.234;
const int T = static_cast<const 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<iostream>

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<const int>(L);
return 0;
}

Best

Kai-Uwe Bux

[Pete Becker]

Daniel T. wrote:

hn********@gmail.com wrote:

>Hi:
I have the following simple program:

#include<iostream>
using namespace std;
int main(int argc, char* argv[]) {
const double L = 1.234;
const int T = static_cast<const 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<const 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**@versatilecoding.comwrote:

Daniel T. wrote:

hn********@gmail.com wrote:

Hi:
I have the following simple program:
>
#include<iostream>
using namespace std;
int main(int argc, char* argv[]) {
const double L = 1.234;
const int T = static_cast<const 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<const 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********@gmail.com wrote:

Hi:
I have the following simple program:

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

const double L = 1.234;
const int T = static_cast<const 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<iostream>

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

--
Erik Wikström

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

#include<iostream>

int main()
{
#define L 1.234
#define T static_cast<const 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**@versatilecoding.comwrote:

>Daniel T. wrote:

>>hn********@gmail.com wrote:

Hi:
I have the following simple program:

#include<iostream>
using namespace std;
int main(int argc, char* argv[]) {
const double L = 1.234;
const int T = static_cast<const 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<const 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<const 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 <[e-mail]:

Erik Wikström wrote:

On 2007-01-20 17:08, [e-mail] wrote:

Hi:
I have the following simple program:

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

const double L = 1.234;
const int T = static_cast<const 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<iostream>

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, [e-mail] wrote:

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

#include<iostream>

int main()
{
#define L 1.234
#define T static_cast<const 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<iostream>

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

--
Erik Wikström

[hn.ft.pris]

"Erik Wikström 写�：
"

On Jan 21, 10:51 am, [e-mail] wrote:

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

#include<iostream>

int main()
{
#define L 1.234
#define T static_cast<const 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<iostream>

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

Yes, I get it.
But how to explain the following odd behavior of MS VS8:

int main(int argc, char** argv){
#define L 1.234
#define T static_cast<int>(L)
int arr[T];
arr[0] = 1;
cout << arr[0] << endl;
return 0;
}
Fine, without any warnings or errors.

int main(int argc, char** argv){
double L = 1.234;
int T = static_cast<int>(L);
int arr[T];
arr[0] = 1;
cout << arr[0] << endl;
return 0;
}
error C2057: expected constant expression; error C2466: cannot allocate
an array of constant size 0, on MS VS8
while gcc 3.4.2 approve it.

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<const int>(1.234)];
delete[] arr; // don't forget this when you are done
return 0;
}

--
Erik Wikström

[Kai-Uwe Bux]

Erik Wikström wrote:

On Jan 20, 8:10 pm, Kai-Uwe Bux <[e-mail]:

>Erik Wikström wrote:

On 2007-01-20 17:08, [e-mail] wrote:
Hi:
I have the following simple program:

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

>const double L = 1.234;
const int T = static_cast<const 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.

The static_cast explanation is wanting:

unsigned char const c = 'c';
unsigned int const i = static_cast<int>( c );

typedef char array [i];

int main ( void ) {
array A;
}

The above compiles fine and is, as far as I can tell standard conforming
even though it uses a static cast. The crux of the matter seems to be that
non-integer arithmetic expressions do not qualify as compile time constants
(allowing the compiler and the target processor to differ in their opinion
on how to approximate pi properly).

>

which it needs to be for the
compiler to know how much space to allocate on the stack.

[non-disputed facts snipped]
Best

Kai-Uwe Bux

[Greg]

Kai-Uwe Bux wrote:

>
unsigned char const c = 'c';
unsigned int const i = static_cast<int>( c );

typedef char array [i];

int main ( void ) {
array A;
}

The above compiles fine and is, as far as I can tell standard conforming
even though it uses a static cast. The crux of the matter seems to be that
non-integer arithmetic expressions do not qualify as compile time constants
(allowing the compiler and the target processor to differ in their opinion
on how to approximate pi properly).

Although this program is legal:

unsigned int const i = 3.14159;

typedef char array[i];

int main()
{
array A;
}

Greg

[Kai-Uwe Bux]

Greg wrote:

>
Kai-Uwe Bux wrote:

>>
unsigned char const c = 'c';
unsigned int const i = static_cast<int>( c );

typedef char array [i];

int main ( void ) {
array A;
}

The above compiles fine and is, as far as I can tell standard conforming
even though it uses a static cast. The crux of the matter seems to be
that non-integer arithmetic expressions do not qualify as compile time
constants (allowing the compiler and the target processor to differ in
their opinion on how to approximate pi properly).

Although this program is legal:

unsigned int const i = 3.14159;

typedef char array[i];

int main()
{
array A;
}

This is making me dizzy. Here is what the standard has to say [5.19/1]:

[...] An integral constant-expression can involve only literals (2.13),
enumerators, const variables or static data members of integral or
enumeration types initialized with constant expressions (8.5), non-type
template parameters of integral or enumeration types, and sizeof
expressions. Floating literals (2.13.3) can appear only if they are cast
to integral or enumeration types. Only type conversions to integral or
enumeration types can be used. In particular, except in sizeof
expressions, functions, class objects, pointers, or references shall not
be used, and assignment, increment, decrement, function-call, or comma
operators shall not be used.

So, it appears that the only legal way that a float can make it into an
integral constant-expression is as a floating literal converted to an
integral or enumeration type. In particular, const variables of floating
point types do not qualify even if they are initialized with a (floating
point) constant-expression.

Makes you wonder what the rational might be.
Best

Kai-Uwe Bux