constant-expression

[Dan Smithers]

What constitutes a constant-expression? I know that it is something that
can be determined at compile time.

I am trying to use template code and keep getting compiler errors
"error: cannot appear in a constant-expression"

template <int s>
class CFoo
{
private:
int m_val[s];
};

struct SParams
{
const int m_sz;
SParams(int sz) : m_sz(sz) {;}
};

int main(int argc, char *argv[])
{
CFoo<2foo;

const int sz(2);
CFoo<szfoo2;

int sz2(3);
CFoo<sz2foo3;

const SParams params(4);
CFoo<params.m_szfoo4;
}

When I compile this code I get the following output.

dwhs1@triton:~/test$ g++ -o template template.cpp
template.cpp:22: error: ‘sz2’ cannot appear in a constant-expression
template.cpp:22: error: template argument 1 is invalid
template.cpp:22: error: invalid type in declaration before ‘;’ token
template.cpp: In function ‘int main(int, char**)’:
template.cpp:25: error: ‘params’ cannot appear in a constant-expression
template.cpp:25: error: `.' cannot appear in a constant-expression
template.cpp:25: error: template argument 1 is invalid
template.cpp:25: error: invalid type in declaration before ‘;’ token

I understand that in line 22 I am using a local variable that is not
const, but on line 25 there is a const member of a const structure.

What's going on here?

thanks

dan

[Victor Bazarov]

Dan Smithers wrote:

What constitutes a constant-expression? I know that it is something
that can be determined at compile time.

I am trying to use template code and keep getting compiler errors
"error: cannot appear in a constant-expression"

template <int s>
class CFoo
{
private:
int m_val[s];
};

struct SParams
{
const int m_sz;
SParams(int sz) : m_sz(sz) {;}
};

int main(int argc, char *argv[])
{
CFoo<2foo;

const int sz(2);
CFoo<szfoo2;

int sz2(3);
CFoo<sz2foo3;

const SParams params(4);
CFoo<params.m_szfoo4;
}

When I compile this code I get the following output.

dwhs1@triton:~/test$ g++ -o template template.cpp
template.cpp:22: error: 'sz2' cannot appear in a constant-expression
template.cpp:22: error: template argument 1 is invalid
template.cpp:22: error: invalid type in declaration before ';' token
template.cpp: In function 'int main(int, char**)':
template.cpp:25: error: 'params' cannot appear in a
constant-expression template.cpp:25: error: `.' cannot appear in a
constant-expression template.cpp:25: error: template argument 1 is
invalid
template.cpp:25: error: invalid type in declaration before ';' token

I understand that in line 22 I am using a local variable that is not
const, but on line 25 there is a const member of a const structure.

What's going on here?

I think a function call (and you have a constructor defined in your
'SParams' struct) interferes with the "const-ness" of 'params' object.
Since the initialisation requires a constructor call (never mind that
it actually can be optimized), the object is considered initialised at
run-time, and therefore cannot be part of the compile-time const expr.

That's my take on it, I didn't actually verify with the Standard.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[terminator]

On May 9, 7:28*pm, Dan Smithers <[e-mail]:

What constitutes a constant-expression? I know that it is something that
can be determined at compile time.

I am trying to use template code and keep getting compiler errors
"error: cannot appear in a constant-expression"

template <int s>
class CFoo
{
private:
* int m_val[s];

};

struct SParams
{
* const int m_sz;
* SParams(int sz) : m_sz(sz) {;}

};

int main(int argc, char *argv[])
{
* CFoo<2foo;

* const int sz(2);
* CFoo<szfoo2;

* int sz2(3);
* CFoo<sz2foo3;

* const SParams params(4);
* CFoo<params.m_szfoo4;

}

When I compile this code I get the following output.

dwhs1@triton:~/test$ g++ -o template template.cpp
template.cpp:22: error: ‘sz2’ cannot appear in a constant-expression
template.cpp:22: error: template argument 1 is invalid
template.cpp:22: error: invalid type in declaration before ‘;’ token
template.cpp: In function ‘int main(int, char**)’:
template.cpp:25: error: ‘params’ cannot appear in a constant-expression
template.cpp:25: error: `.' cannot appear in a constant-expression
template.cpp:25: error: template argument 1 is invalid
template.cpp:25: error: invalid type in declaration before ‘;’ token

I understand that in line 22 I am using a local variable that is not
const, but on line 25 there is a const member of a const structure.

What's going on here?

thanks

dan

constant exp and const object are different matters.the former is
usually refered to as literal or internally-linked value while the
second is called a read-only object.
A const object is that which is constructed at run-time and constant
since construction until destruction.
A constant exp is a value determined at compile-time.

regards,
FM.

[James Kanze]

On 10 mai, 06:57, terminator <[e-mail]:

On May 9, 7:28 pm, Dan Smithers <[e-mail]:

What constitutes a constant-expression? I know that it is
something that can be determined at compile time.

I am trying to use template code and keep getting compiler
errors "error: cannot appear in a constant-expression"

template <int s>
class CFoo
{
private:
int m_val[s];
};

struct SParams
{
const int m_sz;
SParams(int sz) : m_sz(sz) {;}
};

int main(int argc, char *argv[])
{
CFoo<2foo;

const int sz(2);
CFoo<szfoo2;

int sz2(3);
CFoo<sz2foo3;

const SParams params(4);
CFoo<params.m_szfoo4;
}

When I compile this code I get the following output.

dwhs1@triton:~/test$ g++ -o template template.cpp
template.cpp:22: error: ‘sz2’ cannot appear in a constant-expression
template.cpp:22: error: template argument 1 is invalid
template.cpp:22: error: invalid type in declaration before ‘;’ token
template.cpp: In function ‘int main(int, char**)’:
template.cpp:25: error: ‘params’ cannot appear in a constant-expression
template.cpp:25: error: `.' cannot appear in a constant-expression
template.cpp:25: error: template argument 1 is invalid
template.cpp:25: error: invalid type in declaration before ‘;’ token

I understand that in line 22 I am using a local variable
that is not const, but on line 25 there is a const member of
a const structure.

constant exp and const object are different matters.the former is
usually refered to as literal or internally-linked value while the
second is called a read-only object.
A const object is that which is constructed at run-time and constant
since construction until destruction.
A constant exp is a value determined at compile-time.

I might add that in C++, a const object of integral type *can*
be used in a constant expression, but only if its initializers
are visible and constant expressions.

--
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]