newbie question

[mrkafk]

Hello everyone,

I'm very much (greenish) new to C++, learning it with S. Lippman's "C+
+ Primer" in hand.

I tried to extend one example:


#include <iostream>
using namespace std;

class IntArray
{
public:
explicit IntArray(int sz = DefaultArraySize);
IntArray(int *array, int array_size);
int siz();
private:
static const int DefaultArraySize = 12;
int _size;
int* ia;
};

IntArray::IntArray(int sz)
{
_size = sz;
ia = new int[_size];
for (int i=0; i<_size; i++)
ia=0;
}

IntArray::siz()
{
return _size;
}

int main()
{
IntArray a(5);
cout << "a.siz() : " << a.siz() << endl;
}

Yet, g++ complains:

e.C:25: error: ISO C++ forbids declaration of 'siz' with no type

? I don't get it? After all, I have declared siz() to be of int type
in the class declaration? I can't use IntArray::int siz(), can I?

(of course I tried it and get e.C:25: error: expected unqualified-id
before 'int' instead, doomed if I do and doomed if I don't)

 

[Klaas Vantournhout]

Hello everyone,

I'm very much (greenish) new to C++, learning it with S. Lippman's "C+
+ Primer" in hand.

I tried to extend one example:


#include <iostream>
using namespace std;

class IntArray
{
public:
explicit IntArray(int sz = DefaultArraySize);
IntArray(int *array, int array_size);
int siz();
private:
static const int DefaultArraySize = 12;
int _size;
int* ia;
};

IntArray::IntArray(int sz)
{
_size = sz;
ia = new int[_size];
for (int i=0; i<_size; i++)
ia=0;
}

IntArray::siz()
{
return _size;
}

int main()
{
IntArray a(5);
cout << "a.siz() : " << a.siz() << endl;
}

Yet, g++ complains:

e.C:25: error: ISO C++ forbids declaration of 'siz' with no type

? I don't get it? After all, I have declared siz() to be of int type
in the class declaration? I can't use IntArray::int siz(), can I?

In your class definition you have indeed defined IntArray::siz(void) to
return return an int, but in your function definition itself you forgot
the int part. you have to state what the function returns.

So place

int IntArray::siz() {
return _size;
}

and the problem should be resolved.

 

[David Côme]

Hello everyone,

I'm very much (greenish) new to C++, learning it with S. Lippman's "C+
+ Primer" in hand.

I tried to extend one example:


#include <iostream>
using namespace std;

class IntArray
{
public:
explicit IntArray(int sz = DefaultArraySize);
IntArray(int *array, int array_size);
int siz();
private:
static const int DefaultArraySize = 12;
int _size;
int* ia;
};

IntArray::IntArray(int sz)
{
_size = sz;
ia = new int[_size];
for (int i=0; i<_size; i++)
ia=0;
}

IntArray::siz()
{
return _size;
}

int main()
{
IntArray a(5);
cout << "a.siz() : " << a.siz() << endl;
}

Yet, g++ complains:

e.C:25: error: ISO C++ forbids declaration of 'siz' with no type

? I don't get it? After all, I have declared siz() to be of int type
in the class declaration? I can't use IntArray::int siz(), can I?

(of course I tried it and get e.C:25: error: expected unqualified-id
before 'int' instead, doomed if I do and doomed if I don't)

All function must have a return-type (except constructor and destrcuteur).
So IntArray::siz() become int IntArray::siz().

 

[Klaas Vantournhout]

Hello everyone,

I'm very much (greenish) new to C++, learning it with S. Lippman's "C+
+ Primer" in hand.

I tried to extend one example:


#include <iostream>
using namespace std;

class IntArray
{
public:
explicit IntArray(int sz = DefaultArraySize);
IntArray(int *array, int array_size);
int siz();
private:
static const int DefaultArraySize = 12;
int _size;
int* ia;
};

IntArray::IntArray(int sz)
{
_size = sz;
ia = new int[_size];
for (int i=0; i<_size; i++)
ia=0;
}

IntArray::siz()
{
return _size;
}

int main()
{
IntArray a(5);
cout << "a.siz() : " << a.siz() << endl;
}

Yet, g++ complains:

e.C:25: error: ISO C++ forbids declaration of 'siz' with no type

? I don't get it? After all, I have declared siz() to be of int type
in the class declaration? I can't use IntArray::int siz(), can I?

In your class definition you have indeed defined IntArray::siz(void) to
return return an int, but in your function definition itself you forgot
the int part. you have to state what the function returns.

So place

int IntArray::siz() {
return _size;
}

and the problem should be resolved.

Something else I noticed is that your main part of the program (the
function int main(void), does not have a return type.

Here you should return the errorcode.

If your program finishes successfull you should finish the main part of
your code with return 0;, If it hangs somwehere in the middle of your
code, give it an other number. This way you can see if your program
finished succesfully after it finished.

Hence
int main() {
IntArray a(5);
cout << "a.siz() : " << a.siz() << endl;
return 0;
}

 

[Abhishek Padmanabh]

Hello everyone,
I'm very much (greenish) new to C++, learning it with S. Lippman's "C+
+ Primer" in hand.
I tried to extend one example:
#include <iostream>
using namespace std;
class IntArray
{
public:
 explicit IntArray(int sz = DefaultArraySize);
 IntArray(int *array, int array_size);
 int siz();
private:
 static const int DefaultArraySize = 12;
 int _size;
 int* ia;
};
IntArray::IntArray(int sz)
{
 _size = sz;
 ia = new int[_size];
 for (int i=0; i<_size; i++)
        ia=0;
}
IntArray::siz()
{
 return _size;
}
int main()
{
 IntArray a(5);
 cout << "a.siz() : " << a.siz() << endl;
}
Yet, g++ complains:
 e.C:25: error: ISO C++ forbids declaration of 'siz' with no type
? I don't get it? After all, I have declared siz() to be of int type
in the class declaration? I can't use IntArray::int siz(), can I?

In your class definition you have indeed defined IntArray::siz(void) to
 return return an int, but in your function definition itself you forgot
the int part.  you have to state what the function returns.

So place

int IntArray::siz() {
return _size;
}

and the problem should be resolved.

Something else I noticed is that your main part of the program (the
function int main(void), does not have a return type.

Here you should return the errorcode.

If your program finishes successfull you should finish the main part of
your code with return 0;, If it hangs somwehere in the middle of your
code, give it an other number.  This way you can see if your program
finished succesfully after it finished.

Hence
int main() {
 IntArray a(5);
 cout << "a.siz() : " << a.siz() << endl;
 return 0;
}

If the return is missing, it is implicitly equivalent to return 0;

 

[Salt_Peter]

Hello everyone,
I'm very much (greenish) new to C++, learning it with S. Lippman's "C+
+ Primer" in hand.
I tried to extend one example:
#include <iostream>
using namespace std;
class IntArray
{
public:
explicit IntArray(int sz = DefaultArraySize);
IntArray(int *array, int array_size);
int siz();
private:
static const int DefaultArraySize = 12;
int _size;
int* ia;
};
IntArray::IntArray(int sz)
{
_size = sz;
ia = new int[_size];
for (int i=0; i<_size; i++)
ia=0;
}
IntArray::siz()
{
return _size;
}
int main()
{
IntArray a(5);
cout << "a.siz() : " << a.siz() << endl;
}
Yet, g++ complains:
e.C:25: error: ISO C++ forbids declaration of 'siz' with no type
? I don't get it? After all, I have declared siz() to be of int type
in the class declaration? I can't use IntArray::int siz(), can I?

In your class definition you have indeed defined IntArray::siz(void) to
return return an int, but in your function definition itself you forgot
the int part. you have to state what the function returns.

So place

int IntArray::siz() {
return _size;
}

and the problem should be resolved.

Something else I noticed is that your main part of the program (the
function int main(void), does not have a return type.

Here you should return the errorcode.

If your program finishes successfull you should finish the main part of
your code with return 0;, If it hangs somwehere in the middle of your
code, give it an other number. This way you can see if your program
finished succesfully after it finished.

Hence
int main() {
IntArray a(5);
cout << "a.siz() : " << a.siz() << endl;
return 0;

}

This is not required. Supplying it doesn't solve anything anyways
since the program's return value is ignored. A much better alternative
is to use exceptions to handle exceptional conditions.

 

[Erik Wikström]

(e-mail address removed) wrote:
Hello everyone,

I'm very much (greenish) new to C++, learning it with S. Lippman's "C+
+ Primer" in hand.

I tried to extend one example:

#include <iostream>
using namespace std;

class IntArray
{
public:
explicit IntArray(int sz = DefaultArraySize);
IntArray(int *array, int array_size);
int siz();
private:
static const int DefaultArraySize = 12;
int _size;
int* ia;
};

IntArray::IntArray(int sz)
{
_size = sz;
ia = new int[_size];
for (int i=0; i<_size; i++)
ia=0;
}

IntArray::siz()
{
return _size;
}

int main()
{
IntArray a(5);
cout << "a.siz() : " << a.siz() << endl;
}

Yet, g++ complains:

e.C:25: error: ISO C++ forbids declaration of 'siz' with no type

? I don't get it? After all, I have declared siz() to be of int type
in the class declaration? I can't use IntArray::int siz(), can I?

In your class definition you have indeed defined IntArray::siz(void) to
return return an int, but in your function definition itself you forgot
the int part. you have to state what the function returns.

So place

int IntArray::siz() {
return _size;
}

and the problem should be resolved.

Something else I noticed is that your main part of the program (the
function int main(void), does not have a return type.

Here you should return the errorcode.

If your program finishes successfull you should finish the main part of
your code with return 0;, If it hangs somwehere in the middle of your
code, give it an other number. This way you can see if your program
finished succesfully after it finished.

Hence
int main() {
IntArray a(5);
cout << "a.siz() : " << a.siz() << endl;
return 0;

}

This is not required. Supplying it doesn't solve anything anyways
since the program's return value is ignored. A much better alternative
is to use exceptions to handle exceptional conditions.

You are speaking about two different things, exceptions should be used
to handle exceptional conditions within the program, but sometimes there
is no other choice than to terminate the program. In those cases a non-
zero return value should be used.

As for the return value being ignored; that is not true, on Unix/Linux
systems the return value can be very useful, especially when writing
command line utilities which might be used by scripts, but also on
Windows systems where batch-files or scripts are used.