C / C++ / MFC

Stuart Dootson15-Apr-09 23:02

15-Apr-09 23:02

To do this, I suspect you need to interface to Outlook using it's COM object model. I don't think ShellExecuteEx will cut it.

This article[^] shows how to talk to Outlook using Javascript. C++ is more involved - to get anywhere, you probably want to be using #import with Outlook's type library - use #import "progid:Outlook.Application" auto_search to get VC++ to generate classes for the Outlook object model.

Alternatively, this article[^] tells you how to do Office Automation with COM without using #import.

Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p

SNI17-Apr-09 0:37

SNI

17-Apr-09 0:37

thanks for your reply. But I want to know the event by which I come to know whether file gets attached or not so i can move further for processing.
Any kind of idea will be helpful.

SNI

Stuart Dootson17-Apr-09 0:49

17-Apr-09 0:49

Yeah, it's a bit different than just launching outlook with a parameter when you use COM.

Here's some VBA that will start Outlook, create and send an e-mail.

Dim o As Object 'will be an Outlook.Application
Dim msg As Object 'will be a MailItem

Set o = CreateObject("Outlook.Application")
Set msg = o.CreateItem(0)
msg.Subject = "this is my subject"

msg.attachments.Add "path to the attachment"

msg.body = "my message"

msg.Recipients.Add "an e-mail address"

msg.send

You can replicate that relatively easily using the #import approach to interfacing with Outlook.

Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p

Stuart Dootson17-Apr-09 1:06

Puzzle about c++ inherent

17-Apr-09 1:06

C++ code for the VBA I posted - note that it's pretty much a direct translation...

#import "progid:Outlook.Application" auto_search

int main(int, char**)
{
   try
   {
      CoInitializeEx(0, COINIT_APARTMENTTHREADED);
      Outlook::_ApplicationPtr ol(__uuidof(Outlook::Application));
      Outlook::_MailItemPtr msg(ol->CreateItem(Outlook::olMailItem));

      msg->Subject = "this is my subject";
      
      msg->Attachments->Add("path to the attachment");

      // If execution gets here, the attachment's been added to msg, and you can go on and do other stuff!
      
      msg->Body = "my message";
      
      msg->Recipients->Add("an e-mail address");
      
      msg->Send();
   }
   catch (_com_error& e)
   {
      std::cerr << (char*)e.Description() << std::endl;
   }
   return 0;
}

Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p

sharion15-Apr-09 19:08

sharion

15-Apr-09 19:08

Here is the code snippet.

class Super
{
public:
    Super(){}
    virtual void someMethod() {cout<<"in Super class"<<endl;}
};

class Sub : public Super
{
public:
    Sub();
    virtual void someMethod(){cout<<"in Sub class"<<endl;}
};

Sub mySub;
Super& ref =mySub;
ref.someMethod();

I don't really understand how the compiler works when it comes to the statement ref.someMethod(); . I guess the compiler first find someMethod() method in the Sub class. If doesn't find, then enter base class---Super class to find it.
above explanation is correct?
------------
If altering someMethod() in Sub class like this:

virtual void someMethod(int i){cout<<"in Sub class"<<endl;}<br />

what happens when the compiler runs ref.someMethod(); statement?
Obviously the output is "in Super class".
------------
If the someMethod() in Sub class is

virtual void someMethod(int i = 1){cout<<"in Sub class"<<endl} <br />

when the compiler runs ref.someMethod(); statement,what is the output?
In VS 2005,the output is "in Super class". I feel it is hard to understand,I believe the right result is "in Sub class".

Anyone could give me some idea,thanks in advance.

sharion

Re: Puzzle about c++ inherent [modified]

_AnsHUMAN_ 15-Apr-09 19:27

_AnsHUMAN_

15-Apr-09 19:27

This feature is called run-time linking.
Based on the current type of the object the function call is resolved. So in first case the output is printed as "in Sub class" ie when function signatures in super and sub are same

When the function signatures are different, and the call is with a int parameter, it would look for a match which is defined in super and function is only available from the base part. So it calls the base class function

So obviously the compiler would say, I can not see it and I dont know what to call for and calls the base class exact match function for the remaining two cases.

You need to google first, if you have "It's urgent please" mentioned in your question.
Wink | ;-)

_AnShUmAn_

modified on Thursday, April 16, 2009 1:34 AM

Re: Puzzle about c++ inherent

ParagPatel15-Apr-09 21:50

ParagPatel

15-Apr-09 21:50

Hi,

sharion wrote:
If doesn't find, then enter base class---Super class to find it.
above explanation is correct?

- Here you are littlebit right, but virtual keyword doing magic here,

sharion wrote:
when the compiler runs ref.someMethod();

Basically there is two type of binding

1. Run-time binding (or late binding)
- if same function define in parent class with same signature and it virtual, then compiler ignore it for run-time, and run time function will be called using VPTR and VPTABLE.

2. Compile-time binding (or erly binding)
- if above is not a case then compiler resolve binding at compile time only.

Parag Patel
Sr. Software Eng, Varaha Systems

Re: Puzzle about c++ inherent

Stuart Dootson15-Apr-09 22:51

How do I remove thousand seperator.

15-Apr-09 22:51

sharion wrote:
If the someMethod() in Sub class is

virtual void someMethod(int i = 1){cout<<"in Sub class"<<endl}

when the compiler runs ref.someMethod(); statement,what is the output?
In VS 2005,the output is "in Super class". I feel it is hard to understand,I believe the right result is "in Sub class".

No, the compiler is correct. In this case, there are two possible functions matching ref.someMethod(). The C++ Standard defines rules by which the compiler can decide which is the best match. These rules specify that a function with no parameters is a better match than a function with a parameter which has a default value. The fact that one of the methods is implemented in the derived class and one in the base class has no impact.

Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p

cozyu15-Apr-09 16:06

cozyu

15-Apr-09 16:06

I have a problem of number formatting

I made a function to convert from number to string.
and I used "stringstream" class for converting.
like this

string strText;
stringstream ssBuf;	
ssBuf<<55555;
ssBuf>>strText;

I expect "55555" as result. but the result is "55,555".
How do I remove thousand seperator "," in string?

Thanks.

cozyu

modified on Wednesday, April 15, 2009 10:25 PM

Stuart Dootson15-Apr-09 22:47

15-Apr-09 22:47

Your locale[^] is getting in the way.

You need to tell the stringstream to use a locale without number punctuation. The simplest way to do that is to just use the std::locale::classic() locale like this:

std::string strText;
std::stringstream ssBuf;
ssBuf.imbue(std::locale::classic()); // This line tells the stringstream to use the classic locale
ssBuf<<55555;
ssBuf>>strText;

However, this loses any other locale-specific properties, so you might just want to use the number punctuation bit of the classic locale, like this:

std::string strText;
std::stringstream ssBuf;
std::locale l;
l = l.combine<std::numpunct<char> >(std::locale::classic());
ssBuf.imbue(l);
ssBuf<<55555;
ssBuf>>strText;

The combine method call copies your global locale and replaces the copy's number punctuation facet with the classic locale's number punctuation facet.

Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p

Cedric Moonen16-Apr-09 0:00

Cedric Moonen

16-Apr-09 0:00

Great reply, got my 5 Smile | :)

Don't you consider writing an article for CP ? I think you could bring really valuable things Smile | :)

Cédric Moonen
Software developer

Charting control [v2.0 - Updated]
OpenGL game tutorial in C++

Stuart Dootson16-Apr-09 0:08

[OT] Re: How do I remove thousand seperator.

16-Apr-09 0:08

Cedric Moonen wrote:
Great reply, got my 5

Cheers

Cedric Moonen wrote:
Don't you consider writing an article for CP ?

There are a few things I've got that I've thought about writing up. Trouble is, I'm lazy Smile | :)

It's a good attribute for a developer, as it means you're more likely to use someone else's code than write your own, but not so great for article writing!

Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p

David Crow16-Apr-09 3:28

David Crow

16-Apr-09 3:28

Stuart Dootson wrote:
Your locale[^] is getting in the way.

Mine is set to use commas, yet I see 55555 when using the OP's code snippet.

"Old age is like a bank account. You withdraw later in life what you have deposited along the way." - Unknown

"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons

Re: [OT] Re: How do I remove thousand seperator.

Stuart Dootson16-Apr-09 3:32

16-Apr-09 3:32

Yeah, so did I, but as locales are the only thing that affect stream output like that, it seems likely that locales were the issue. Anyway - I suspect your default C++ locale (which isn't necessarily the same as what your regional settings say!) is set to "C", a.k.a the 'classic' locale, like mine.

No, I don't know what sets the default C++ locale...

Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p