Solved

How to pass VARIANT to function thats an UINT32

Posted on 2004-09-01
1
918 Views
Last Modified: 2008-02-01
How would I pass a VARIANT thats an UINT to a function?  This is what I have so far.

VARIANT v;
UINT nPerm = 0;

v.vt = VT_UINT;
v.puintVal = &nPerm;
                  
prop->PutValue(&v);

help! :)
0
Comment
Question by:jploettner
1 Comment
 
LVL 19

Accepted Solution

by:
drichards earned 300 total points
ID: 11956252
No, it would probably look like this:

    v.vt = VT_UINT;
    v.uintVal = nPerm;

If you're trying to pass a pointer (not likely), it should look like this:

    v.vt = VT_UINT | VT_BYREF;
    v.puintVal = &nPerm;

0

Featured Post

Gigs: Get Your Project Delivered by an Expert

Select from freelancers specializing in everything from database administration to programming, who have proven themselves as experts in their field. Hire the best, collaborate easily, pay securely and get projects done right.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

Title # Comments Views Activity
wordmultiple challenge 12 131
SUM 2 INTEGER ARRAYS INTO 1 10 100
Java Loop 6 59
Help Required 2 30
Go is an acronym of golang, is a programming language developed Google in 2007. Go is a new language that is mostly in the C family, with significant input from Pascal/Modula/Oberon family. Hence Go arisen as low-level language with fast compilation…
This article is meant to give a basic understanding of how to use R Sweave as a way to merge LaTeX and R code seamlessly into one presentable document.
In this fifth video of the Xpdf series, we discuss and demonstrate the PDFdetach utility, which is able to list and, more importantly, extract attachments that are embedded in PDF files. It does this via a command line interface, making it suitable …
With the power of JIRA, there's an unlimited number of ways you can customize it, use it and benefit from it. With that in mind, there's bound to be things that I wasn't able to cover in this course. With this summary we'll look at some places to go…

776 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question