I has many user define function with arguments and without. I use CUdfHelper from this article http://www.jkp-ads.com/articles/RegisterUDF00.asp for register function.
Registered function ask arguments for arguments, even if they are not.
Example my fuction without arguments:
Public Function getProjects()
getProjects = Utils.execute("getProjects", "getWSEntitiesData")
End Function
On MyFunction.c
#include <windows.h>
#define DLL_EXPORT __declspec(dllexport)
DLL_EXPORT void getProjects() {
return;
}
compile on MyFunction.dll
I register the function with these parameters.
SetGlobalName = Application.ExecuteExcel4Macro("MyFunction.dll", "getProjets", "P", "getProjects", "", 1, "MyFunctionCategory", "", "", "Return list projects")
If I register as
SetGlobalName = Application.ExecuteExcel4Macro("MyFunction.dll", "getProjets", "P", "getProjects",, 1, "MyFunctionCategory", "", "", "Return list projects")
Function argument dialog is displayed all the same.
If I register as
SetGlobalName = Application.ExecuteExcel4Macro("MyFunction.dll", "getProjets", "P", "getProjects",, 1, "MyFunctionCategory")
Function argument dialog isn't displayed, but the description is no longer available.
REGISTER() Arguments
Path and name of the dll
Name of the function you wish to call
Type string
The name you want to use in Excel cells
A list of arguments to use in the function wizard
The Macro type (2 for a function, 2 for a command)
Which function wizard category to add the function to
Short cut text if the function being registered is a command
Path to help file
Function help to show in the function wizard
11-30 onwards help text for each argument in the function wizard.
I think the problem is in the arguments, as by default it is set to an empty string, and I can not figure out how to change the parameters to the function.
On CUdfHelper
' structure definition
Private Type REGARG
sDllName As String
sDllProc As String
sArgType As String
sFunText As String
**sArgText As String**
iMacType As Integer
vCatName As Variant
sKeyText As String
sHlpPath As String
**sFunHelp As String**
aArgHelp(1 To 20) As String
End Type
How to correctly set the parameters so that the window does not appear, and stores the description?
If register function with ExecuteExcel4Macro and set Function Description. In an XLL add-in, however, you may run into a bug in the Excel API. The bug shows itself if a parameterless function is mentioned in an ADXExcelFunctionDescriptor that has a non-empty string in the Description property. If this is the case, you'll get another version of the Function Arguments dialog.
That is, to bypass that issue, you need to leave the ADXExcelFunctionDescriptor.Description property empty.
Find on https://www.add-in-express.com/docs/net-excel-udf-tips.php
Related
I'm not sure how to explain this, but when I run
console.log`1`
In google chrome, I get output like
console.log`1`
VM12380:2 ["1", raw: Array[1]]
Why is the backtick calling the log function, and why is it making a index of raw: Array[1]?
Question brought up in the JS room by Catgocat, but no answers made sense besides something about templating strings that didn't really fit why this is happening.
It is called Tagged Template in ES-6 more could be read about them Here, funny I found the link in the starred section of the very chat.
But the relevant part of the code is below (you can basically create a filtered sort).
function tag(strings, ...values) {
assert(strings[0] === 'a');
assert(strings[1] === 'b');
assert(values[0] === 42);
return 'whatever';
}
tag `a${ 42 }b` // "whatever"
Basically, its merely tagging the "1" with console.log function, as it would do with any other function. The tagging functions accept parsed values of template strings and the values separately upon which further tasks can be performed.
Babel transpiles the above code to
var _taggedTemplateLiteralLoose = function (strings, raw) { strings.raw = raw; return strings; };
console.log(_taggedTemplateLiteralLoose(["1"], ["1"]));
As you can see it in the example above, after being transpiled by babel, the tagging function (console.log) is being passed the return value of the following es6->5 transpiled code.
_taggedTemplateLiteralLoose( ["1"], ["1"] );
The return value of this function is passed to console.log which will then print the array.
Tagged template literal:
The following syntax:
function`your template ${foo}`;
Is called the tagged template literal.
The function which is called as a tagged template literal receives the its arguments in the following manner:
function taggedTemplate(strings, arg1, arg2, arg3, arg4) {
console.log(strings);
console.log(arg1, arg2, arg3, arg4);
}
taggedTemplate`a${1}b${2}c${3}`;
The first argument is an array of all the individual string characters
The remaining argument correspond with the values of the variables which we receive via string interpolation. Notice in the example that there is no value for arg4 (because there are only 3 times string interpolation) and thus undefined is logged when we try to log arg4
Using the rest parameter syntax:
If we don't know beforehand how many times string interpolation will take place in the template string it is often useful to use the rest parameter syntax. This syntax stores the remaining arguments which the function receives into an array. For example:
function taggedTemplate(strings, ...rest) {
console.log(rest);
}
taggedTemplate `a${1}b${2}c${3}`;
taggedTemplate `a${1}b${2}c${3}d${4}`;
Late to the party but, TBH, none of the answers give an explanation to 50% of the original question ("why the raw: Array[1]")
1. Why is it possible to call the function without parenthesis, using backticks?
console.log`1`
As others have pointed out, this is called Tagged Template (more details also here).
Using this syntax, the function will receive the following arguments:
First argument: an array containing the different parts of the string that are not expressions.
Rest of arguments: each of the values that are being interpolated (ie. those which are expressions).
Basically, the following are 'almost' equivalent:
// Tagged Template
fn`My uncle ${uncleName} is ${uncleAge} years old!`
// function call
fn(["My uncle ", " is ", " years old!"], uncleName, uncleAge);
(see point 2. to understand why they're not exactly the same)
2. Why the ["1", raw: Array[1]] ???
The array being passed as the first argument contains a property raw, wich allows accessing the raw strings as they were entered (without processing escape sequences).
Example use case:
let fileName = "asdf";
fn`In the folder C:\Documents\Foo, create a new file ${fileName}`
function fn(a, ...rest) {
console.log(a); //In the folder C:DocumentsFoo, create a new file
console.log(a.raw); //In the folder C:\Documents\Foo, create a new file
}
What, an array with a property ??? ???
Yes, since JavaScript arrays are actually objects, they can store properties.
Example:
const arr = [1, 2, 3];
arr.property = "value";
console.log(arr); //[1, 2, 3, property: "value"]
Access 2013
I'm calling a formula to modify a string and it's changing the values w/in the parent sub.
Example:
Debug.Print Str 'Hello World my name is bob
BOBexists = InStringChceck(Str,"bob")
Debug.Print Str 'HELLO WORLD MY NAME IS BOB
Debug.Print BOBexists 'TRUE
I've used this function, InStringCheck, in Excel VBA before (and it's just an example, all of my string tools are doing this same thing now and I don't know why)
Function InStringCheck(Phrase as string, Term as string) as Boolean
Phrase = UCase(Phrase)
Term = UCase(Term)
if instr(1, Phrase, Term) then InStringCheck = True else InStringCheck = False
end function
In several of my functions I manipulate the input variables, to arrive at a solution, but I don't want those manipulations to persist outside of the function unless I pass them back up - some how they're being passed up, but they're not dimed as public variables
VBA parameters are implicitly passed by reference (ByRef). This means you're passing a reference to the value, not the value itself: mutating that value inside the procedure will result in that mutated value being visible to the calling code.
This is often used as a trick to return multiple values from a function/procedure:
Public Sub DoSomething(ByVal inValue1 As Integer, ByRef outResult1 As Integer, ...)
You have two options:
Pass the parameters by value (ByVal)
Introduce local variables and mutate them instead of mutating the paramters (and heck, pass the parameters ByRef explicitly)
If you have lots of occurrences of parameters being implicitly passed ByRef in your project, fixing them everywhere can easily get tedious. With Rubberduck you can easily locate all occurrences, navigate there, and apply appropriate fixes:
Disclaimer: I'm heavily involved in the Rubberduck project.
Building a little on #Sorcer's answer, VBA has default Sub/Functions parameters passing "by reference" (i. e.: "ByRef" keyword assumed if not specified) so that if you don't want their "inside" modifications survive outside them you have to explicitly type "ByVal" keyword before them in the arguments list.
But you have the option to avoid such modifications take place altoghether by using StrComp():
Function InStringCheck(Phrase as string, Term as string) as Boolean
InStringCheck = StrComp(Phrase, Term, vbTextCompare) = 0
End Function
Which could also lead you to avoid the use of InStringCheck() in favour of a direct use of StrComp() in your code
I have this problem with a very simple function written in an XLL, using VS2012. I have tried reading up in MSDN and Steve Dalton's book, and I cannot see what I am doing wrong.
The tricky bit is that I need my function to read values in worksheet cells other than the one from which it is called. The function takes no arguments, and returns an integer. I have declared it as J# (the # signifying that it can call XLM functionality as advised by Dalton...although I still get the same problem without the #). I have not included the declaration of my function to save space, but it is simple and I do not think it is the cause of the problem.
This first block of code works fine. I wrote it just to build confidence.
//This block works correctly. A trial copied from the old Excel 97 documentation
XLOPER12 xlInput1, xlOutput2;
/* Evaluate the string "2+3" */
xlInput1.xltype = xltypeStr;
xlInput1.val.str = L"\0032+3"; //prefix with string length in Octal
Excel12(xlfEvaluate, &xlOutput2, 1, (LPXLOPER12) &xlInput1);
//works OK, and xlOutput2 contains 5
But this second block does not work. I cannot see why. I am trying to read a value from a cell, which is a different cell from that from which the function was called. What I get is an return XLOPER12 that contains an error (xltypeErr) and junk values in the val.num field (the worksheet cell does contain an integer value).
//This block does not work
XLOPER12 xlInput3, xlOutput3;
/* Look up the name Tst on the active sheet called Sht */
xlInput3.xltype = xltypeStr;
xlInput3.val.str = L"\003Tst"; //this also gives problems regardless of whether the string is defined as \004!Tst or \007Sht!Tst
Excel12(xlfEvaluate, &xlOutput3, 1, (LPXLOPER12) &xlInput3); //xlOutput3 now has a type of xltypeErr, rather than the correct integer value on the worksheet
Can you kindly explain what is going wrong?
If you're trying to read a value from a cell that is different from the cell calling the function you'll need a parameter to refer to that different cell. For example in A1 you may have '=myfunc(A2)'. Then your C++ extension func will need to be declared 'JP#', with the P corresponding to the A2 cell reference parameter. If Excel can resolve the 'A2' reference it will pass in an XLOPER with that value as xltypeNum, xltypeInt or xltypeStr depending on the the contents of A2. If not you may get an xltypeSRef.
xlfEvaluate: here's the MS doc https://msdn.microsoft.com/en-us/library/office/bb687913(v=office.15).aspx
Note that MS specify that the string passed to xlfEvaluate must 'contain only functions, not command equivalents'. I suspect L"\003Tst" doesn't correspond to any function known to your Excel. There's no built in function called Tst in my Excel 2013. It's possible you have an addin that supplies a function called Tst, but I'm guessing not. So try changing xlInput3.val.str to L"\006RAND()" and see what happens.
So this is a robust problem. I have a function which accepts 2 args (string_name, macros). Here it is so I can further explain.
function ParseStrings(string_name, macros)
return my_table[string_name]
-- All it does it returns the string_name's value
end
The problem is that the second arg is a table, and if it's a table then in the string there are going to be various parts that have the format "String stuff $MACRO_KEY; more string text" and the content between the $ and ; is the key to look up in the macro table sent with it. Now anytime a value like that appears in the string there will always be a second arg that's a table, so no problems their. I need to be able to count up the number of instances of macros in a string and then replace each macro component with it's respective macros' table value. So here's how the func is called in this instance...
local my_table = {
my_string = "My string content $MACRO_COMPONENT; more string stuff $MACRO_COMPONENT_SUB;$MACRO_COMPONENT_ALT;"
}
local macro = {
MACRO_COMPONENT = "F",
MACRO_COMPONENT_SUB = "Random Text",
MACRO_COMPONENT_ALT = "14598"
}
function ParseStrings(string_name, macros)
return my_table[string_name]
-- All it does it returns the string_name's value
end
ParseStrings("my_string", macro)
So I am thinking:
string.gsub(my_table[my_string]:match("%b$;"):sub(2,my_table[my_string]:match("%b$;"):len() - 1)
but this is a long and overtly complex answer (AFAIK) and from my tests it only does 1 replacement (because the pattern is only found once) and that's doesn't work well if there are multiple instances in the string. So ideas?
I create a Public vba function with 2 parameters(module).
When I call the function I type "=InvoiceAmount2(A9;B9)".
The first parameter turns blue. The second black.
I remake the same function using one parameter, the second I use into the function, that way it´s ok. But I need two parameters
This is how you call a user defined function with two parameters:
=MyFunction(A1,B1)
Sample Code:
Function MyFunction(rCellA As Range, rCellB As Range)
MyFunction = rCellA.Value + rCellB.Value
End Function
I found the problem.
I call the udf function in formula constructor. I informed the parameters (A9 and B9)
finally, the udf function filled the cell. "=InvoiceAmount2(A9;B9)".