Perhaps it was mentioned here already, but: Additionally, MStatus information no matter how it is passed out from the method you call regarding the API documentation, will be thrown as exception.
Thus I personally ignore MStatus pointers and return values of any kind if writing python, and just decide to catch them whereever probability of failure is high. In c++, the API never throws to transmit status information. Regards, Sebastian On Mon, Feb 16, 2009 at 9:40 PM, ryant <[email protected]> wrote: > > Judd told me that my statement about the pointer was incorrect. I > stand corrected, there are cases you can use arguments that are > declared as a pointer. It is only the MString and MStatus that you > cant use. > > RyanT > Technical Artist > www.rtrowbridge.com > Naughty Dog Inc. > > On Feb 16, 12:17 pm, ryant <[email protected]> wrote: > > Also: > > > > int numVertices (MStatus *ReturnStatus=NULL) const > > > > This function says the MStatus argument is a "*" pointer. I believe > > there is not a single case in the Maya API where Python can pass a > > object to a argument declared as a pointer. All MStatus and MString > > arugments are declared as pointers so you can not use them with > > Python. You can pass a regular Python string in the place of a MString > > though. So that kind of contradicts what I said, but the point is you > > cant use that they say you can use. Also if you attempt to pass > > generic bool, double, floats and other types to a argument and it > > fails it is probably a documentation error. I have found that > > sometimes using MScriptUtil to declare a pointer of that type will > > allow me to use the function. Even if that function does not declare > > that the argument is a reference. > > > > RyanT > > Technical Artistwww.rtrowbridge.com/blog > > Naughty Dog Inc > > > > On Feb 15, 6:27 pm, Jason Osipa <[email protected]> wrote: > > > > > "Why do my eyes hurt?" > > > > > "Because you've never used them before" > > > > > I'm feeling a little THAT right now. Thank you a 1000 times over. > > > > > On Sun, Feb 15, 2009 at 6:19 PM, Dean Edmonds <[email protected] > >wrote: > > > > > > On Sun, Feb 15, 2009 at 14:16, jasonosipa <[email protected]> > wrote: > > > > > > > What do const, &, and * each mean? > > > > > > > When you look at documentation for an object, each argument can > have a > > > > > lot of different markup, and I'm not confident with guesses at what > > > > > those mean, so I'll just ask: > > > > > > > What do const, &, and * each mean? What does "blank" mean, where > > > > > there is no const at the head of the line, and no * or & ? How > about > > > > > the stuff after the = in the argument? Is that just a "recommended > > > > > default"? Does const only show up for things that are MObjects? > > > > > > As Tim explained, '&' and '*' are slightly different ways of passing > > > > the value itself. If neither of those is present than what gets > passed > > > > is a copy of the value. > > > > > > In terms of translating all that markup into Pythonic terms, you > first > > > > have to differentiate between simple types, like integers and floats, > > > > and object types. > > > > > > In Python, if I pass an object to a function the function can > directly > > > > manipulate it and the caller will be able to see the results of that > > > > manipulation. For example: > > > > > > class myObj: > > > > someValue = 12 > > > > > > def changeObj(theObj): > > > > theObj.someValue = 15 > > > > > > testObj = myObj() > > > > > > print testObj.someValue > > > > 12 > > > > > > changeObj(testObj) > > > > print testObj.someValue > > > > 15 > > > > > > The same is *not* true when passing a simple value to a function: > > > > > > def changeValue(theValue): > > > > theValue = 15 > > > > > > someValue = 12 > > > > print someValue > > > > 12 > > > > > > changeValue(someValue) > > > > print someValue > > > > 12 > > > > > > In C++ there is no such differentiation between objects and simple > > > > values. If a parameter is passed by pointer ('*') or reference ('&') > > > > then the function can modify it such that the caller will see the > > > > changes. If neither of those markers is present then the function is > > > > passed a copy of the value and any changes that the function makes to > > > > that copy will not be visible to the caller. > > > > > > In addition, if a parameter is tagged as 'const' that means that the > > > > method is promising not modify it, even it has been passed by pointer > > > > or reference. > > > > > > So, if you are calling an API method from Python and one of its > > > > parameters takes an object type, such as MDagPath, the following > > > > markups mean that the caller's copy of the object will not be changed > > > > by the method: > > > > > > MDagPath > > > > const MDagPath* > > > > const MDagPath& > > > > > > while the following mean that the caller's copy of the object can be > > > > changed by the method: > > > > > > MDagPath* > > > > MDagPath& > > > > > > In terms of the syntax you use in Python, you always just pass the > > > > object without doing anything special. E.g: > > > > > > somePath = om.MDagPath() > > > > om.MDagPath.getAPathTo(someNode, somePath) > > > > > > If the markup indicates that the method can change the object that > you > > > > pass to it and you don't want it to change your copy of the object, > > > > then you would need to pass the method a separate copy of the > object:. > > > > In the example below 'tempPath' will be modified by the call to > > > > getAPathTo() but 'pathToKeep' will remain unchanged: > > > > > > tempPath = om.MDagPath(pathToKeep) > > > > om.MDagPath.getAPathTo(someNode, tempPath) > > > > > > For simple types we run into a bit of a problem because Python does > > > > not provide any way for a function to modify a simple type which is > > > > passed to it such that the caller sees the change. For example, > > > > MFnMesh has a getUV() method which is declared as follows: > > > > > > MStatus getUV (int uvId, float &u, float &v, const MString > > > > *uvSet=NULL) const; > > > > > > The intent is that you pass the id of the UV you want in the 'uvId' > > > > parameter and get back its U and V coordinates in the 'u' and 'v' > > > > parameters. Unfortunately there is no direct way to do that in > Python, > > > > so the Maya API provides the MScriptUtil class to get around that. > > > > E.g: > > > > > > uptr = om.MScriptUtil().asFloatPtr() > > > > vptr = om.MScriptUtil().asFloatPtr() > > > > > > meshFn.getUV(13, uptr, vptr) > > > > > > u = om.MScriptUtil.getFloat(uptr) > > > > v = om.MScriptUtil.getFloat(vptr) > > > > > > Finally, if 'const' appears at the end of the method declaration that > > > > means that the method promises not to change any of the object's > > > > member variables. For example, MFnMesh's numVertices() method is > > > > declared as follows: > > > > > > int numVertices (MStatus *ReturnStatus=NULL) const > > > > > > This means that if you call numVertices() you can be assured that the > > > > MFnMesh object will not be changed by call. > > > > > > -- > > > > -deane > > > > > -- > > > Jason Osipa, Manager > > > Osipa Entertainment, LLCwww.osipaentertainment.com > > > 415.374.9651 > > > > > > > --~--~---------~--~----~------------~-------~--~----~ Yours, Maya-Python Club Team. -~----------~----~----~----~------~----~------~--~---
