Thread: how to pass symbollically the name of a struc member to a function ?

Hi all,

Sorry in advance if this question has already been raised.

I have a struct with several members, for instance:

insert

Code:

typedef struct {
float u;
float v;
float w;
} mystruct;

main ()

mystruct  test;

...

I have a piece of code that should operate on test.u that looks like this:

switch (j) {

case 0:
 f = test.u + ...;
 break;

case 1:
 f  = X - test.u;
 break;

case 2:
 f = test.u + 2;
 break;

}

Now I would like to use a similar piece of code but for the other members, test.v and test.w. In principle I could copy the piece of code above and just replace test.u by test.v or test.w. Since it is the same code for all members, I would like to avoid this (copying many times the same piece of code) and replace it by a call to function. The question is, how to pass the name of the struct member I am considering to the function ? How can I tell the function, operates on member .u or .v ?
Would it be possible to have a generic piece of code

insert

Code:

function (member y,...)

switch (j) {

case 0:
 f = test.y + ...;
 break;

case 1:
 f  = X - test.y;
 break;

case 2:
 f = test.y + 2;
 break;

}

where the function could be called with function(u) or function(w) and would replace automatically .y by .u or .w ?

Thanks in advance.
Xtof

Originally Posted by vart

why not pass the member by value?

Code:

function (float y,...)
 
switch (j) {
 
case 0:
 f = y + ...;
 break;
 
case 1:
 f  = X - y;
 break;
 
case 2:
 f = y + 2;
 break;
 
}

I forgot to mention that f is also a struct and has the same members. Actually, it should be something like:

Code:

function (member y,...)


switch (j) {


case 0:
 f.y = test.y + ...;
 break;


case 1:
 f.y  = X - test.y;
 break;


case 2:
 f.y = test.y + 2;
 break;


}

That is, if I pass the member .u, f.u should be calculated, if I pass .v, f.v should be calculated, and so on. Maybe I could pass simply f.u by reference.

Code:

function(&f.u,test.u);
function(&f.v,test.v);
function(&f.w,test.w);

Do you think it would work like that ?