structures and pointers

I'm not sure what I'm doing...

Let say I have a structure

struct Inner{
    exampleType a;
    int b;
}

struct Outer{
    int current;
    int total;
    Inner reco开发者_如何学Pythonrds[MAXNUMBER];
}

struct Outer2{
    Outer outer;
}

And I have the following functions:

void try3( Outer2& outer, type var, type2 var2 ){
}

void try2( Outer2* outer ){
    try3(*outer, var, var2);
}

Inside main:

int myMain( int argc, char *argv[] ){
    Outer2 outer2;
    try2 (&outer2);
}

Here's the question. Can I increment the value of current by sticking the following line in try3:

++outer.outer.current;

errr, no, try3 has no knowledge of a thing caller outer2.

you can go outer.outer.current++; in try3

Sure you can, why not? Here is a working example for you:

typedef short exampleType;

struct Inner
{
    exampleType a;
    int b;
};

struct Outer
{
    enum { MAXNUMBER = 2 };

    int current;
    int total;
    Inner records[MAXNUMBER];
};

struct Outer2
{
    Outer outer;
};

typedef int type;
typedef int type2;

void try3 (Outer2& outer, type var, type2 var2)
{
    ++outer.outer.current;
}

void try2 (Outer2* outer)
{
    int var = 1, var2 = 2;
    ++outer->outer.current;
    try3(*outer, var, var2);
}

int main ()
{
    Outer2 outer2;
    outer2.outer.current = 1986;
    try2 (&outer2);
}

From try3, you would need to use ++outer.outer.current since outer is the name of the variable at the point. To answer the actual question though, look at the two functions. try2 takes a pointer to an Outer2, so no copy is made. try2 passes this to try3 which take a reference to an Outer2, so again no copy is made. This means that yes, ++outer.outer.current will in fact affect the original outer2 declared in myMain.

Edit (per Brian's comment)

However, outer.outer.current is never initialized, unless that's not your true myMain, so the final value of current is undefined since the value was never defined before the increment.