accessing inner struct members

[Walter Deodiaus]

I have
typedef struct {
union _union{
....
struct {
int i;
}u1;
....
}Union;
} Struct ;
now I want to define a method's signaure whose first arg will be "u1"

e.g. foo(Struct::Uni on.u1)

Is this possible without changing the layout to
struct _u1{
int b;
} ;
typedef struct {
union _union{
....
_u1 u1;
....
}Union;
} Struct ;

method's signature
foo(u1 the_u1);

[mlimber]

Walter Deodiaus wrote:

I have
typedef struct {
union _union{
...
struct {
int i;
}u1;
...
}Union;
} Struct ;
now I want to define a method's signaure whose first arg will be "u1"

e.g. foo(Struct::Uni on.u1)

Is this possible without changing the layout to
struct _u1{
int b;
} ;
typedef struct {
union _union{
...
_u1 u1;
...
}Union;
} Struct ;

method's signature
foo(u1 the_u1);

You want something like:

struct S1
{
union U
{
char c[4];
struct S2
{
int i;
} s2;
} u;
};

void foo(S1::U::S2);

int main()
{
foo( S1::U::S2() );
}

Cheers! --M

[Old Wolf]

Walter Deodiaus wrote:

struct {
int i;
}u1;

now I want to define a method's signaure whose first arg will be "u1"

You didn't give the struct a type name, so it is impossible to
refer to its type.