Reputation: 1976
let's say we have a union:
typedef union someunion {
int a;
double b;
} myunion;
Is it possible to check what type is in union after I set e.g. a=123? My approach is to add this union to some structure and set uniontype to 1 when it's int and 2 when it's double.
typedef struct somestruct {
int uniontype
myunion numbers;
} mystruct;
Is there any better solution?
Upvotes: 50
Views: 23230
Reputation: 31
There isn't a direct way to inspect the type of data stored in the union. So, you are responsible for tracking the type in the union.
For a better readablity, using an Enum over int can be a good idea.
typedef union someunion {
int a;
double b;
} myunion;
typedef enum someEnum{
INT,
DOUBLE
} myuniontype;
typedef struct somestruct {
myuniontype uniontype;
myunion numbers;
} mystruct;
int main() {
mystruct my_container;
my_container.uniontype = myuniontype::INT;
my_container.numbers.a=5;
switch (my_container.uniontype) {
case myuniontype::INT:
std::cout<<my_container.numbers.a<<std::endl;
break;
case myuniontype::DOUBLE:
std::cout<<my_container.numbers.b<<std::endl;
break;
}
return 0;
}
Upvotes: 1
Reputation: 1
Maybe my variant is helping
struct Table
{
char mas[10];
int width;
int high;
union stat
{
int st;
char v;
} un;
};
Table tble[2];
strcpy(tble[0].mas, "box");
tble[0].high = 12;
tble[0].width = 14;
tble[0].un.v = 'S';
strcpy(tble[1].mas, "bag");
tble[1].high = 12;
tble[1].width = 14;
tble[1].un.st = 40;
//struct Table *ptbl = &tble[0];
//ptbl++;
for (int i = 0; i < 2; i++)
{
void *pt = &tble[i].un;
if(*((char*)pt) == 'S')
sort(put_on_bag_line);
else
sort(put_on_box_line);
}
Upvotes: 0
Reputation: 726479
Is there any better solution?
No, the solution that you showed is the best (and the only) one. union
s are pretty simplistic - they do not "track" what you've assigned to what. All they do is let you reuse the same memory range for all their members. They do not provide anything else beyond that, so enclosing them in a struct
and using a "type" field for tracking is precisely the correct thing to do.
Upvotes: 44
Reputation: 2474
Warning: the following is just for learning purpose:
You could use some ugly tricks to do so (as long as the data types in your union have different sizes, which is the present case):
#include <stdio.h>
typedef union someunion {
int a;
double b;
} myunion;
typedef struct somestruct {
int uniontype;
myunion numbers;
} mystruct;
#define UPDATE_CONTENT(container, value) if ( \
((sizeof(value) == sizeof(double)) \
? (container.uniontype = ((container.numbers.b = value), 2)) \
: (container.uniontype = ((container.numbers.a = value), 1))))
int main()
{
mystruct my_container;
UPDATE_CONTENT(my_container, 42);
printf("%d\n", my_container.uniontype);
UPDATE_CONTENT(my_container, 37.1);
printf("%d\n", my_container.uniontype);
return (0);
}
But I advise you never do this.
Upvotes: 2
Reputation: 7343
Depending on the application, if it is a short lived object you may be able to encode the type in the control flow, ie. have separate blocks/functions for both cases
struct value {
const char *name;
myunion u;
};
void throwBall(Ball* ball)
{
...
struct value v;
v.name = "Ball"; v.u.b = 1.2;
process_value_double(&v); //double
struct value v2;
v2.name = "Age";
v2.u.a = 19;
check_if_can_drive(&v2); //int
...
}
void countOranges()
{
struct value v;
v.name = "counter";
v.u.a = ORANGE;
count_objects(&v); //int
}
Upvotes: 2
Reputation: 40613
There is no way to directly query the type currently stored in a union
.
The only ways to know the type stored in a union
are to have an explicit flag (as in your mystruct
example), or to ensure that control only flows to certain parts of the code when the union has a known active element.
Upvotes: 6
Reputation: 1924
C does not automatically keep track of which field in a union is currently in use. (In fact, I believe reading from the "wrong" field results in implementation defined behavior.) As such, it is up to your code to keep track of which one is currently used / filled out.
Your approach to keeping a separate 'uniontype' variable is a very common approach to this, and should work well.
Upvotes: 9