Can I 'extend' a struct in C?
Evidently this feature has been added to C11, but alas I don't have access to a C compiler of recent vintage (>= GCC 4.6.2).
typedef struct foo {
int a;
} foo;
typedef struct bar {
struct foo;
int b;
} bar;
int main() {
bar b;
b.a = 42;
b.b = 99;
return 0;
}
You can, using pointers, because a pointer to a structure object is guaranteed to point its first member. See e.g. this article.
#include <stdlib.h>
#include <stdio.h>
typedef struct foo_s {
int a;
} foo;
typedef struct bar_s {
foo super;
int b;
} bar;
int fooGetA(foo *x) {
return x->a;
}
void fooSetA(foo *x, int a) {
x->a = a;
}
int main() {
bar* derived = (bar*) calloc(1, sizeof(bar));
fooSetA((foo*) derived, 5);
derived->b = 3;
printf("result: %d\n", fooGetA((foo*) derived));
return 0;
}
If you ment
typedef struct foo_s {
int a;
} foo;
typedef struct bar_s {
foo my_foo;
int b;
} bar;
so you can do:
bar b; b.my_foo.a = 3;
Otherwise, There's no way of doing it in C since the sizeof(bar_s)
is detriment on compile time. It's not a good practice but you can save a void * ptr;
pointer within bar_s, and another enum which describes the ptr
type, and cast by the type.
i.e:
typedef enum internalType{
INTERNAL_TYPE_FOO = 0,
}internalType_t;
typedef struct bar_s {
internalType_t ptrType;
void* ptr;
int b;
} bar;
and then:
bar b; foo f;
b.ptrType = INTERNAL_TYPE_FOO;
b.ptr = &f;
and some where else in the code:
if (b.ptrType == INTERNAL_TYPE_FOO) {
foo* myFooPtr = (foo *)b.ptr;
}
Not possible in C
the way you did. But you can mimic inheritance having a foo
member variable in bar
.
typedef struct bar_s {
foo obj;
int b;
} bar;
bar b;
b.obj.a = 10;