Hi i want to make a function which calculate the size of structure without padding bytes.
Example :
struct test{
int x;
char y;
int z;
};
If i calculate the size of above structure i must get 9 byte (i.e. without padding bytes)
And consider the case where i might not know the variable present in struct.
Example :
struct test{
int x;
....
....
....
int z;
};
so if i calculate the size it should give correct size.
Is it possible do write such function?I tried reading on structure's but i dint find any solution.I saw there is some compiler option are present from that i can get but i dont want any in build compiler option.
No, C doesn't have enough introspection to make this possible. At run-time, there is no information left that the program can use to "know" which fields are in a structure, or what their types are.
The sizeof compiler construct should give you the size of the object "as the compiler sees it". Most compilers will align structures on 4 byte boundaries for efficient memory access at the expense of space. Try
#include <stdio.h>
int main() {
struct test{
int x;
char a;
int y;
};
struct test myStruct;
printf("size of myStruct is %ld\n", sizeof(myStruct));
}
When I run it, I get
size of myStruct is 12
Which shows that my compiler, in this situation, is allocating four bytes to 'a'.
I think that's your answer... unless I really misunderstood your question, and in particular your "without padding bytes" comment.
Rewriting the definition of the struct as:
struct test{
int x;
char a;
int y;
}__attribute__((packed));
The output becomes
size of myStruct is 9
This only works at compile time - so it is not really a "function". And it doesn't apply if you don't know your struct at compile time. As such, I think my answer and the one given by @unwind above are not inconsistent.
