CODEWITHSUNDEEP
cstructureunions
name_masked
name_masked

Reputation: 9793

Structures and Unions in C, determining size and accessing members

All,

Here is an example on Unions which I find confusing.

struct s1
{
    int a;
    char b;
    union
    {
       struct
       {
          char *c;
          long d;
       }
       long e;
     }var;
};

Considering that char is 1 byte, int is 2 bytes and long is 4 bytes. What would be the size of the entire struct here ? Will the union size be {size of char*}+ {size of double} ? I am confused because of the struct wrapped in the union.

Also, how can I access the variable d in the struct. var.d ?

Upvotes: 3

Views: 3203

Answers (3)

C.J.
C.J.

Reputation: 16111

Unions are dangerous and risky to use without strict discipline. And the fact you put it in a struct is really dangerous because by default all struct members are public: that exposes the possibility of client code making changes to your union, without informing your program what type of data it stuffed in there. If you use a union you should put it in a class where at least you can hide it by making it private.

We had a dev years ago who drank the koolaid of unions, and put it in all his data structures. As a result, the features he wrote with it, are now one of the most despised parts of our entire application, since they are unmodifiable, unfixable and incomprehensible.

Also Unions throw away all the type safety that modern c/c++ compilers give you. Surely if you lie to the compiler it will get back at you someday. Well actually it will get back at your customer when your app crashes.

Upvotes: -3

RBerteig
RBerteig

Reputation: 43356

With no padding, and assuming sizeof(int)==sizeof(char *)==sizeof(long)==4, the size of the outer struct will be 13.

Breaking it down, the union var overlaps an anonymous struct with a single long. That inner struct is larger (a pointer and a long) so its size controls the size of the union, making the union consume 8 bytes. The other members are 4 bytes and 1 byte, so the total is 13.

In any sensible implementation with the size assumptions I made above, this struct will be padded to either 2 byte or 4 byte boundaries, adding at least 1 or 3 additional bytes to the size.

Edit: In general, since the sizes of all of the member types are themselves implementation defined, and the padding is implementation defined, you need to refer to the documentation for your implementation and the platform to know for sure.

The implementation is allowed to insert padding after essentially any element of a struct. Sensible implementations use as little padding as required to comply with platform requirements (e.g. RISC processors often require that a value is aligned to the size of that value) or for performance.

If using a struct to map fields to the layout of values assumed by file format specification, a coprocessor in shared memory, a hardware device, or any similar case where the packing and layout actually matter, then you might want to be concerned that you are testing at either compile time or run time that your assumptions of the member layout are true. This can be done by verifying the size of the whole structure, as well as the offsets of its members.

See this question among others for a discussion of compile-time assertion tricks.

Upvotes: 1

Matthew Flaschen
Matthew Flaschen

Reputation: 284927

The sizes are implementation-defined, because of padding. A union will be at least the size of the largest member, while a struct will be at least the sum of the members' sizes. The inner struct will be at least sizeof(char *) to sizeof(long), so the union will be at least that big. The outer struct will be at least sizeof(int) + 1 + sizeof(char *) + sizeof(long). All of the structs and unions can have padding.

You are using an extension to the standard, unnamed fields. In ISO C, there would be no way to access the inner struct. But in GCC (and I believe MSVC), you can do var.d.

Also, you're missing the semi-colon after the inner struct.

Upvotes: 5

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