sizeof a union in C/C++

Question

What is the sizeof the union in C/C++? Is it the sizeof the largest datatype inside it? If so, how does the compiler calculate how to move the stack pointer if one of the smaller datatype of the union is active?

Answer 1

A union always takes up as much space as the largest member. It doesn't matter what is currently in use.

union {   short x;   int y;   long long z; } 

An instance of the above union will always take at least a long long for storage.

Side note: As noted by Stefano, the actual space any type (union, struct, class) will take does depend on other issues such as alignment by the compiler. I didn't go through this for simplicity as I just wanted to tell that a union takes the biggest item into account. It's important to know that the actual size does depend on alignment.

Answer 2

The Standard answers all questions in section 9.5 of the C++ standard, or section 6.5.2.3 paragraph 5 of the C99 standard (or paragraph 6 of the C11 standard, or section 6.7.2.1 paragraph 16 of the C18 standard):

In a union, at most one of the data members can be active at any time, that is, the value of at most one of the data members can be stored in a union at any time. [Note: one special guarantee is made in order to simplify the use of unions: If a POD-union contains several POD-structs that share a common initial sequence (9.2), and if an object of this POD-union type contains one of the POD-structs, it is permitted to inspect the common initial sequence of any of POD-struct members; see 9.2. ] The size of a union is sufficient to contain the largest of its data members. Each data member is allocated as if it were the sole member of a struct.

That means each member share the same memory region. There is at most one member active, but you can't find out which one. You will have to store that information about the currently active member yourself somewhere else. Storing such a flag in addition to the union (for example having a struct with an integer as the type-flag and an union as the data-store) will give you a so called "discriminated union": An union which knows what type in it is currently the "active one".

One common use is in lexers, where you can have different tokens, but depending on the token, you have different informations to store (putting line into each struct to show what a common initial sequence is):

struct tokeni {     int token; /* type tag */     union {         struct { int line; } noVal;         struct { int line; int val; } intVal;         struct { int line; struct string val; } stringVal;     } data; }; 

The Standard allows you to access line of each member, because that's the common initial sequence of each one.

There exist compiler extensions that allow accessing all members disregarding which one currently has its value stored. That allows efficient reinterpretation of stored bits with different types among each of the members. For example, the following may be used to dissect a float variable into 2 unsigned shorts:

union float_cast { unsigned short s[2]; float f; }; 

That can come quite handy when writing low-level code. If the compiler does not support that extension, but you do it anyway, you write code whose results are not defined. So be certain your compiler has support for it if you use that trick.