Difference between 'struct' and 'typedef struct' in C++?


In C++, is there any difference between:

struct Foo { ... };


typedef struct { ... } Foo;
3/4/2009 8:41:12 PM

Accepted Answer

In C++, there is only a subtle difference. It's a holdover from C, in which it makes a difference.

The C language standard (C89 §, C99 §6.2.3, and C11 §6.2.3) mandates separate namespaces for different categories of identifiers, including tag identifiers (for struct/union/enum) and ordinary identifiers (for typedef and other identifiers).

If you just said:

struct Foo { ... };
Foo x;

you would get a compiler error, because Foo is only defined in the tag namespace.

You'd have to declare it as:

struct Foo x;

Any time you want to refer to a Foo, you'd always have to call it a struct Foo. This gets annoying fast, so you can add a typedef:

struct Foo { ... };
typedef struct Foo Foo;

Now struct Foo (in the tag namespace) and just plain Foo (in the ordinary identifier namespace) both refer to the same thing, and you can freely declare objects of type Foo without the struct keyword.

The construct:

typedef struct Foo { ... } Foo;

is just an abbreviation for the declaration and typedef.


typedef struct { ... } Foo;

declares an anonymous structure and creates a typedef for it. Thus, with this construct, it doesn't have a name in the tag namespace, only a name in the typedef namespace. This means it also cannot be forward-declared. If you want to make a forward declaration, you have to give it a name in the tag namespace.

In C++, all struct/union/enum/class declarations act like they are implicitly typedef'ed, as long as the name is not hidden by another declaration with the same name. See Michael Burr's answer for the full details.

7/14/2018 5:08:20 PM

In this DDJ article, Dan Saks explains one small area where bugs can creep through if you do not typedef your structs (and classes!):

If you want, you can imagine that C++ generates a typedef for every tag name, such as

typedef class string string;

Unfortunately, this is not entirely accurate. I wish it were that simple, but it's not. C++ can't generate such typedefs for structs, unions, or enums without introducing incompatibilities with C.

For example, suppose a C program declares both a function and a struct named status:

int status(); struct status;

Again, this may be bad practice, but it is C. In this program, status (by itself) refers to the function; struct status refers to the type.

If C++ did automatically generate typedefs for tags, then when you compiled this program as C++, the compiler would generate:

typedef struct status status;

Unfortunately, this type name would conflict with the function name, and the program would not compile. That's why C++ can't simply generate a typedef for each tag.

In C++, tags act just like typedef names, except that a program can declare an object, function, or enumerator with the same name and the same scope as a tag. In that case, the object, function, or enumerator name hides the tag name. The program can refer to the tag name only by using the keyword class, struct, union, or enum (as appropriate) in front of the tag name. A type name consisting of one of these keywords followed by a tag is an elaborated-type-specifier. For instance, struct status and enum month are elaborated-type-specifiers.

Thus, a C program that contains both:

int status(); struct status;

behaves the same when compiled as C++. The name status alone refers to the function. The program can refer to the type only by using the elaborated-type-specifier struct status.

So how does this allow bugs to creep into programs? Consider the program in Listing 1. This program defines a class foo with a default constructor, and a conversion operator that converts a foo object to char const *. The expression

p = foo();

in main should construct a foo object and apply the conversion operator. The subsequent output statement

cout << p << '\n';

should display class foo, but it doesn't. It displays function foo.

This surprising result occurs because the program includes header lib.h shown in Listing 2. This header defines a function also named foo. The function name foo hides the class name foo, so the reference to foo in main refers to the function, not the class. main can refer to the class only by using an elaborated-type-specifier, as in

p = class foo();

The way to avoid such confusion throughout the program is to add the following typedef for the class name foo:

typedef class foo foo;

immediately before or after the class definition. This typedef causes a conflict between the type name foo and the function name foo (from the library) that will trigger a compile-time error.

I know of no one who actually writes these typedefs as a matter of course. It requires a lot of discipline. Since the incidence of errors such as the one in Listing 1 is probably pretty small, you many never run afoul of this problem. But if an error in your software might cause bodily injury, then you should write the typedefs no matter how unlikely the error.

I can't imagine why anyone would ever want to hide a class name with a function or object name in the same scope as the class. The hiding rules in C were a mistake, and they should not have been extended to classes in C++. Indeed, you can correct the mistake, but it requires extra programming discipline and effort that should not be necessary.

Licensed under: CC-BY-SA with attribution
Not affiliated with: Stack Overflow