using struct keyword in variable declaration in C++

Question

I have a feeling this may be related to C syntax, but I started my programming life with C++ so I am not sure.

Basically I have seen this:

struct tm t;
memset( &t, 0, sizeof(struct tm) );

I am a bit confused with this syntax, as normally I would expect the above to instead look like this:

tm t;
memset( &t, 0, sizeof(tm) );

What is the difference between the two, and why is the former used instead?

Update

The structure tm that I am referring to is in wchar.h, and the definition is as follows:

struct tm {
        int tm_sec;     /* seconds after the minute - [0,59] */
        int tm_min;     /* minutes after the hour - [0,59] */
        int tm_hour;    /* hours since midnight - [0,23] */
        int tm_mday;    /* day of the month - [1,31] */
        int tm_mon;     /* months since January - [0,11] */
        int tm_year;    /* years since 1900 */
        int tm_wday;    /* days since Sunday - [0,6] */
        int tm_yday;    /* days since January 1 - [0,365] */
        int tm_isdst;   /* daylight savings time flag */
        };

struct tm is from the standard C library and should be found in time.h not wchar.h — Nordic Mainframe
– Nordic Mainframe, Commented Oct 11, 2011 at 17:13
It probably is, but it's also in wchar.h on my platform as well (MSVC). The struct definition is wrapped in ifndef guard checking for _TM_DEFINED, but this is irrelevant to my original question. — void.pointer
– void.pointer, Commented Oct 11, 2011 at 17:15
@Praetorian It's only a duplicate in hindsight. My original confusion comes from a different perspective, which is still valuable and a valid & useful alternative to finding arguably the same information. — void.pointer
– void.pointer, Commented Oct 11, 2011 at 17:22

Community · Accepted Answer · 2017-05-23 10:30:16Z

24

The simple answer is that the struct keyword there is present to restrict the lookup of the identifier tm to only user defined class types. It is probably left for compatibility with C, where it is required.

Contrary to what others say, there is no such thing as auto-typedef, nor do C and C++ differ with respect to how the identifiers for user defined types are managed. The only difference is in lookup.

You can read more here

edited May 23, 2017 at 10:30

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answered Oct 11, 2011 at 17:11

David Rodríguez - dribeas

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7 Comments

James Kanze Over a year ago

C and C++ do differ with regards to how identifiers for user defined types are managed, in two ways. First, in C the names of structs, unions and enums are in a separate namespace, which is used instead of the normal namespace when looking up names behind these keywords. Such names can never conflict with other names in C, and such names will never be found except when they follow one of these keywords. C++ has special rules to allow the conflict, and resolve it, but only if there is a conflict.

James Kanze Over a year ago

And unlike C++, in C, even if a struct is defined in another struct, its name is as if it were defined outside the encapsulating struct. Thus, in C++ struct A { struct B {}; }; defines two structs: A and A::B. In C++, it also defines two structs: struct A and struct B.

David Rodríguez - dribeas Over a year ago

@JamesKanze: Conceptually C++ still maintains the two namespaces with the difference that the user defined types will be looked up if the lookup in the general namespace fails. As in C, the names of user defined types cannot conflict with other identifiers (variables, functions) --the typedef would cause an identifier to be added to the general namespace and that can cause problems. Again, as I said before, the difference is in the lookup not. The other difference (nested structs) is true and related to the absence of mangling of identifiers in the C language.

James Kanze Over a year ago

That's not the way the C++ standard describes it. §3.3.10/2 speaks of the class name or enumeration name being "hidden" when something else is declared in the same scope, and §3.3.1/4 speaks of declaring a class name or an enumeration name in the same scope as other names.

James Kanze Over a year ago

§3.4/1 also says "The name lookup rules apply uniformly to all names (including typedef-names (7.1.3), namespace-names (7.3), and class-names (9.1)) wherever the grammar allows such names in the context discussed by a particular rule." The rule, as I understand it, is that the other name will hide the class or enumeration name, except in contexts where the other name cannot legally appear (in practice, only in an elaborated type specifier).

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pmg · Accepted Answer · 2011-10-11 17:06:55Z

16

In C, the struct tag names do not form identifiers on the global name space

struct not_a_global_identifier { /* ... */ };

To refer to that struct you have to use the keyword struct (to specify the name space)

struct not_a_global_identifer object;

or create a new identifier, in the global name space, with typedef

typedef struct not_a_global_identifer { /* ... */ } global_name_space_identifier;

There are 4 namespaces in C, see 6.2.3 in the C99 Standard:

label names
the tags of structures, unions, and enumerations
the members of structures or unions (not a single name space ... as many as structures or unions are defined)
global name space, for all other identifiers

This is a legal C program :-)

int main(void) {
  typedef struct foobar { int foobar; } foobar;
  foobar boo;
  boo.foobar = 42;
  if (boo.foobar) goto foobar;
foobar:
  return 0;
}

answered Oct 11, 2011 at 17:06

pmg

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4 Comments

void.pointer Over a year ago

So basically with that in mind, in C I can declare a global struct named struct foo {}; and also right beside it a global variable with the same name, int foo; and there will be no conflict?

pmg Over a year ago

Yes. You can have the same identifier in different name spaces without problem. In the code above, foobar is at the same time a type (global name space), a structure member (member name space), a structure tag (tag name space), and a label (label name space).

David Rodríguez - dribeas Over a year ago

@RobertDailey: Not only in C, you can also do that in C++, and then you will hit the problem of having to add struct (or class) to declare a variable: struct X {}; int X; struct X x; Removing the struct keyword from the definition of x before will cause a compiler error.

Kindred Over a year ago

Thank you so much, very clear now. I suddenly found that it's kind of two-level idea of this... If it's not in current one the goto the drawer of struct and find foobar.

jwodder · Accepted Answer · 2011-10-11 16:57:53Z

1

Using struct tm t; is for compatibility with C, in which declaring a struct named "tm" defines a type named "struct tm" but not one named "tm" (as opposed to C++, in which both names for the type are declared).

answered Oct 11, 2011 at 16:57

jwodder

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Comments

Jaffa · Accepted Answer · 2011-10-11 17:12:50Z

1

The user-defined types have their own identifier space, i.e. when the compiler parse the file it stores each identifier in its corresponding space.

When you refer to tm, the C compiler (as the C++ one) will search for this identifier in the global identifier space. The C++ compiler will then lookup in the user-defined types identifier space if it hadn't found the symbol before.

Basically, if you want to have the same behavior as in C++, add this line :

typedef struct tm tm;

You can combine struct declaration and typedef like that :

typedef struct tm { int field } tm;

Or using a anonymous struct :

typedef struct { int field } tm;

The same behavior applies for enum and union :

typedef enum { VALUE } myEnum;
typedef union { int integer; char charArray[4]; } myUnion;

edited Oct 11, 2011 at 17:12

answered Oct 11, 2011 at 16:57

Jaffa

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5 Comments

Jaffa Over a year ago

The auto typedef just do what I did, but automatically. As C++ is more oriented on type, the type is directly available. C is more oriented on the meaning of variable, thus it is important to know that t is a struct tm, not just a tm. Notice that the same thing happens with enums and unions

David Rodríguez - dribeas Over a year ago

There is no auto typedef in C++ either. The effect of a typedef is declaring a new identifier in the global identifier space to refer to the type in the user-defined-types identifier space. That is, typedef struct tm tm; creates a new identifier tm to refer to struct tm. The behavior of typedef is exactly the same in C and C++. The actual difference is in lookup, where C++ will automatically lookup identifiers in the user-defined-types identifier space if the identifier was not previously found.

void.pointer Over a year ago

@DavidRodríguez-dribeas Could you please add some more detail to your last statement:

will automatically lookup identifiers in the user-defined-types identifier space if the identifier was not previously found

. Not sure what the "user-defined-types identifier space" is, nor what the exact lookup logic is that you are referring to.

David Rodríguez - dribeas Over a year ago

@RobertDailey: You can read more here

void.pointer Over a year ago

@DavidRodríguez-dribeas Thanks David. You should post an actual answer with all of this information!

BenMorel · Accepted Answer · 2014-06-14 14:06:03Z

1

In your example tm can have been a typecasted structure.

e.g.

typedef struct tm_t
{
  int x; 
}tm;

and then you can do

tm t;

edited Jun 14, 2014 at 14:06

BenMorel

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answered Oct 11, 2011 at 16:58

Jay D

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Comments

zar · Accepted Answer · 2011-10-11 18:00:44Z

0

You can see the reference linked below and quoting from there "In C, you must explicitly use the struct keyword to declare a structure. In C++, this is unnecessary once the type has been defined." See the link for more information and examples.

http://msdn.microsoft.com/en-us/library/64973255%28v=vs.80%29.aspx

answered Oct 11, 2011 at 18:00

zar

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