Compilation difference C and C++

Question

I have a piece of code which looks like

#include <stdio.h>
#include <netdb.h>
#include <arpa/inet.h>


int main(int argc, char *argv[])
{
    struct hostent *server;
    server = gethostbyname(argv[1]);
    struct in_addr * address=(in_addr * )server->h_addr;
        fprintf(stderr,"[%d] [%s] server [%s] \n",__LINE__,__func__,inet_ntoa(*address));
}

When i use g++ compiler it gets compiled but the same gives the following error with gcc compiler

myclient.c: In function ‘main’:
myclient.c:10:31: error: ‘in_addr’ undeclared (first use in this function)
myclient.c:10:31: note: each undeclared identifier is reported only once for each function it appears in
myclient.c:10:41: error: expected expression before ‘)’ token

Am I missing something here?

`struct in_addr * address=(struct in_addr * )server->h_addr;` — user703016, Jan 24 '12 at 11:36
@Cicada: ahh I got it, but why is g++ is giving is as valid compilation? — Global Warrior, Jan 24 '12 at 11:41
C and C++ are not the same language, why do you even expect it to work out of the box? — PlasmaHH, Jan 24 '12 at 11:43
When I go to China and greet people with "Good morning" I get blank looks ... but it works great in South Africa. Am I missing something? — pmg, Jan 24 '12 at 11:44

score 4 · Answer 1 · answered Jan 24 '12 at 11:35

4

You should use struct in_addr, not just in_addr in c code (when casting too).

answered Jan 24 '12 at 11:35

Michael Krelin - hacker

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sidyll · Accepted Answer · 2012-01-24T15:10:53.560

3

It's important to note that C has multiple namespaces. The general namespace includes typedefs and other identifiers, but there is a separate namespace for structs, enums, and unions tags; another one for labels, and also each struct has it's own namespace. In code like this:

struct name {
    int name;
} name;

name means 3 completely different things and this is perfectly legal. To avoid typing the struct keyword, programmers often typedef a new name for the struct and its tag:

typedef struct name {
    int name;
} my_struct;

my_struct s1; /* easier than: struct name s1 */
my_struct s2;

In this last example name can even be omitted, unless you want to use a pointer to the struct inside itself. This practice was so common that in C++ the compiler considers the tag a defined type for structs. This is perfectly legal in C++:

struct my_struct {
    int name;
};

my_struct s1;
my_struct s2;

And indeed is what's happening in your program. You're using the struct tag as a type, but it's only a valid type in C++. You have two alternatives:

1.Typedef the struct in its declaration

typedef struct in_addr {
    /* ... */
} in_addr;

2.Use the `struct` keyword to refer to the struct type

(struct in_addr *)

The first one is not applicable since you didn't declare the struct. It already comes that way. So I'd stick with the second.

edited Jan 24 '12 at 15:10

answered Jan 24 '12 at 12:19

sidyll

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1

In c++ the two different namespaces still exist and are separate. The difference is in lookup, where for each scope the compiler will first check as in C the general identifier space, and if that fails it will also look in the user defined types identifier space. Check for example: [here](http://stackoverflow.com/a/7970257/36565) or [here](http://stackoverflow.com/a/1675446/36565) – David Rodríguez - dribeas Jan 24 '12 at 12:52
Thanks for the comment @DavidRodríguez-dribeas I must admit that my C++ knowledge is poor. I edited the answer slightly. Just to confirm: typedef'ed version with the same name for the struct tag and new type is then perfectly valid in C++, right? – sidyll Jan 24 '12 at 15:14
I am not sure about what you exactly mean, but this is ok: `typedef struct A {} B; struct B {};` although confusing. In that context `B` refers to the first type and `struct B` to the second. – David Rodríguez - dribeas Jan 25 '12 at 02:12
@DavidRodríguez-dribeas I was refering to the "1st alternative" I posted in the answer, more in the lines of `typedef struct A {} A;`. But your comment answered that. Thanks. – sidyll Jan 25 '12 at 11:41

score 2 · Answer 3 · answered Jan 24 '12 at 11:36

2

Change:

struct in_addr * address=(in_addr * )server->h_addr;

to:

struct in_addr * address=(struct in_addr * )server->h_addr;

Also, C (before C99) only allows variable declarations at the top of a block, and struct in_addr * address isn't.

answered Jan 24 '12 at 11:36

NPE

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score 0 · Answer 4 · answered Jan 24 '12 at 12:06

C and C++ have different rules for struct types; in C++, the keyword struct creates a class type where all members are public by default. This type name can stand on its own, unlike in C.

C++ is not merely a superset of C, and there's plenty of code that will compile as one but not the other, or will behave differently because of different semantics.

Compilation difference C and C++

4 Answers4

1.Typedef the struct in its declaration

2.Use the struct keyword to refer to the struct type

2.Use the `struct` keyword to refer to the struct type