Passing structures in C functions

Hi All.

I'm struggling to get back to grips with C following a 15 year absence...

I have a structure, as follows:-

Code:

typedef struct  {
        long mtype;
        int useragent;
        time_t mdatetime;
        char mtext[4096];
} busMessage;

And I want to populate it through a function/procedure by passing it to the function along with some variables as follows:-

Code:

void buildMessage( busMessage BM, long MT, int UA ) {
        BM.mtype     = MT;
        BM.useragent = UA;
        BM.mdatetime  = 1005454;
        strcpy(BM.mtext, "A test");
}

But when I use this function in the following way:-

Code:

int main(void) {
        busMessage myMessage;

        buildMessage (myMessage, 4, 4);
        printMessage (myMessage);
        exit(0);
}

.... all the fields in myMessage are zero.

The function printMessage just does this:-

Code:

void printMessage( busMessage BM ) {

	printf(ctime(&BM.mdatetime)); // Prints 1st Jan 1970 (correct, as mdatetime must be 0)

        printf("Time as int : %d\n", BM.mdatetime); // Zero

        printf("UA as int : %d\n", BM.useragent); // Zero

        printf("MT as int : %d\n", BM.mtype); // Zero

        printf("Payload :  %s\n", BM.mtext); // Blank
}

I have a funny feeling this may be a scope issue(?) but can't quite get my head around it.

Any help would be gratefully received!

Thanks.

If you want the function to modify it, you must pass it via a pointer.

Many thanks Elysia.

Something like (been a long time since I did this):-

Code:

busMessage myMessage;

doSomething( &myMessage );

void doSomething( *busMessage BM ) {
 (*BM).field1 = 123;
 // etc....
}

?

Thanks!

Close,

Code:

void doSomething( busMessage * BM ) {
 (*BM).field1 = 123;
 // etc....
}

BTW,

Code:

(*BM).field1 = 123;

is the same as

Code:

BM->field1 = 123;

The latter is easier on the eyes .
Don't forget if you dereference a NULL pointer bad things happen.

Perhaps check against NULL

Code:

void doSomething( busMessage * BM ) {
    if(BM == NULL)
        return;            /* don't continue */
    BM->field1 = 123;
    // etc....
}

Fantastic - thanks for your help.

It's all coming back to me slowly...

I would rather think an assert is better, since it's clearly a programmer error:

Code:

void doSomething( busMessage * BM ) {
    assert(BM);
    BM->field1 = 123;
    // etc....
}

Shouldnt the structure be defined like this:

Code:

typedef struct busMessage {
        long mtype;
        int useragent;
        time_t mdatetime;
        char mtext[4096];
};

Doing it the way you did it would create an instance of the structure...?

No, that would generate a compile error - the struct is a typedef, so you must specify a new name for the type (it's not creating an instance).
Otherwise, loose the "typedef" keyword.

nvm...............

EDIT: Sorry im used to C++ where adding typedef in front of a struct has no effect/not the same effect. In C++ doing this will create a new instance:

Code:

struct  {
        long mtype;
        int useragent;
        time_t mdatetime;
        char mtext[4096];
}busMessage;

busMessage.mtype = 1;

Originally Posted by 39ster

EDIT: Sorry im used to C++ where adding typedef in front of a struct has no effect/not the same effect.

No, it does not:

Code:

	typedef struct {
		long mtype;
		int useragent;
		time_t mdatetime;
		char mtext[4096];
	} a;

	a busMessage;
	busMessage.mtype = 1;

Originally Posted by Elysia

No, it does not:

Code:

	typedef struct {
		long mtype;
		int useragent;
		time_t mdatetime;
		char mtext[4096];
	} a;

	a busMessage;
	busMessage.mtype = 1;

ahuh...Well i use structures in C++ like this:

Defining a structure:

Code:

struct SomeStruct
{
  int a, b, c;
};

Create an instance of a struct:

Code:

struct
{
  int a, b, c;
} blah;
blah.a = blah.b = blah.c = 32;

Do both:

Code:

struct SomeStruct
{
  int a, b, c;
} blah;
blah.a = blah.b = blah.c = 32;

Yes, this is more correct.