CODEWITHSUNDEEP
cstruct
user2930701
user2930701

Reputation: 163

Struct in C misunderstanding

I am trying to understand the concept of struct and typedef in C. SO i created this simple program, but for some reason it is not working. 

#include <stdio.h>
#include <stdlib.h>

typedef struct testT{

    int number;

} testT;

int main()
{
    testT.number=10;

    printf("%d", testT->number);
}

However, it is giving me error: expected identifier or '(' before '.' token Why is this showing up?

Thanks

Upvotes: 0

Views: 129

Answers (6)

HennyH
HennyH

Reputation: 7944

typedef struct testT {
    int number;
} testT;

Defines a type called testT, which when used like so:

testT aTestStruct;

Is functionally equivalent to (an anonymous struct type):

struct { int number } aTestStruct;

The typdef just allows us to save ourselfs from having to repeat the anonymous struct declaration syntax of struct { ... } each time.

Upvotes: 0

Eregrith
Eregrith

Reputation: 4366

No, typedef struct xxx { ... } yyy; is not 'a C++ syntax', it is a correct C syntax.

What you are doing here is defining a type called testT which represents a struct testT being a struct containing a field of type int called number.

What you need to do to use this type is declare a variable of that type:

int main(void)
{
  testT variable;

  variable.number = 10;
  printf("%d\n", variable.number);
}

-> notation is used to replace (*).. For example :

//With variable declared like this :
testT *variable;
//You can write:
variable->number
//Or:
(*variable).number

Upvotes: 1

haccks
haccks

Reputation: 106012

testT is a type. First declare a variable of type testT 

testT t;

And do not forget to change 

printf("%d", testT->number);

to 

printf("%d", t.number);

Upvotes: 0

Sergey L.
Sergey L.

Reputation: 22542

Both struct <name> and typedef <type> <name> only define types. You have to actually allocate memory to hold an instance of that type:

int main()
{
  struct testT a;
  a.number=10;

  printf("%d", a.number);
}

Also you are misusing the -> operator. This is only useful with pointers to structs.

int main()
{
  struct testT a;
  struct testT *b = &a;
  a.number=10;

  printf("%d", b->number);
}

b->number is equivalent to (*b).number;

Upvotes: 1

mattnedrich
mattnedrich

Reputation: 8022

You need to create an instance of your struct. You've defined it, but you need to instantiate it.

testT mytestT;
mytestT.number=10;
...

Upvotes: 1

Yu Hao
Yu Hao

Reputation: 122383

testT is a type, just like int, you need to use a variable:

testT t;
t.number=10;
printf("%d", t.number);

Also note that you should use dot operator . because the arrow operator -> is used on a pointer to struct.

Upvotes: 5

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