CODEWITHSUNDEEP
cstruct
Bracketz
Bracketz

Reputation: 327

How to declare an uint8_t "string" in a struct?

Creating an uint8_t "string" works fine for me:

uint8_t myString[] = "Foo";

However when I try to declare it in a struct, it does not work:

#ifndef PLAYER_H_
#define PLAYER_H_

#include <stdint.h>

typedef struct{
    uint32_t id;
    uint8_t name[];
    uint16_t life;
    uint16_t mana;
    uint16_t fist;
    uint16_t shielding;
} player;

#endif // PLAYER_H_

I get a compilation message saying: Field has incomplete type uint8_t[].

What am I missing here guys?

Upvotes: 2

Views: 4036

Answers (3)

Bo Persson
Bo Persson

Reputation: 92211

You have to give name a size inside the string. Otherwise the compiler doesn't know how large the struct is.

When declaring a variable like

uint8_t myString[] = "Foo";

the compiler can figure the size out from the size of "Foo", but if you declare an array without an initial value, you have to tell the size.

(The comments question if a uint8_t array can really be initialized with a string. On systems where char is unsigned, uint8_t could be a typedef for char.)

Upvotes: 2

ameyCU
ameyCU

Reputation: 16607

  uint8_t name[];

You have not mentioned size here , either mention size of name. You cannot create array of unknown size in somewhere middle of your structure. 

uint8_t name[50];                // this could do

And you could initialize it like this -

player p1={.name="hello"};

Or use memcpy .

Upvotes: 2

Jasen
Jasen

Reputation: 12402

if you want to have an array of indeterminate size in a struct it must be the last element of the struct, and when allocating memory for the struct you must allocate extra for the array contents.

Upvotes: 1

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