Initializing typedef arrays in structs

Question

I'm making a Camera struct. the struct use a vec3 which is defined with typedef float vec3[3].

To initialize a vec3 I do vec3 vector = {0.,0.,0.};

My Cameras struct is like this:

typedef struct Cameras {
    vec3 eye;
    vec3 target
} Camera;

but when I do:

Camera cam;
cam.eye = { .5, 1., 0. };

it crashes and the compiler tells me: expression must be modifiable.

I thought it was an error with pointer but no, and replacing vec3 eye to vec3 * eye and cam->eye = {.5,1.,0.} changes nothing.

Am I creating the struct the wrong way or is a really common issue C and I'm just blind?

here my goal is not only to initalise the array, but also accessing the data and modifing/passing into function after the creation.

Possible duplicate of [Initialize values of array in a struct](https://stackoverflow.com/questions/9856045/initialize-values-of-array-in-a-struct) — static_cast, Dec 21 '18 at 13:53
@static_cast not fully a dubplicate because i want also to acces the var after creating it, seem like the only option i've found it doing array[0], array[1], ect.. — Cewein, Dec 22 '18 at 21:06

score 6 · Accepted Answer · answered Dec 21 '18 at 13:53

6

It's because after the line cam is defined, subsequent "assignments" are no longer initializations.

To initialize an instance, you must put the "assignment" at the same place where it's defined:

Camera cam = { {.5, 1., 0.} };

or specifying member:

Camera cam = { .eye = {.5, 1., 0.} };

Note that this will also automatically zero-initialize cam.target as no initial values are specified for it.

answered Dec 21 '18 at 13:53

iBug

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is it the same when you want to use struct after the init ? – Cewein Dec 21 '18 at 14:29
@Cewein No. You can only assign like this during initialization, not at any point afterwards. – iBug Dec 21 '18 at 14:37
to use it afer, i'm try to cast it, seem to not work – Cewein Dec 21 '18 at 14:48
regarding: `{1.0, 2.0, 3.0}` This is defining 3 `double` literals. To define 3 float literals use: `{1.0f, 2.0f, 3.0f}` Note the `f` trailing each literal value – user3629249 Dec 22 '18 at 07:53

score 1 · Answer 2 · edited Jun 20 '20 at 09:12

1

The error you get is obvious, type of vec3 is float [3] - array type and you certainly cannot "assign" to array types.

You can use the initialization as mentioned by iBug in another answer. However, with a small trick, you'll be able to use the assignment, also. This would need the usage of a pointer and compound literal.

You need to change the typedef to vec3

 typedef float* vec3;

so as to make it a float * and then, you can use compound literal to assign the values, like

cam.eye = (float [3]){ .5, 1., 0. };
cam.target = (float [3]){ .45, 2.5, 0.9 }; // just some other values.

The primary benefit of using this approach is, you are not limited to only "initialization", you can perform the assignment at any time.

P.S. - Compound literals are modifiable, so you do not lose any operation capabilities.

P.P.S.

Quoting C11, chapter §6.5.2.5/ P12

"/tmp/fileXXXXXX" 
 (char []){"/tmp/fileXXXXXX"} 
 (const char []){"/tmp/fileXXXXXX"} 
The first always has static storage duration and has type array of char, but need not be modifiable; the last two have automatic storage duration when they occur within the body of a function, and the first of these two is modifiable.

edited Jun 20 '20 at 09:12

Community

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answered Dec 21 '18 at 13:56

Sourav Ghosh

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The only problem is that the compound literal will be placed in the RO memory. Any attempt to modify them will be the UB (same as string literals) – 0___________ Dec 21 '18 at 13:59
@P__J__ What? How? IIRC, CLs are modifiable Lvalue. – Sourav Ghosh Dec 21 '18 at 13:59
@P__J__ So, you're saying the standard is wrong? :) Chapter §6.5.2.5/P12. – Sourav Ghosh Dec 21 '18 at 14:10
I just look at the real compiler output. As I see it is in the .rodata segment. – 0___________ Dec 21 '18 at 14:12
@P__J__ That can be answered / reasoned by compiler docs. I'm talking from Spec perspective. – Sourav Ghosh Dec 21 '18 at 14:13
And I from the real programmer use. It is one of the reasons I do not like them (I only use them if I am not going to modify it). Better to avoid - they are not in the common use and as you see it hard to relay on them. – 0___________ Dec 21 '18 at 14:16

score 0 · Answer 3 · answered Dec 21 '18 at 13:54

I personally prefer more "descriptive" initialization - I like members to be specified - the code is easier to read for humans.

typedef float vec3[3];

typedef struct Cameras {
    vec3 eye;
    vec3 target;
} Camera;

Camera c = 
{
  .eye = {1.0, 2.0, 3.0},   
  .target = {4.0, 5.0, 6.0},   
};

Camera c1 = 
{
  .target = {4.0, 5.0, 6.0},   
};

Camera c2 = 
{
  .eye = {1.0, 2.0, 3.0},   
};

Initializing typedef arrays in structs

3 Answers3