typedef fixed length array

I have to define a 24-bit data type.I am using char[3] to represent the type. Can I typedef char[3] to type24? I tried it in a code sample. I put typedef char[3] type24; in my header file. The compiler did not complain about it. But when I defined a function void foo(type24 val) {} in my C file, it did complain. I would like to be able to define functions like type24_to_int32(type24 val) instead of type24_to_int32(char value[3]).

The typedef would be

typedef char type24[3];

However, this is probably a very bad idea, because the resulting type is an array type, but users of it won't see that it's an array type. If used as a function argument, it will be passed by reference, not by value, and the sizeof for it will then be wrong.

A better solution would be

typedef struct type24 { char x[3]; } type24;

You probably also want to be using unsigned char instead of char, since the latter has implementation-defined signedness.

You want

typedef char type24[3];

C type declarations are strange that way. You put the type exactly where the variable name would go if you were declaring a variable of that type.

From R..'s answer:

> However, this is probably a very bad idea, because the resulting type > is an array type, but users of it won't see that it's an array type. > If used as a function argument, it will be passed by reference, not by > value, and the sizeof for it will then be wrong.

Users who don't see that it's an array will most likely write something like this (which fails):

#include <stdio.h>

typedef int twoInts[2];

void print(twoInts *twoIntsPtr);
void intermediate (twoInts twoIntsAppearsByValue);

int main () {
    twoInts a;
    a[0] = 0;
    a[1] = 1;
    print(&a);
    intermediate(a);
    return 0;
}
void intermediate(twoInts b) {
    print(&b);
}

void print(twoInts *c){
    printf("%d\n%d\n", (*c)[0], (*c)[1]);
}

It will compile with the following warnings:

In function ‘intermediate’:
warning: passing argument 1 of ‘print’ from incompatible pointer type [enabled by default]
    print(&b);
     ^
note: expected ‘int (*)[2]’ but argument is of type ‘int **’
    void print(twoInts *twoIntsPtr);
         ^

And produces the following output:

0
1
-453308976
32767

Arrays can't be passed as function parameters by value in C.

You can put the array in a struct:

typedef struct type24 {
    char byte[3];
} type24;

and then pass that by value, but of course then it's less convenient to use: x.byte[0] instead of x[0].

Your function type24_to_int32(char value[3]) actually passes by pointer, not by value. It's exactly equivalent to type24_to_int32(char *value), and the 3 is ignored.

If you're happy passing by pointer, you could stick with the array and do:

type24_to_int32(const type24 *value);

This will pass a pointer-to-array, not pointer-to-first-element, so you use it as:

(*value)[0]

I'm not sure that's really a gain, since if you accidentally write value[1] then something stupid happens.

To use the array type properly as a function argument or template parameter, make a struct instead of a typedef, then add an operator[] to the struct so you can keep the array like functionality like so:

typedef struct type24 {
  char& operator[](int i) { return byte[i]; }
  char byte[3];
} type24;

type24 x;
x[2] = 'r';
char c = x[2];

Here's a short example of why typedef array can be confusingly inconsistent. The other answers provide a workaround.

#include <stdio.h>
typedef char type24[3];

int func(type24 a) {
        type24 b;
        printf("sizeof(a) is %zu\n",sizeof(a));
        printf("sizeof(b) is %zu\n",sizeof(b));
        return 0;
}

int main(void) {
        type24 a;
        return func(a);
}

This produces the output

sizeof(a) is 8
sizeof(b) is 3

because type24 as a parameter is a pointer. (In C, arrays are always passed as pointers.) The gcc8 compiler will issue a warning by default, thankfully.

Building off the accepted answer, a multi-dimensional array type, that is a fixed-length array of fixed-length arrays, can't be declared with

typedef char[M] T[N];  // wrong!

instead, the intermediate 1D array type can be declared and used as in the accepted answer:

typedef char T_t[M];
typedef T_t T[N];

or, T can be declared in a single (arguably confusing) statement:

typedef char T[N][M];

which defines a type of N arrays of M chars (be careful about the order, here).