Array Initialisation in C using Typedefs

Question

Getting an odd error trying to initialise an array in C - anyone know why this might happen?

I have a global variable:

static my_type foo[6];

And in an included header file, I have:

typedef uint32_t my_type[5];

I then in a function in the same file as the global variable try to do:

foo = {{1, 2, 3, 4, 5}, {1, 2, 3, 4, 6}, {1, 2, 3, 4, 7}, {1, 2, 3, 4, 8}, {1, 2, 3, 4, 9}, {1, 2, 3, 4, 10}};

The compiler (GCC4) gives the error 'expected expression before '{' token'.

Anyone know what's gone wrong and how to fix it?

Cheers!

Fred Foo · Accepted Answer · 2013-10-29 10:45:14Z

4

That's not initialization, that's assignment. Initialization has to be a single statement:

static my_type foo[6] = {{1, 2, 3, 4, 5},
                         {1, 2, 3, 4, 6},
                         {1, 2, 3, 4, 7},
                         {1, 2, 3, 4, 8},
                         {1, 2, 3, 4, 9},
                         {1, 2, 3, 4, 10}};

You cannot assign to a whole array in C89 with this syntax. What you can do is memcpy from a const:

void initialize_foo()
{
    static const my_type init[6] =
                        {{1, 2, 3, 4, 5},
                         {1, 2, 3, 4, 6},
                         {1, 2, 3, 4, 7},
                         {1, 2, 3, 4, 8},
                         {1, 2, 3, 4, 9},
                         {1, 2, 3, 4, 10}};
    assert(sizeof(foo) == sizeof(init));
    memcpy(foo, init, sizeof(foo));
}

edited Oct 29, 2013 at 10:45

answered Oct 29, 2013 at 10:37

Fred Foo

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5 Comments

mjs Over a year ago

He does say "in a function" though. Is the issue that he is attempting to initialize a variable other than at its declaration?

Chris B Over a year ago

Yeah, basically I have a function lower down in the file (my global variables at the top) and I'm trying to set the array to what I wrote in my function. Surely this is possible? Or if not, is there another way of doing what I'd like to do?

mjs Over a year ago

I think there might be a GCC extension, but I am not sure. In this case though, you might be best to assign into a temp array, and memcopy... but I dont like that.

Chris B Over a year ago

So I can only set an array like this at the time I declare the array, and can't then change the contents of the array later? Hmm, that's a bit pants, but okay, thanks!

Fred Foo Over a year ago

@ChrisB: I added a memcpy example.

David Ranieri · Accepted Answer · 2013-10-30 08:59:05Z

4

If you are under C99:

ISO C99 supports compound literals. A compound literal looks like a cast containing an initializer. Its value is an object of the type specified in the cast, containing the elements specified in the initializer; it is an lvalue. As an extension, GCC supports compound literals in C89 mode and in C++.

But foo must be a pointer

#include <stdio.h>
#include <stdint.h>

typedef uint32_t my_type[5];

int main(void)
{
    int i, j;
    my_type *foo;

    foo = ((my_type[]) {
      {1, 2, 3, 4, 5},
      {1, 2, 3, 4, 6},
      {1, 2, 3, 4, 7},
      {1, 2, 3, 4, 8},
      {1, 2, 3, 4, 9},
      {1, 2, 3, 4, 10}
    });
    for (i = 0; i < 6; i++) {
        for (j = 0; j < 5; j++) {
            printf("%d ", foo[i][j]);
        }
        printf("\n");
    }
    return 0;
}

edited Oct 30, 2013 at 8:59

answered Oct 29, 2013 at 10:46

David Ranieri

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4 Comments

dyp Over a year ago

Are you sure foo = ABC is allowed? foo is an lvalue, but array lvalues are not modifiable.

dyp Over a year ago

As per 6.2.5/6 " If the compound literal occurs outside the body of a function, the object has static storage duration; otherwise, it has automatic storage duration associated with the enclosing block." I wouldn't make foo static, the assigned literal only has automatic storage duration; also, what's the benefit of a static variable in main?

David Ranieri Over a year ago

You are right (again), there is no benefit in this case, but OP is using static in his example. Anyway edited.

dyp Over a year ago

"but OP is using static in his example." Oh, yeah, but the OP also said: "I have a global variable". Then, (s)he tries to assign inside a function. That won't work with compound literals, I fear.

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