CODEWITHSUNDEEP
c
b15
b15

Reputation: 2361

global variable going null outside function

I'm extremely new to C, so I'm sure this is really easy.

I'm trying to declare an array[10] of fractions in a header file and define it as a static variable in my example.c file. I initialize it in my function init_heap(). All elements of the array are null when that function returns, however. How do I do this properly? I need the changes to myArray to stick.

Header snippet:

struct fraction
{
    signed char sign;
    unsigned int numerator;
    unsigned int denominator;
};

extern struct fraction *myArray[10];

example.c:

//includes...

static struct fraction *myArray[10];


void init_heap()
{
    struct fraction myArray[] = {
        {0,0,1},
        {0,0,2},
        {0,0,3},
        {0,0,4},
        {0,0,5},
        {0,0,6},
        {0,0,7},
        {0,0,8},
        {0,0,9},
        {0,0,10}
    };
    beginFreeIndex = 0;
}
//etc...

Thanks in advance..

Upvotes: 3

Views: 1030

Answers (1)

Tushar
Tushar

Reputation: 8049

struct fraction myArray[] = {
    {0,0,1},
    {0,0,2},
    {0,0,3},
    {0,0,4},
    {0,0,5},
    {0,0,6},
    {0,0,7},
    {0,0,8},
    {0,0,9},
    {0,0,10}
};

You're making a local array called myArray which is hiding the global myArray. Then, when your function ends, the local myArray goes out of scope and you lose everything. Meanwhile, the global myArray is still null.

Try:

    myArray = {
    {0,0,1},
    {0,0,2},
    {0,0,3},
    {0,0,4},
    {0,0,5},
    {0,0,6},
    {0,0,7},
    {0,0,8},
    {0,0,9},
    {0,0,10}
    };

EDIT:

As @David Heffernan points out, you're declaring an array of pointers to struct fraction in this line: extern struct fraction *myArray[10];. I think you're trying to get just an array of struct fraction, so you should try this instead in place of that line: extern struct fraction myArray[10]

Upvotes: 2

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