Segmentation fault when attempting to malloc arrays inside pointer to struct

Question

I'm trying to allocate memory for an array inside a struct like this one:

    typedef struct
    {
        int a;
        int *b;
    } MyStruct;

I have a function to initialize the struct like this one:

    void init( MyStruct * myStruct, int size )
    {
        myStruct = malloc( sizeof( MyStruct ) );
        myStruct->a = size;
        myStruct->b = malloc( size * sizeof( int ) );
    }

But when I try to allocate memory for the fields inside the array b, I get the segmentation fault error

    int main( void )
    {
        int i, size = 5;
        MyStruct *myStruct;
        init( myStruct, size );

        for( i = 0; i < size; i++ )
        {
            myStruct->b[i] = malloc( sizeof( int ) ); //fails here
            myStruct->b[i] = i*i;
        }
    }

I tried searching over and over, but I was not able to solve this problem. Someone knows why this is happening?

collapsar · Accepted Answer · 2015-11-21 02:08:25Z

5

Parameter passing in c is call-by-value, so the memory allocated in your init function is never assigned to the myStruct Pointer variable in main but only to its copy that is passed into init.

answered Nov 21, 2015 at 2:08

collapsar

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cdonts · Accepted Answer · 2015-11-21 02:09:26Z

1

First, you've already allocated five "slots" to put ints in there, so why would you need to call malloc again inside the loop? Second, there's no reason to create a pointer to MyStruct. So, third, there's no need to allocate memory for the whole structure.

void init( MyStruct * myStruct, int size )
{
    myStruct->a = size;
    myStruct->b = malloc(size * sizeof( int ));
}

int main(int argc, char *argv[])
{
    int i, size = 5;
    MyStruct myStruct;
    init( &myStruct, size );

    for( i = 0; i < size; i++ )
    {
        myStruct.b[i] = i*i;
    }

    free(myStruct.b);

    return 0;
}

answered Nov 21, 2015 at 2:09

cdonts

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