Passing a string array to a function using pointers

Question

So I am trying to a pass a char array into a function and print it out and then return the array back.

Below is my main function. I am doing Binary Search Trees however Im just stuck on passing a char array.

int main()
{
    BST tree = new_bst();
    int noRecords;
    int tfn;
    char name[10];
    char *test;

    FILE *fileptr;
    fileptr = fopen("welfare1.txt", "r");
    fscanf(fileptr, "%d", &noRecords);
    fscanf(fileptr, "%s", name);
    fscanf(fileptr, "%d", &tfn);

    insert_bst(&tree, tfn, &name);
}

And here is the insert BST function so far. I haven't implemented it yet as I can't pass the name field to my function, im using the printf as a test to see if it worked.

BSTNodePtr insert_bst_node(BSTNodePtr self, int n, char *name)
{

    if (self == NULL)
    {
        TaxRecordPtr newRecord = new_record();


        self = malloc(sizeof *self);
        self->data = newRecord;
        self->left = self->right = NULL;

        printf("%c", name[0]);
}

arrays are passed by reference only so, simply insert_bst(&tree, tfn, name); and BSTNodePtr insert_bst_node(BSTNodePtr self, int n, char name[]) should work for you — Shubham Khatri
Using insert_bst(&tree, tfn, name); and BSTNodePtr insert_bst_node(BSTNodePtr self, int n, char name[]) It steps through but the printf function fails and I get this Exception thrown at 0x0F63125C (ucrtbased.dll) in Assign1.exe: 0xC0000005: Access violation reading location 0x0000007C. — Tarik Stoneman
You have mismatched parens in insert_bst_node and the name does not match the call to insert_bst — stark

John Bode John Bode · Accepted Answer · 2017-04-27T13:27:16

Change

insert_bst(&tree, tfn, &name);

to

insert_bst(&tree, tfn, name);

Except when it is the operand of the sizeof or unary & operators, or is a string literal used to initialize a character array in a declaration, an expression of type "N-elemement array of T" will be converted ("decay") to an expression of type "pointer to T", and the value of the expression will be the address of the first element of the array.

When you call

insert_bst(&tree, tfn, name);

the expression name isn't the operand of the sizeof or unary & operators, so it "decays" to type char *, which matches the type of the argument in the function definition.

When you call

insert_bst(&tree, tfn, &name);

the expression name is the operand of the unary & operator, so the conversion doesn't happen; instead, the type of the expression is char (*)[10], or "pointer to 10-element array of char", which doesn't match the type of the argument in the function definition.

Passing a string array to a function using pointers

2 Answers