I'm working on a function that inserts a value in a binary search tree, I wrote this code:
// this is my decalarations if its relevent
typedef struct element* tree;
typedef struct element
{
type data;
tree right;
tree left;
} noed;
tree rightSon(tree head)
{
return (head->right);
}
void insert(tree* a,int val)
{
if(!empty(*a)) {
if((*a)->data>val) {
if(!empty(leftSon(*a)))
insert(&leftSon(*a),val); // error here leftson return a tree
else {
(*a)->left=newNoed(val);
}
}
else if((*a)->data<val) {
if(!empty(rightSon(*a)))
insert(&rightSon(*a),val); //same error here
else
(*a)->right=newNoed(val);
}
else printf("value already exist!\n");
}
else {
*a=newNoed(val);
}
}
I solved the problem by declaring local variables of type tree then assigning the values to them like this:
tree lson;
lson=leftson(*a);
insert(&lson,val);
But I still don't get what was the problem in the first place.
It seems the function leftSon (and rightSon) has the return type tree. That is it returns a temporary object. You may not apply the operator & to a temporary object like in this statement
insert(&leftSon(*a),val);
You could write the code without calling the functions leftSon or rightSon like
if(!empty(rightSon(*a)))
insert( &(*a)->right ,val);
In fact the if-else statements like this
if(!empty(leftSon(*a)))
insert(&leftSon(*a),val); // error here leftson return a tree
else
{
(*a)->left=newNoed(val);
}
are redundant. Instead of them you could just write
insert( &(*a)->left, val );
and
insert( &(*a)->right, val );
Your rightSon function is returning the value of the pointer object stored at head->right. You can't take the address of a value.
What you want to do instead is have rightSon return the address of head->right which can then be passed directly to insert.
So change rightSon to return the address:
tree *rightSon(tree head)
{
return &head->right;
}
And call it like this:
insert(rightSon(*a),val);
leftSon presumably has the same issue, so make a similar change for that.
Also, hiding a pointer behind a typedef is considered bad practice as it's no longer obvious by just looking at the code that a variable of that type is a pointer and can cause confusion to the reader.