C struct and malloc problem (C)

Question

It's amazing how even the littlest program can cause so much trouble in C.

#include <stdio.h> 
#include <stdlib.h> 

typedef struct node {
    int value;
    struct node *leftChild;
    struct node *rightChild;
} node;

typedef struct tree {
    int numNodes;
    struct node** nodes;
} tree;

tree *initTree() {
    tree* tree = (tree*) malloc(sizeof(tree));
    node *node = (node*) malloc(sizeof(node));
    tree->nodes[0] = node;
    return tree;
}

int main() {
    return 0;
}

The compiler says:

main.c: In function 'initTree':
main.c:17: error: expected expression before ')' token 
main.c:18: error: expected expression before ')' token

Can you please help?

Solution 1

You're using two variables named tree and node, but you also have structs typedefed as tree and node.

Change your variable names:

#include <stdio.h>
#include <stdlib.h>

typedef struct node {
    int value;
    struct node *leftChild;
    struct node *rightChild;
} node;

typedef struct tree {
    int numNodes;
    struct node** nodes;
} tree;

tree *initTree() {
   /* in C code (not C++), don't have to cast malloc's return pointer, it's implicitly converted from void* */
   tree* atree = malloc(sizeof(tree)); /* different names for variables */
   node* anode = malloc(sizeof(node));
   atree->nodes[0] = anode;
   return atree;
}

int main() {
    return 0;
}

Solution 2

tree and node is your case are type names and should not be used as variable names later on.

tree *initTree() {
    tree *myTree = (tree*) malloc(sizeof(tree));
    node *myNode = (node*) malloc(sizeof(node));
    myTree->nodes[0] = myNode;
    return myTree;
}

Solution 3

Change (tree*) and (node*) to (struct tree*) and (struct node*). You can't just say tree because that's also a variable.

Solution 4

Change the body of initTree as follows:

tree* myTree = (tree *)malloc(sizeof(tree));
node *myNode = (node *)malloc(sizeof(node));
myTree->nodes[0] = myNode;
return myTree;

Solution 5

Don't use typedef'ed names as variable names, and there is not need to cast malloc(); in C.

#include <stdio.h> 
#include <stdlib.h> 

typedef struct node {
    int value;
    struct node *leftChild;
    struct node *rightChild;
} node;

typedef struct tree {
    int numNodes;
    struct node** nodes;
} tree;

tree *initTree() {
    tree->nodes[0] = malloc(sizeof(node));
    return malloc(sizeof(tree));
}

int main() {
    return 0;
}

View more solutions

Comments

[Admin]

Not necessary (typedef struct tree {} tree).

[Admin]

I tried it now, and it seems to compile. However, it works without struct (but only when the variable has a name not clashing with the type name).

[user541686]

@delnan: Exactly my point... hence why he needs to say struct if he doesn't change the rest of his code.

[Admin]

If he doesn't change the rest of the code, yes. But the majority (look at the votes) seems to think that renaming the variables is the real solution (it also allows dropping the arguably unnecessary struct before the typename).

[user541686]

@delnan: I certainly agree that that's a better solution, but no one said all answers to the same question have to give the same solution. :)

[Admin]

But if there is a better solution, why not propose the better solution (and perhaps mention inferior ones)?

[user541686]

@delnan: This is just the first fix that jumped to my mind, and it's a bit pointless to change anything now.

[Gal]

So should I never try to cast a void returned from malloc?

[wkl]

@sombe - you don't have to (in C), the conversion from void * to your pointer type is implicit.

[Jens Gustedt]

@sombe, yes. If you do, you might hide some implicit casts to int if the prototype of malloc is not included.