C local variable address return warning

Question

I have this function in C:

static Node* newNode(void* e){
Node n={e,NULL,NULL};
return &n;
}

And while compiling I get the following warning that I would like to understand why it happens:

warning: function returns address of local variable [enabled by default]

What kind of dangers are lurking behind this?

Thank you

possible duplicate of [Pointer to local variable](http://stackoverflow.com/questions/4570366/pointer-to-local-variable) — Bo Persson, Jul 01 '12 at 11:43

score 5 · Accepted Answer · answered Jul 01 '12 at 10:18

5

Local variables are destroyed when you return from a function. Accessing them after the function returned is undefined behavior, don't do this.

answered Jul 01 '12 at 10:18

ouah

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The code is just totally broken. It returns a pointer to the location the Node was in, but the Node is no longer there. So the returned pointer is garbage. – David Schwartz Jul 01 '12 at 10:27
So I should make a copy of the value insted of passing its reference? – Leonardo Marques Jul 01 '12 at 14:21
@Reonarudo use `malloc` in your `newNode` function to create the node: `Node *n = malloc(sizeof *n);` – ouah Jul 01 '12 at 14:23
@ouah Like this then: static Node* newNode(void* e){ Node* n=(Node *)malloc(sizeof(*n)); n->e=e; return n; } – Leonardo Marques Jul 01 '12 at 14:40

score 0 · Answer 2 · answered Jul 01 '12 at 11:16

0

The warning is because the scope of the variable is local to the function - once the function is returned, that variable is no longer in scope, and it's value is undefined.

answered Jul 01 '12 at 11:16

user893569

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C local variable address return warning

2 Answers2