C ++中的char []和char [10]有什么区别？

Question

Is there any difference between char[] and char[10] (or any other arbitrary constant)?

for example:

char[] = "here";
char[10] = "there";

When I ran such a program:

struct TreeNode
{
    struct TreeNode* left;
    struct TreeNode* right;
    char elem;
};
void BinaryTreeFromOrderings(char* ,char* ,int);

int main()
{
    char a[] = "";
    char b[] = "";
    cin >> a >> b;
    BinaryTreeFromOrderings(b, a, strlen(a));
    return 0;
}

void BinaryTreeFromOrderings(char* inorder, char* preorder, int length)
{
    if(length == 0) return;
    TreeNode* node = new TreeNode;
    node->elem = *preorder;
    int rootIndex = 0;
    for(;rootIndex < length ; rootIndex ++)
    {
        if(inorder[rootIndex] == *preorder)
            break;
    }
    //left
    BinaryTreeFromOrderings(inorder,preorder+1,rootIndex);
    //right
    BinaryTreeFromOrderings(inorder + rootIndex +1 ,preorder + rootIndex +1,length - (rootIndex + 1));
    cout << node->elem;
    delete [] node;
    return;
}

The result seems right, but the program will dump just before exit.

Then I made an experiment:

int main()
{
    char a[] = "";
    cin >> a;
    cout << a;
    return 0;
}

It will run successfully when I input less than 9 characters. (gcc version 4.6.3 (Ubuntu/Linaro 4.6.3-1ubuntu5))

And if i initialize a[] with:

char a[] = "123456789";

It will success less than 25 charactres.

I guess there's something that compiler cannot determine the size of a. But what's the detailed reasons?

Answer 1

 char[] = "here";

This is an array is size 5, automatically deduced from the 4 letters, plus an implicit null terminator ('\\0') tacked onto the end. You are allowed to write and read from positions 0-4. Anything else is undefined behavior.

char[10] = "there";

This is an array is size 10, contents "there\\0\\0\\0\\0\\0" . You are allowed to write and read from positions 0-9. Anything else is undefined behavior.

char a[] = "";

This is an array of size 1, just a null terminator. When you input 9 characters into it, that's undefined behavior. (actually, using standard string input functions, you can't even safely input 1 character, because the standard string input functions automatically tack on a null terminator.

char a[] = "123456789";

This is an array of size 10, and when you input 25 characters into it, that's undefined behavior.

http://en.wikipedia.org/wiki/Undefined_behavior

Answer 2

char a[] = "here";

The compiler will determine the size of the char array a , which is 4 characters + 1 ending character \\0 .

char a[10] = "there";

The size of char array a is 10 including the \\0 , so you can put at most 9 chars into int. Otherwise, you are writing to memory that does not belong to the array. If you do the above way, character 5-9 are null initialized. See a live example here: http://ideone.com/O7c8Zp