问题
void push(stack *head, int valuee)
{
if(head->next==NULL && head->value==-1)
{
head->value = valuee;
printf("First element %d inserted\n",valuee);
}
else
{
stack *temp = new stack;
temp->value = valuee;
temp->next = head;
head = temp;
printf("Element %d inserted\n",valuee);
}
}
First element is inserted properly but when i continue inserting elements, none of the elements are inserted after the first one. Read somewhere that i have to pass pointer to pointer of stack but I did this same thing during a postfix infix question and it was working over there. Help me with this problem here. Many thanks for any help in advance
Previous example of infix postfix which was working fiine
void push(char c, stack *node)
{
stack *B = (stack *)malloc(sizeof(stack));
if (node->next == NULL)
{
node->next = B;
B->value = c;
B->next =NULL;
}
else
{
B->next = node->next;
node->next = B;
B->value = c;
}
}
回答1:
You can change the function like this
stack* push(stack *head, int valuee)
//return void to stack *
return head;
//In the end return the new head
and it will work. Call push like this
head = push(head,value);
回答2:
Why u need pointer to pointer ?
You want to modify the contents of a pointer so if you pass only a pointer to the function it is pass by copy so you can not modify it actually.
In the case of infix-prefix scenario you must have not modified the string you would have read it only, so a pointer to pointer is not required.
回答3:
For this line of code in the else portion of the if statement:
head = temp;
Your intention is to mutate head, in other words, change what head is pointing to. However, pointers are passed in as values, just like other variables. In other words, suppose I call the push function somewhere else. For simplicity, suppose I call it in the main function, something like this:
int main()
{
stack *headOfStack = new stack;
// suppose this next push triggers the else portion of the push code
push(headOfStack, 6);
}
Now, after the push(headOfStack, 6); statement has been executed, your intention is to expect headOfStack to point to a new "stack node" which contains the value 6. Now, headOfStack is a pointer to a variable of type stack. It stores a memory address. You can think of a memory address as some integer. When we call push, we are copying the content of headOfStack (the content of headOfStack is a memory address) into the local variable head of the push function. Therefore, when:
head = temp;
is executed, we are assigning the contents of temp to head. What is temp? It is a pointer to a variable of type stack. In other words, the value of temp is a memory address. So head = temp; simply assigns the memory address contained in temp to the local variable head. The local variable head in the push function and our headOfStack variable in the main function are two completely different variables with different memory addresses. If my explanation has been clear so far, this means that when we modify head in the push function, the headOfStack variable in main is totally unchanged.
What you want to do in this case is:
void push(stack **headPtr, int valuee)
{
// this will get the actual pointer we are interested in
stack *head = *headPtr;
if(head->next==NULL && head->value==-1)
{
head->value = valuee;
printf("First element %d inserted\n",valuee);
}
else
{
stack *temp = new stack;
temp->value = valuee;
temp->next = head;
// mutation is done here
*headPtr = temp;
printf("Element %d inserted\n",valuee);
}
}
And its usage, using our fictional main function:
int main()
{
stack *headOfStack = new stack;
// notice the use of &headOfStack instead of headOfStack
push(&headOfStack, 6);
}
Just remember that pointers store memory addresses, and that pointers are just variables, and they have memory addresses as well. To mutate a pointer (change what a pointer is pointing to), just pass in its memory address to the function.
Hope that helps!
EDIT for new edit in question
void push(char c, stack *node)
{
stack *B = (stack *)malloc(sizeof(stack));
if (node->next == NULL)
{
node->next = B;
B->value = c;
B->next =NULL;
}
else
{
B->next = node->next;
node->next = B;
B->value = c;
}
}
For this version of push, what it's doing is essentially:
If
node->next == NULL, sonodehas no successor, then set its successor to a newly allocated node with valuecOtherwise,
node->next != NULLandnodehas some successor. Then we would set the newly allocated nodeBtonode's successor, and setnode's original successor to beB's successor. Or in other words, it splices a new nodeB(with valuec) in betweennodeand its successor.
I am finding it quite hard to explain this, but a simple explanation is, this push does not change what node is pointing to. At nowhere did we show intention to mutate node. I think the stuff involving B should be quite understandable, so let's focus on the node->next assignments.
I am assuming that the stack is a struct that looks something like this:
struct stack {
char value;
struct stack *next;
};
Now, suppose in our main function, we have a stack:
stack x;
Notice that x is not a pointer. I think we all agree that doing x.value = something and x.next = something will mutate those fields.
Now, let's look at this:
stack *y = malloc(sizeof(struct stack));
We know that y stores an address to an actual struct stack (the actual struct stack is at *y). So y->value = something and y->next = something will mutate those fields.
So hopefully you can see why the node->value assignments work. Essentially node contains an address to an actual struct stack, whose value is *node. By pointer syntax, node->value and node->next assignments will change the contents of node.
Not a very good explanation I know. But just write more code. Pointers confused the hell out of me when I was first starting out with C. I think that these days, I can still be confused by 2 or 3 layers of indirection, and I've encountered some very nasty pointer bugs. Just practice more... some day you will really get it. I know it's what they all say, but it's true.
来源：https://stackoverflow.com/questions/20613273/push-of-stack-not-inserting-the-new-value-c