dequeueing function working with 2 queues encounters segfault? - c

Question

The function below is supposed to dequeue a structure that is made of 2 queues. Each time we dequeue the first queue, its rear needs to become the front of the second queue. Essentially moving the first element in the second queue to the first queue to be its rear. I came up with the algorithm below:
int dequeue(queue* Q1,queue* Q2){
node* temp;
if(Q1->rear=NULL){
return 0;
}
if(count<3){
temp=Q1->front;
Q1->front=Q1->front->next;
free(temp);
count--;
return 1;
}
if(count>=3){
temp=Q1->front;
Q1->front=Q1->front->next;
free(temp);
Q1->rear->next=Q2->front;
Q1->rear=Q1->rear->next;
Q2->front=Q2->front->next;
Q1->rear->next=NULL;
if(Q2->front=NULL){
Q2->rear=NULL;
}
count--;
return 1;
}
}
It gives a segfault at Q1->rear->next=Q2->front;
is there an alternative way to achieve this?

Answer 1

For starters there is a typo in the condition of the if statement
if(Q2->front=NULL){
Q2->rear=NULL;
}
You are using assignment instead of comparison.
There is also a bug in this if statement
if(count<3){
temp=Q1->front;
Q1->front=Q1->front->next;
free(temp);
count--;
return 1;
}
q1-front after this statement
Q1->front=Q1->front->next;
can be equal to NULL. In this case you need to set also Q1->rare to NULL.
But in any case your approach with the if statements
if(count<3){
temp=Q1->front;
Q1->front=Q1->front->next;
free(temp);
count--;
return 1;
}
if(count>=3){
temp=Q1->front;
Q1->front=Q1->front->next;
free(temp);
//...
makes the code less clear and as it occurred error prone.
I would write the function the following way
int dequeue( queue *q1, queue *q2 )
{
int success = q1->front != NULL;
if ( success )
{
if ( q2->front != NULL )
{
q1->rear->next = q2->front;
q1->rear = q1->rear->next;
q2->front = q2->front->next;
if ( q2->front == NULL ) q2->rear = NULL;
q1->rear->next = NULL;
}
node *tmp = q1->front;
q1->front = q1->front->next;
if ( q1->front == NULL ) q1->rear = NULL;
free( tmp );
--count;
}
return success;
}
Pay attention to that it is a bad programming practice when a function depends on a global variable (I mean the variable count). You could avoid such a situation wrapping queues and the variable count in a structure.
Here is a demonstration program that shows the function in action.
#include <stdio.h>
#include <stdlib.h>
typedef struct node
{
int data;
struct node *next;
} node;
typedef struct queue
{
node *front;
node *rear;
} queue;
size_t count;
int dequeue( queue *q1, queue *q2 )
{
int success = q1->front != NULL;
if ( success )
{
if ( q2->front != NULL )
{
q1->rear->next = q2->front;
q1->rear = q1->rear->next;
q2->front = q2->front->next;
if ( q2->front == NULL ) q2->rear = NULL;
q1->rear->next = NULL;
}
node *tmp = q1->front;
q1->front = q1->front->next;
if ( q1->front == NULL ) q1->rear = NULL;
free( tmp );
--count;
}
return success;
}
int push( queue *q, int data )
{
node *new_node = malloc( sizeof( node ) );
int success = new_node != NULL;
if ( success )
{
new_node->data = data;
new_node->next = NULL;
if ( q->rear == NULL )
{
q->front = new_node;
}
else
{
q->rear->next = new_node;
}
q->rear = new_node;
++count;
}
return success;
}
int empty( const queue *q )
{
return q->front == NULL;
}
int front( queue *q )
{
return q->front->data;
}
int main( void )
{
queue q1 = { .front = NULL, .rear = NULL };
queue q2 = { .front = NULL, .rear = NULL };
const int N = 10;
for ( int i = 0; i < N; i++ )
{
if ( i < N / 2 ) push( &q1, i );
else push( &q2, i );
}
while ( !empty( &q1 ) )
{
printf( "%d ", front( &q1 ) );
dequeue( &q1, &q2 );
}
putchar( '\n' );
for ( int i = 0; i < N; i++ )
{
if ( i < N / 2 ) push( &q1, i );
else push( &q2, i );
}
while ( !empty( &q1 ) )
{
printf( "%d ", front( &q1 ) );
dequeue( &q1, &q2 );
}
putchar( '\n' );
}
The program output is
0 1 2 3 4 5 6 7 8 9
0 1 2 3 4 5 6 7 8 9
There are the same code is executed two times to show that the function dequeue works correctly.