为什么我的C程序在每次执行时都更改输出而没有任何输入或代码更改？

Question

I'm working on my first program on C and it is about two process communication. The messages send from one process to another can vary in length and I need to store them until the receiver process need it. I thought that would be good to use FIFO structure which length can vary so I decided to program it. This is the code:

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

typedef struct Queue{
        int length;
        struct Member *first;
        struct Member *last;
}Queue;

typedef struct Member{
        char ch;
        struct Member *next;
}Member;

Queue create(){
        struct Queue cola;
        cola.length = 0;
        return cola;
} 
void add(struct Queue *q, char c){
        if(q->length == 0){
                struct Member m;
                m.ch = c;
                q->first = &m;
                q->last = &m;

        }else{
                struct Member m;
                m.ch = c;
        q->last->next = &m;
        q->last = &m;
        }   
        q->length++;
}
void print(struct Queue n){
    struct Member *p;
    p = n.first;
    for(int i = n.length; i > 0; i--){
        printf("%c,", p->ch);
        p = p->next;
    }
    printf("\n");
}

int main(){
        struct Queue cola = create();
        add(&cola, 'a');
        printf("%c\n",cola.first->ch);
        printf("%c\n",cola.last->ch);
        print(cola);
}

The problem is that, first of all, when I try to print the queue it do not works as expected (I don't know if it is problem of the print() function or the structure itself), but this is something that I can find out by myself. The biggest problem, and the one that do not allow me to find the first problem, is that the answer of the program changes. This is a copy of my terminal:

    $./list
a
a
p,
    $./list
a
a
�,
    $./list
a
a
 ,
    $./list
a
a
0,
    $./list
a

,
    $./list
a
a
�,

The desired output would be:

    $./list
a
a
a,

I don't know why this happens and I would like to know it. Thanks!

Answer 1

   q->first = &m;
   q->last = &m;

These are poiting to a memory with automatic storage duration which will not be there once the scope on which it is declared ends - accessing it outside the scope like you did is undefined behavior.

Solution would be to dynamically allocate the memory using *alloc ( malloc / calloc etc) and then you can pass them around. The only thing is you need to free them once you are done working with it. Something like this:- (Yes you can/should modify it as you need)

struct Member* m = malloc(sizeof *m);
if(!m){ perror("malloc");exit(EXIT_FAILURE);}
m->ch = c;
q->first = m;
q->last = m;
...