【发布时间】:2015-04-14 09:42:40
【问题描述】:
//Colin James P. Naranjo
//CMSC123 CD-1L
//This program demonstrates postfix evaluation through pop and push functions
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
typedef struct node{ //Uses a combination of typedef and tagged structure for the singly linked list
char value;
struct node *next;
}stack;
char evaluate(char a, char b, char c){
int ans;
if(c == '*'){
ans = (int)a * (int)b;
}
if(c == '+'){
ans = (int)a + (int)b;
}
if(c == '-'){
ans = (int)a - (int)b;
}
if(c == '/'){
ans = (int)a / (int)b;
}
if(c == '%'){
ans = (int)a % (int)b;
}
return (char)ans;
}
char pop(stack *head){ //For popping a value in the stack, create a temporary variable to take over the head
stack *temp;
char x;
printf("Your sequence is mostly likely not in order.\n");
temp = head; //Then the new head will be the value next to it. Save its value in x then free the temporary variable and return x
printf("Your sequence is mostly likely not in order.\n");
head = head->next;
x = temp->value;
free(temp);
return x;
}
void push(stack *head, char op){ //For pushing a value to the stack, create a temporary variable to store the new value
stack *temp, *h;
temp=(stack*)malloc(sizeof(stack));
temp->value = op; //Tthe temporary value will be the new head, and the previous head will be placed next to it
if (head == NULL){
head=temp;
head->next=NULL;
}else{
temp->next=head;
head=temp;
}
h = head;
while(h!=NULL){
printf("%c-->",h->value);
h = h->next;
}
printf("\n");
}
main(){
int i = 0;
char op[50], a, b, c, answer;
stack *head = NULL;
printf("Enter the operators and operands: \n"); //Asks for the sequence and checks if there's an invalid character
scanf("%s", op);
while(op[i] != 0){
printf("%c\n", op[i]);
if(op[i] < 48 || op[i] > 57){
if(op[i] != '*' && op[i] != '+' && op[i] != '-' && op[i] != '/' && op[i] != '%'){
printf("You may have entered an invalid character.\n");
exit(0);
}
}
i++;
}
i = 0;
while(op[i] != 0){
if(op[i] >= 48 && op[i] <= 57){
printf("test: %c \n", op[i]);
push (head, op[i]);
printf("\n");
}else if(op[i] == '*' || op[i] == '+' || op[i] == '-' || op[i] == '/' || op[i] == '%'){
push (head, op[i]);
if((op[i-1] >= 48 && op[i-1] <= 57) && (op[i-2] >= 48 && op[i-2] <= 57)){
printf("test: %c \n", op[i]);
c = pop (head);
b = pop (head);
a = pop (head);
answer = evaluate (a, b, c);
printf("test: %d + %d = %d\n", a, b, answer);
push (head, answer);
}else{
printf("Your sequence is mostly likely not in order or is missing something.\n");
}
}
i++;
}
answer = pop(head);
printf("%d\n", answer);
}
这是一个堆栈程序,它的功能是使用链表在头部插入和删除。我的问题是它不断给出状态访问冲突,我发现我的程序不断删除我输入的先前节点。在我不知道这是怎么回事之前,我已经这样做了。
void push(stack *head, char op){ //For pushing a value to the stack, create a temporary variable to store the new value
stack *temp, *h;
temp=(stack*)malloc(sizeof(stack));
temp->value = op; //Tthe temporary value will be the new head, and the previous head will be placed next to it
h = head;
if (h == NULL){
h=temp;
h->next=NULL;
}else{
temp->next=h;
h=temp;
}
h = head;
while(h!=NULL){
printf("%c-->",h->value);
h = h->next;
}
printf("\n");
}
这是已编辑的部分,它仍然无法正常工作。唯一打印的是我添加的最后一个节点,其余的都没有了。
编辑: 我不能再 90 分钟发布另一个问题,所以我要在这里问。在评估部分,我需要对字符变量做一个算术方程。将它们设为整数并暂时将它们更改为 char 或其他方式是否更好。代码在上面,在评估部分
【问题讨论】:
-
Please don't cast the result of
malloc()。ans变量和evaluate()中的转换也是完全多余的。 -
这就是我们被教导要做的事情。这就是使之前的节点消失的原因吗?它实际上是一个后缀算术表达式的堆栈,因此所有节点值都是字符但数字暂时切换为整数
-
不 - 这是代码稳健性和可读性的问题。我正在加载您的代码,看看是否可以重现该错误。
标签: c stack singly-linked-list