堆栈的链表实现我根据调用 top->data 时的代码块分段转储编写了以下函数,并将其通过返回整数的 ICP 函数我不知道是什么导致了分段转储,就像我使用指针的方式一样或许?
struct node
{
char data;
struct node* link;
};
//create pointer top to indicate top of stack
struct node* top=NULL;
void push(char x)
{
struct node* temp = (struct node*)malloc(sizeof(struct node*));
//requires all of this to be done when inserting a node at the end
//set temp.data to character x
temp->data = x;
//set link of node to address of current top
temp->link = top;
//set top of list to newly created node
top = temp;
}
char pop()
{
struct node *temp;
if(top==NULL)
return;
else
{
//pointer temp node pointing to top node
temp = top;
//set address of top to the next node
top=top->link;
//returns character stored in top
return temp->data;
//frees memory space
free(temp);
}
}
int ICP(char z)
{
/*checks if z is + or -, returns ICP*/
if(z=='+'||z=='-')
{return(1);}
if(z=='*'||z=='/')
/*checks if z is * or /, returns ICP*/
{return(3);}
if(z=='^')
/*checks if z is ^, returns ICP*/
{return(6);}
}
int ISP(char z)
{
if(z=='(')
/*checks if z is "(", returns ISP*/
{return(0);}
if(z=='+'||z=='-')
/*checks if z is + or -, returns ISP*/
{return(2);}
if(z=='*'||z=='/')
/*checks if z is * or /, returns ISP*/
{return(4);}
if(z=='^')
/*checks if z is ^, returns ICP*/
{return(5);}
}
int convert(char input[],char output[],int rank)
{
char x;
char TOKEN;
int a=0;
int m=0;
for(m=0;input[m]!='\0';m++)
{
TOKEN=input[m];
if(isalnum(input[m]))
{output[a]=TOKEN;rank++;a++;}
else
{
if(TOKEN=='(')
{push('(');printf("%d",m);}
else
if (TOKEN==')')
{
while((x=pop())!='(')
{
output[a]=rank;rank=rank-1;a++;
}
}
else
{
while(ICP(TOKEN)<ISP(top->data)) **//seg core dumps here**
{
x=pop();
output[a]=x;
rank=rank-1;
a++;
}
push(TOKEN);
}
}
}
return (rank);
}
看起来您正在尝试使用堆栈将前缀转换为后缀(反之亦然)...只是在代码中需要注意的一些事情已经在注释中提到了。
void push(char x)
{
struct node* temp = (struct node*)malloc(sizeof(struct node*));
// Other code here...
}
应该是:
void push(char x)
{
struct node* temp = malloc(sizeof(struct node));
// Other code here...
}
我们不需要
sizeof(struct node*)
,因为这里不需要 node*
的大小。我们只需要节点的大小即可为节点动态分配该空间量。
此外,你的 pop 函数需要重新考虑。
char pop()
{
struct node *temp = NULL; // Initialize your temp pointers to null.
// Check if list is empty
if(top==NULL)
return; // Cannot return void here; You will need to return something as you function type is a char.
else
{
//pointer temp node pointing to top node
temp = top;
//set address of top to the next node
top=top->link;
//frees memory space
free(temp);
//returns character stored in top
return temp->data;
}
}
您需要先删除数据
temp->data
,然后再点击return
的pop()
部分,否则程序将不会弹出temp->data
中保存的内容。请告诉我这是否有帮助,谢谢!