该代码用于使用的算术表达式评估 即使在我的代码计算表达式之后，我也遇到了分段错误。它在在线编译器上显示运行时错误...但是在离线编译器上运行时也是如此。不同操作的优先级也与在前面提到的相同。码。 有人可以帮我找出错误吗？
# include <stdio.h>
# include <stdlib.h>
# include <string.h>
//-------IMPLEMENTATION OF STACK USING LINK LIST--------------
// define a structure of type struct node
struct node
{
char* p;
struct node* next;
};
//function to create a node for a given char
struct node* CreateNode(char* c)
{
struct node* nd;
nd = calloc(1,sizeof(struct node));
(nd->p) = calloc(2,sizeof(char));
*(nd->p) = *c;
return nd;
}
// function to create an empty stack and returns it
struct node* CreateEmptyStack()
{
struct node* new;
new = calloc(1, sizeof(struct node));
new = NULL;
return new;
}
// function to check a given stack is empty or not
int IsEmpty(struct node* stack)
{
if (stack==NULL)
{
return 1;
}
else
{
return 0;
}
}
// function to return the topmost element of stack
char* Top(struct node* stack)
{
if (!IsEmpty(stack))
{
struct node* temp = stack;
return temp->p;
}
}
// function to push an element at top of stack
struct node* Push(char* c ,struct node* stack)
{
struct node* temp = CreateNode(c);
temp->next = stack;
stack = temp;
return stack;
}
//function to pop an element form a stack
struct node* Pop(struct node* stack)
{
struct node* temp = stack;
stack = temp->next;
free(temp);
return stack;
}
//--------algorithm to evaluate arithmetic expression------------
// assigning outside stack priorities to incoming characters
int OutsideStackPriority(char c)
{
switch (c)
{
case '#' : return 0;
case '+' : return 2;
case '-' : return 2;
case '*' : return 3;
case '/' : return 4;
case '^' : return 6;
case '(' : return 7;
case ')' : return 0;
}
}
// assigning inside stack priorites to different characters
int InsideStackPriority(char c)
{
switch (c)
{
case '$' : return 0;
case '+' : return 2;
case '-' : return 1;
case '*' : return 4;
case '/' : return 3;
case '^' : return 5;
case '(' : return 0;
}
}
// function to calculate a^b
int power(int a, int b)
{
int i , result=1 ;
for (i = 1; i <= b; ++i)
{
result = result*a;
}
return b;
}
// Operate given two numbers and a given operator
// this function assumes that there is only integer division
int Operation(int n1 , int n2 , char o)
{
switch (o)
{
case '+' : return n2 + n1;
case '-' : return n2 - n1;
case '*' : return n2 * n1;
case '/' : return n2 / n1; // return integer division only
case '^' : return power(n2,n1);
}
}
// function to determine a given char is a number or not
// here is one assumption that the given expression is valid
int IsNum(char ch)
{
switch (ch)
{
case '0':
case '1':
case '2':
case '3':
case '4':
case '5':
case '6':
case '7':
case '8':
case '9': return 1;
default : return 0;
}
}
// function to handle overall logic of calculation
void function(struct node* NStack , struct node* OStack)
{
char* i;
char* c1 ;
char* c2 ;
char* c3 ;
char* temp;
int x=0;
c3 = calloc(10,sizeof(char));
char* o ;
int n1 , n2 , n3;
i = calloc(1,sizeof(char));
*i = '$';
OStack = Push(i,OStack);
temp = Top(OStack);
printf("%c\n",*temp);
*i = getchar();
printf("scanned character is %c\n",*i);
while(x!=1)
{
if(IsNum(*i))
{
// now it is clear that i am definitely going to push the number into NStack
NStack = Push(i,NStack);
temp = Top(NStack);
printf("%c\n",*temp);
/**i = getchar();
printf("scanned character is %c\n",*i);
int k = 1;
while(IsNum(*i))
{
NStack->p = realloc(NStack->p,(k+2)*sizeof(char));
*(NStack->p + k) = *i;
k++;
*i = getchar();
printf("scanned character is %c\n",*i);*/
// }
}
// what to do when x is a operator
// one assumption is still there i assume the incoming character is valid
else
{
// now do the case analysis on what character i got
switch (*i)
{
case ')': { while(*(Top(OStack)) != '(')
{
// Evaluate(NStack,OStack);
c1 = Top(NStack);
n1 = atoi(c1);
NStack = Pop(NStack);
temp = Top(NStack);
printf("%c\n",*temp);
c2 = Top(NStack);
n2 = atoi(c2);
NStack = Pop(NStack);
// temp = Top(NStack);
// printf("%c\n",*temp);
o = Top(OStack);
n3 = Operation(n1,n2,*o) ;
OStack = Pop(OStack);
sprintf(c3,"%d",n3);
NStack = Push(c3,NStack);
temp = Top(NStack);
printf("%c\n",*temp);
}
OStack = Pop(OStack);
}
case '#' : { while(*(Top(OStack)) != '$')
{
// Evaluate(NStack,OStack
c1 = Top(NStack);
n1 = atoi(c1);
NStack = Pop(NStack);
temp = Top(NStack);
printf("%c\n",*temp);
c2 = Top(NStack);
n2 = atoi(c2);
NStack = Pop(NStack);
// temp = Top(NStack);
// printf("%c\n",*temp);
o = Top(OStack);
n3 = Operation(n1,n2,*o) ;
OStack = Pop(OStack);
sprintf(c3,"%d",n3);
NStack = Push(c3,NStack);
temp = Top(NStack);
printf("%c\n",*temp);
}
x=1;
printf("%s\n",Top(OStack));
OStack = Pop(OStack) ;
}
default : { while(InsideStackPriority(*(Top(OStack))) >= OutsideStackPriority(*i))
{
// Evaluate(NStack,OStack);
c1 = Top(NStack);
n1 = atoi(c1);
NStack = Pop(NStack);
temp = Top(NStack);
printf("%c\n",*temp);
c2 = Top(NStack);
n2 = atoi(c2);
NStack = Pop(NStack);
//temp = Top(NStack);
//printf("%c\n",*temp);
o = Top(OStack);
n3 = Operation(n1,n2,*o) ;
OStack = Pop(OStack);
sprintf(c3,"%d",n3);
NStack = Push(c3,NStack);
temp = Top(NStack);
printf("%c\n",*temp);
}
OStack = Push(i,OStack);
}
}
}
if(x==0)
{
*i = getchar();
printf("scanned character is %c\n",*i);
}
}
arkdown and HTML are turned off in code blocks:
<i>This is // char* ans = Top(NStack);
// printf("%s\n",ans);
// NStack = Pop(NStack);
// free(NStack);
// free(OStack);
}
int main()
{
struct node* NStack = CreateEmptyStack();
struct node* OStack = CreateEmptyStack();
function(NStack,OStack);
return 0;
}
答案 0 :(得分：1)
我想也许你没有在输入中处理换行符：
Program received signal SIGSEGV, Segmentation fault.
0x0000000000400d87 in function (NStack=0x0, OStack=0x602090) at b1.c:260
260 printf("%c\n",*temp);
我添加了以下内容并且segfault停止了：
switch (*i)
{
case ' ':
case '\n':
break;
为了好玩，您可能会喜欢Dijkstra的两个堆栈算法版本，我将其作为一个示例用例：http://ccodearchive.net/info/deque.html