我编写了一个程序，将表达式从算术中缀表示法转换为后缀。这部分程序没有问题。转换后，程序应评估表达式并使用堆栈给出数字答案。
当引用push2时，问题出现在第232行，这会产生分段错误（SIGSEGV）。
我该如何解决这个问题？
#include <stdio.h>
#include <stdbool.h>
#include <ctype.h>
#include <stdlib.h>
#include <string.h>
#define MAX 100 // maximum number of input characters
bool open_close(char opening, char closing);
bool check_balanced_par(char * exp, int length);
struct stack
{
char my_stack[MAX];
int pointer;
};
struct stack2
{
float my_stack[MAX];
int pointer;
};
int precedence(char x);
void push(struct stack *S, char *x);
char pop(struct stack *S);
void push2(struct stack2 *S, float *x);
char pop2(struct stack2 *S);
int precedence(char x)
{
if(x == '#')
return 0;
if(x == '(')
return 1;
if(x == '+' || x == '-')
return 2;
if(x == '*' || x == '/')
return 3;
else
return 4;
}
void push(struct stack *S, char *x)
{
S->pointer++;
S->my_stack[S->pointer] = *x;
}
char pop(struct stack *S)
{
char data = S->my_stack[S->pointer];
S->pointer--;
return data;
}
void push2(struct stack2 *S, float *x)
{
S->pointer++;
S->my_stack[S->pointer] = *x;
}
char pop2(struct stack2 *S)
{
char data = S->my_stack[S->pointer];
S->pointer--;
return data;
}
// checks to see if ()/[]/{} is the case
bool open_close(char opening, char closing)
{
if (opening == '(' && closing == ')') return true;
if (opening == '{' && closing == '}') return true;
if (opening == '[' && closing == ']') return true;
return false;
}
bool check_balanced_par(char * exp, int length)
{
struct stack S;
S.pointer = 0;
int i;
for(i=0; i<length; i++)
{
if (exp[i] == '(' || exp[i] == '{' || exp[i] == '[')
{
S.pointer = S.pointer +1;
S.my_stack[S.pointer] = exp[i];
}
else if (exp[i] == ')' || exp[i] == '}' || exp[i] == ']')
{
if(S.pointer == 0 || !open_close(S.my_stack[S.pointer],exp[i]))
return false;
else
S.pointer = S.pointer-1;
}
else if (exp[i] == '*' || exp[i] == '/' || exp[i] == '+' || exp[i] == '-' || isdigit(exp[i]))
{
continue;
}
else
return false; // if input is not as expected
}
if (S.pointer == 0) // checks if stack is empty
return true;
else
return false;
}
int main(void)
{
char c;
int size = 0;
char arr[MAX];
int o=0; // counter for arr[]
printf("Enter an expression to check:\n");
fflush(stdout);
fgets(arr, MAX, stdin);
arr[strlen(arr)-1] = '\0'; // clears \n from array
while((c=arr[o++]) != '\0')
{
size++;
}
if(check_balanced_par(arr, size))
{
printf("Balanced.\n");
printf("Postfix notation is: ");
struct stack S;
struct stack *ptr; // pointer to stack
ptr = &S; // setting the pointer
S.pointer = -1;
int z = 0; // counter to traverse input array
char output[MAX]; // stores output
int out_count = 0; // stores no. of characters of output
char x = ' ';
char y = '#';
push(ptr, &y);
while('\0' != arr[z]) // going through the input array and incrementing z everytime
{
if(isdigit(arr[z])) // if an operand is found print to screen
{
output[out_count] = arr[z];
out_count++;
}
else if(arr[z] == '(') // if ( is found push on stack
{
push(ptr, &arr[z]);
}
else if(arr[z] == ')') // if ) is found pop stack until ( is found
{
while(!(S.my_stack[S.pointer] == '('))
{
output[out_count] = pop(ptr);
out_count++;
}
S.pointer--; // pop the extra (
}
else if(arr[z] == '*' || arr[z] == '/' || arr[z] == '+' || arr[z] == '-')
{
push(ptr, &x); // used for correct output layout
while(precedence(S.my_stack[S.pointer]) >= precedence(arr[z]))
{
// pop stack
if(arr[z] == '(' || arr[z] == ')')
continue;
output[out_count] = pop(ptr);
out_count++;
}
// push arr[z] on stack
push(ptr, &arr[z]);
}
z++;
}
int j = S.pointer;
while(S.my_stack[j] != '#')
{
if(S.my_stack[j] == '(')
continue;
output[out_count] = S.my_stack[j];
out_count++;
j--;
}
output[out_count] = '\0';
int k=0;
while(output[k] != '\0') // go through output array and print
{
printf("%c", output[k]);
k++;
}
// start evaluating the postfix format [THIS IS WHERE THE ISSUE IS]
struct stack2 T;
struct stack2 *ptr2; // pointer to stack
ptr2 = &T; // setting the pointer
int i=0, l, g; // i counter for output[], l counter for nums[]
float a,b;
float toPush;
float *ptr_toPush;
char nums[MAX];
char operator;
float total;
while(output[i++] != '\0')
{
if(isdigit(output[i]))
{
l=0;
int y=i;
while(output[y] != ' ') // if character is not a space
{
nums[l] = output[y]; // place character in nums[]
y++;
l++;
}
i=y; // setting the counter for output[] to the character we have reached
toPush = atof(nums);
for(g=0; g<k; g++)
nums[g] = '\0';
ptr_toPush = &toPush;
push2(ptr2, ptr_toPush);
}
else if(output[i] == ' ') // if a space is found continue
{
continue;
}
else // operator
{
a = pop(ptr);
b = pop(ptr);
operator = output[i];
switch(operator)
{
case '+':
total = a+b;
push2(ptr2,&total);
break;
case '-':
total = a-b;
push2(ptr2,&total);
break;
case '*':
total = a*b;
push2(ptr2,&total);
break;
case '/':
total = a/b;
push2(ptr2,&total);
break;
default: printf("Error.\n");
}
}
}
printf("Result is: %d", pop2(ptr2));
}
else
printf("Not balanced.");
return 0;
}
答案 0 :(得分：0)
定义ptr2（和任何其他堆栈指针）后，使用malloc正确设置它。还要保持指向堆栈顶部的指针，以便在弹出时进行错误检查以确保没有弹出太多。因此，当您不再需要它时，您可以正确地释放它。