通過使用反向波蘭記法和使用堆棧計算器

Question

大家好我有一個分段錯誤，你能幫助嗎？如果我有這個操作符「3 5 +」，這意味着3 + 5 ，並像「9 8 * 5 + 4 +罪」，「罪（（（9 * 8）+5）+4）」 所以我的想法檢查第一個和第二個數字是否是數字，然後在堆棧中推送它們，然後當我有操作員時，彈出數字並進行計算，然後再次推送答案。 `

typedef struct st_node { 
    float val; 
    struct st_node *next; 
} t_node; 

typedef t_node t_stack; 


// a function to allocate memory for a stack and returns the stack 
t_stack* fnewCell() { 
    t_stack* ret; 
    ret = (t_stack*) malloc(sizeof(t_stack)); 
    return ret; 
} 
// a function to allocate memory for a stack, fills it with value v and pointer n , and returns the stack 
t_stack* fnewCellFilled(float v, t_stack* n) { 
    t_stack* ret; 
    ret = fnewCell(); 
    ret->val = v; 
    ret->next =n; 
    return ret; 
} 

//function to initialize stack 
void initstack(t_stack** stack) { 
    fnewCellFilled(0,NULL); 
} 

// add new cell 
void insrtHead(t_stack** head,float val) { 
    *head = fnewCellFilled(val,*head); 
} 

//function to push the value v into the stack s 
void push(t_stack **s, float val) { 
    insrtHead(s,val); 
} 

//function to pop a value from the stack and returns it 
int pop(t_stack **s) { 
    t_stack* tmp; 
    int ret; 
    tmp = (*s)->next; 
    ret = (*s)->val; 
    free(*s); 
    (*s) = tmp; 
    return ret; 
} 

int isempty (t_stack *t) { 
    return t == NULL; 
} 


//function to transfer a string(str) to int (value) 
//returns -1 when success , i otherwise 
int str2int(char *str,int *value) { 
    int i; 
    *value = 0; 
    int sign=(str[0]=='-' ? -1 : 1); 
    for(i=(str[0]=='-' ? 1 : 0);str[i]!=0;i++) { 
     if(!(str[i]>=48 && str[i]<=57)) // Ascii char 0 to 9 
      return i; 
     *value= *value*10+(str[i]-48); 
    } 
    *value = *value * sign; 
    return -1; 
} 

//a function that takes a string, transfer it into integer and make operation using a stack 
void function(t_stack *stack, char *str) 
{ 
    char x[10]=" "; 
    int y,j,i=0,z; 
    printf("++\n"); 
    if(str[i] != '\0') { 
     strcpy(x, strtok(str, " ")); 
     z= str2int(x, &y); 
     if(z == -1) 
     { 
      push(&stack,y); 
      i=i+2; 
     } 
    } 
    while(str[i] != '\0') 
    { 
     strcpy(x, strtok(NULL, " ")); 
     z= str2int(x, &y); 
     if(z == -1) 
     { 
      printf("yes %d",y); 
      push(&stack,y); 
      i=i+2; 
     } 
     else 
     { 
      y=pop(&stack); 
      j=pop(&stack); 

      if(x[0] == '+') push(&stack,y+j); 
       else if (x[0] == '-') push(&stack,j-y); 
       else if(x[0] == '*') push(&stack,j*y); 
       else if(x[0] == '/') push (&stack ,j/y); 
     } 
    } 
} 

int main() { 
    t_stack *s; 
    initstack(&s); 
    char *str="3 5 +";  
    function(s,str); 
    return 0; 
} 

`

Answer 1

在哪裏，我注意到：

1）

void initstack(t_stack** stack) { 
    fnewCellFilled(0,NULL); 
} 

大概

void initstack(t_stack** stack) { 
    *stack=fnewCellFilled(0,NULL); 
} 

2）

字符串文字是由strtok的

改變

char *str="3 5 +"; 

應該

char str[]="3 5 +";  

3）

while(str[i] != '\0') 
{ 
    strcpy(x, strtok(NULL, " ")); 
    z= str2int(x, &y); 
    if(z == -1) 
    { 
     printf("yes %d",y); 
     push(&stack,y); 
     i=i+2; 
    } 
    else 
    { 
     y=pop(&stack); 
     j=pop(&stack); 
     if(x[0] == '+') push(&stack,y+j); 
     else if(x[0] == '-') push(&stack,j-y); 
     else if(x[0] == '*') push(&stack,j*y); 
     else if(x[0] == '/') push (&stack ,j/y); 
    } 

大概

while(str[i] != '\0') 
{ 
    strcpy(x, strtok(NULL, " ")); 
    z= str2int(x, &y); 
    if(z == -1) 
    { 
     printf("yes %d",y); 
     push(&stack,y); 
     i=i+2; 
    } 
    else 
    { 
     y=pop(&stack); 
     j=pop(&stack); 
     if(x[0] == '+') push(&stack,y+j); 
     else if(x[0] == '-') push(&stack,j-y); 
     else if(x[0] == '*') push(&stack,j*y); 
     else if(x[0] == '/') push (&stack ,j/y); 
     i += 1;//need increment i 
    }