使用逆波兰表示法和堆栈的计算器

Question

大家好，我遇到了段错误，您能帮忙吗？ 如果我有这个运算符“3 5 +”，这意味着 3+5 就像“9 8 * 5 + 4 + sin”，“sin(((9*8)+5)+4)” 所以我的想法是检查第一个和第二个是否是数字并将它们插入堆栈中，然后当我有运算符时我弹出数字并进行计算，然后再次插入答案。 `

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

我注意到的地方:

1)

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

可能

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

2)

字符串文字被 strtok 更改

char *str="3 5 +";

应该是

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);
    } 

可能

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
    }