如何通過引用傳遞結構並更改其值

Question

大家好我有我的C代碼的這個問題。 我正在實現一個堆棧，每當我彈出它改變函數彈出堆棧，但不是原來的堆棧。 請幫忙。

這裏是我的代碼

char pop(Node* top) 
{ 
    char result ; 

    if (size == 0) // The stack is empty. 
    { 
     return '\0' ; 
    } 

    else //The stack contains at least one element . 
    { 
     result = top->opp ; 
     top = top->next ; 
    } 

    size-- ; 

    return result; 
} 

Answer 1

我們需要更多的代碼，但我會嘗試：

我猜你使用該功能是這樣的：

char pop(Node* top) 
{ 
    char result ; 

    if (size == 0) // The stack is empty. 
    { 
     return '\0' ; 
    } 

    else //The stack contains at least one element . 
    { 
     result = top->opp ; 
     top = top->next ; 
    } 

    size-- ; 

    return result; 
} 

int main() 
{ 
    // this is the top of the stack 
    Node *stack; // let's say there already are some elements in this stack 
    pop(stack); 
    return 0; 
} 

的問題是，你想改變指針值，則stack將指向棧頂。爲了做到這一點，你必須使用一個指針的指針，就像這樣：

char pop(Node** top) 
{ 
    char result ; 
    Node *ptr; 

    ptr = *top; 
    if (size == 0) // The stack is empty. 
    { 
     return '\0' ; 
    } 

    else //The stack contains at least one element . 
    { 
     result = (*top)->opp ; 
     (*top) = (*top)->next ; 
     // You should free the pointer, like user2320537 said in his answer. 
     free(ptr); 
    } 

    size-- ; 

    return result; 
} 

int main() 
{ 
    // this is the top of the stack 
    Node *stack; // let's say there already are some elements in this stack 
    pop(&stack); // You give the pointer address 
    return 0; 
} 

Answer 2

嘗試焦炭彈出（節點**頂部）和操作上（*頂部）

Answer 3

你需要釋放頂部的當前位置也...使用免費as

Node *ptr; 

ptr = top; 

if (size == 0) // The stack is empty. 
    { 
     return '\0' ; 
    } 

    else //The stack contains at least one element . 
    { 
     result = top->opp ; 
     top = top->next ; 
    } 

    free(ptr); 

========================================== =======================

稱之爲

int main(){ 
Node front = NULL: 

// place your code of PUSH here. 

pop(&front); // will call the function **pop** 

} 

}

Answer 4

如果你改變你傳遞一個指針的變量（其地址）的值的函數 如

void increment(int *p) { 
     p += 1; 
} 

同樣改變指針的值，你需要一個指針傳遞給函數指針

char pop(Node **top) { 
     char t; 
     Node *p; 
    if(size == 0) { 
     return '\0'; 
    } else { 
     p = *top; 
     t = p->op; 
     (*top) = (*top)-> next; 
     free(p); 
     return t; 
    } 
} 

Answer 5

請把上面的指針引用流行的功能就像是「字符pop（Node ** top）{}「並添加改變你的else塊」top [0] = top [0] - > next;「而不是「top = top-> next」。