Question

我正在实施一个小程序以使用双指针进行锻炼这是主程序：

#include <stdio.h>
#include "serial.h"
#include "stack.h"

int main(void) {
    serial_init();  

    /* Array to hold stack entries */
    int stack[10]={0};
    /* Stack pointer */
    int *stack_p = stack;

    /* Code to call the functions */

    push(&stack_p, 1); 
    push(&stack_p, 2); 
    push(&stack_p, 3); 
    push(&stack_p, 4); 

    printf("popped value: %d\r\n", pop(&stack_p));
    printf("popped value: %d\r\n", pop(&stack_p));
    printf("popped value: %d\r\n", pop(&stack_p));
    printf("popped value: %d\r\n", pop(&stack_p));
}        

void push(int **sp, int value) {
    /*  Implement it */ 
}

int pop(int **sp) {
    /*  Implement it */ 
}

我已经实现了推功能，看来还可以。但是，弹出窗口仅返回最后一个元素，然后返回10

void push(int **sp, int value) {
/* implemented it*/

 int *pt; 
 pt=&value; // Store the value to the pointer 

 printf("Push value is is %d\r\n", *pt);
 sp = &pt; // associate the pointer to the pointer of pointer
 ++(*pt);   
}

int pop(int **sp) {
 /* implemented it */
 int value;
 int *pt;
 pt=&value;

 sp = &pt;
 *pt--;

 return value;
}

Answer 1

您的push和pop函数过于复杂且完全错误：

您想要这个：

void push(int **sp, int value) {
  **sp = value;   // put value onto top of the stack
  (*sp)++;        // increment stack pointer
}

int pop(int **sp) {
  (*sp)--;        // decrement stack pointer
  return **sp;    // return value which is on nthe op of the stack
}

您输入的错误代码带有注释说明：

void push(int **sp, int value) {
  int *pt; 
  pt=&value; // here you put the pointer to the local variable value
             // into pt, but local variables disappear as soon
             // as the function has finished

  //  the printf is the only thing one more or less correct
  //  but you could just print directly 'value' like this:
  //    printf("Pushed value is %d\r\n", value);
  //
  printf("Push value is is %d\r\n", *pt);

  sp = &pt;  // this only assigns the pointer to the local variable pt to 
             // the local variable sp

  ++(*pt);   // here you increment actually the local variable
             // value which is pointless 
}

顺便说一句：尽管在调试过程中可能会有所帮助，但不必将整个堆栈初始化为零。因此，您可以这样编写堆栈的声明：

int stack[10];  // no initialisation necessary

为您运动：

说明不需要将堆栈的所有元素初始化为零的确切原因。

通过使用指针来实现堆栈

1 个答案: