C 指向指针并通过引用传递

Question

我正在尝试学习 C，我对指针的指针有点挂断。 我想我理解你为什么需要它们，但不能完全理解正在发生的事情。

例如，下面的代码似乎不像我期望的那样工作：

#include <stdio.h>

int newp(char **p) {
  char d = 'b';
  *p = &d;
  /*printf("**p = %c\n", **p);*/
  return 1;
}

int main() {
  char c = 'a';
  char *p = &c;
  int result;

  result = newp(&p);
  printf("result = %d\n", result);
  printf("*p = %c\n", *p);
  printf("c = %c\n", c);

  return 0;
}

我得到的结果是这样的：

result = 1
*p = 
c = a

*p 打印为空。 相反，我希望*p = b 。

但是，如果我取消注释第 6 行（ newp函数中的printf ），那么我会得到：

**p = b
result = 1
*p = b
c = a

我错过了什么？

Answer 1

您正在处理未定义的行为。 变量d是本地的（驻留在堆栈上）并且在封闭的 function ( newp ) 返回后不可用。

当在newp之外取消引用p时，堆栈&d上的地址可能会被其他一些局部变量覆盖，或者它可能包含垃圾。

Answer 2

您将局部变量 ( d ) 的地址存储在*p中，然后在变量超出 scope 时取消引用它。 未定义的行为

Answer 3

您可能“缺少”关键字static ，如：

int newp(char **p) {
    static char d = 'b';   /* <--- HERE */
    *p = &d;
    /*printf("**p = %c\n", **p);*/
    return 1;
}

这个static关键字使编译器将局部变量定位在“静态存储”中，该局部变量在调用newp()之后的时间段内继续存在（即保持其值）。 但是，此 memory 只有一个副本 - 包含 function ( newp ) 的每个后续调用都重新使用相同的 memory 位置并可能覆盖当时的值。

如果没有static关键字来限定您的局部变量声明，则存储将是“自动”的，这意味着在包含 function 已返回后，它将自动从当前使用中解除。 newp返回后，以前用于局部变量的 memory 可以用于任何目的。

Answer 4

#include <stdio.h>

// Here *pointer means that the parameter that this function 
// will expect will be a pointer. 
void changeViaPointer(int *pointer);

// Pointer of a pointer.
void changeViaPointerInBetween(int **pointer);

int main(){

    int number;
    number = 20;

    // Here *pointer means that the variable that is declared will be a pointer.
    int *pointer;

    // Actually asigning value to the pointer variable.
    pointer = &number;

    // Pointer of a pointer.
    int **pointerInBetween;

    // Assigning value to the pointer of a pointer. 
    // Assigning the memory location where this pointer points to. 
    // So this is a pointer in between.
    pointerInBetween = &pointer;

    printf("The number before changing is %d\n", number);

    // Pass the pointer variable.
    changeViaPointer(pointer);
    printf("The number after pointer changing is %d\n", number);

    // Pass the pointer of a pointer variable.
    changeViaPointerInBetween(pointerInBetween);
    printf("The number after pointer in between changing is %d\n", number);

    return 0;
}

void changeViaPointer(int *pointer){

    // Okay, at this point we have received a variable called pointer,
    // which points to some value. In order to access this value
    // we need to use *pointer.
    // BUT THIS IS DIFFERENT FROM THE *pointer IN THE FUNCTION DECLARATION!!!
    *pointer = *pointer + 20;
}

void changeViaPointerInBetween(int **pointer){

    // **pointer explanation:
    // Only pointer is the memory location
    // *pointer  is the value of that memory location, which in this specific case is also a memory location
    // **pointer is the value of what the other pointer points to.
    **pointer = **pointer + 20;
}