Question

我正在努力学习C语言，而且我在指针指针上挂了一点。我想我明白你为什么需要它们，但不能完全理解正在发生的事情。

例如，以下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：

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（printf函数中的newp），那么我会得到这个：

**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），并且在封闭函数（newp）返回后不可用。

在p之外取消引用newp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

您正在d中存储局部变量（*p）的地址，然后在变量超出范围时取消引用它。 Undefined Behavior

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;
}

此static关键字使编译器在“静态存储”中定位本地变量，该静态存储在调用newp()后的时间段内继续存在（即保持其值）。但是，此内存只有一个副本 - 每次后续调用包含的函数（newp）都会重新使用相同的内存位置，并可能会覆盖此时的值。 / p>

如果没有static关键字来限定本地变量声明，则存储将是“自动”的，这意味着在包含函数返回后它将自动从当前使用中释放。在newp返回后，以前用于局部变量的内存可以用于任何目的。

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;
}

C指针和指针传递指针

4 个答案: