通过强制转换取消引用 void 指针

Question

我在试验空指针时遇到了以下问题。当我编译下面的代码时一切正常但是当我运行它在我的机器上时它不会给我结果而是提示 The program has stopped working appeared(https://drive.google.com/file/d/0B1mLcnk8kTFUeEtmYnlOaWJ6T3c/view?usp=sharing) ，我不能找出幕后发生的事情，是否有关于取消引用 void 指针的任何问题。

此外，如果我使用注释代码而不是原始代码，它会起作用。

 #include<stdio.h>
 int main()
 {
     void* ptr;
     int dupe = 5;
     *(int* )ptr = dupe;  // ptr = &dupe;
      printf("The value at ptr is %3d",*(int* )ptr);
 }

我在 CodeBlocks 上使用 gcc。

Answer 1

此处您试图取消引用指针而不为其分配内存。这导致 segmentation fault 或 undefined behaviour.

int main(int argc, char *argv[])
{
    void* ptr;
    int dupe = 5;
    ptr = malloc(sizeof(dupe)); /* alloc memory for ptr */
    *(int* )ptr = dupe;
    printf("The value at ptr is %3d",*(int* )ptr);
    free(ptr);
    return 0;

}

如果您不打算分配内存，那么只需在您的代码中将 *(int* )ptr = dupe; 替换为 ptr = &dupe，因为现在 ptr 将指向变量 [=] 的有效内存位置16=]。
有关未定义行为的更多详细信息，请参阅 this

Answer 2

6.       *(int* )ptr = dupe;  // ptr = &dupe;

指针 ptr 未指向有效的内存位置。你在评论里写的，就是你应该做的。

语句ptr = &dupe将使ptr指向变量dupe的内存位置。

Answer 3

1. void* ptr; --->ptr is a void pointer that means it can hold an address of any data type, but currently it is not pointing to any address.
2. *(int* )ptr = dupe ---> ptr point to no where and you are trying to put a value which is not valid.
3. ptr = &dupe; --> in this case ptr will point to address of dump which is a valid one.

Answer 4

一个指针int *ptr，通常是一个地址值，就像一个盒子的编号。这里的int是盒子类型，暗示这个盒子是用来放一个int的。当然，你也可以小心地转换它来容纳其他类型的东西，比如(otherKindType*)ptr.
当尝试用 *ptr = someValue 引用一个指针时，比如将对象放在盒子里，你应该保证它指向某个准确的地方。

Code1（可以工作）：

#include<stdio.h>
int main()
{
    void* ptr; // ptr is a pointer, it should point to somewhere 
    int dupe = 5; // a box (named dupe) fitted for int which hold 5
    ptr = &dupe; // ptr points to that box named dupe
    printf("The value at ptr is %3d",*(int* )ptr); // get content inside the box
}

Code2（无效）：

#include<stdio.h>
int main()
{
    void* ptr;// pointer, point to a random place
    int dupe = 5;
    *(int* )ptr = dupe; // To put 5 to the box that pointed by ptr
                        // But now, ptr still points to a random place
                        // oops! To write to a random box is dangerous!
                        // It will cause error like `BUS error`
                        // Sometimes, to read a random box can output wired value.
    printf("The value at ptr is %3d",*(int* )ptr);
}

Code3（可以工作）：

#include<stdio.h>
int main()
{
    void* ptr;
    int dupe = 5; 

    int new_box;//added         
    ptr = &new_box;//added, now ptr points to new_box        

    *(int* )ptr = dupe; // To put 5(content of dupe) to the box that pointed by ptr
                        // That is to put 5 to new_box
    printf("The value at ptr is %3d",*(int* )ptr); // get content inside the box
}

通过强制转换取消引用 void 指针

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