C语言中数组的动态内存分配(Dynamic Memory Allocation for Arrays in C)

C语言中数组的动态内存分配(Dynamic Memory Allocation for Arrays in C)

所以我试图在C中使用malloc创建一个动态数组，但由于某种原因，它不能正常工作。 这是我的代码：

int* test = (int*) malloc(20 * sizeof(int)); printf("Size: %d\n", sizeof(test));

当我运行这个代码时，控制台输出8，尽管理想情况下它应该输出80，我相信，因为int的大小是4，并且我创建了20个。 那么，为什么这不起作用呢？ 谢谢你的帮助。 :)

So I'm trying to create a dynamic array in C, using malloc, but for some reason it's not working out. Here's my code:

int* test = (int*) malloc(20 * sizeof(int)); printf("Size: %d\n", sizeof(test));

The console outputs 8 when I run this code, although it ideally should output 80, I believe, since the size of an int is 4, and I'm creating 20 of them. So why isn't this working? Thanks for any help. :)

最满意答案

sizeof运算符返回int*的大小，这只是一个指针。 在C中，当你使用malloc动态分配内存时，对malloc的调用返回一个指向新分配的内存块的指针（通常在堆上）。指针本身只是一个内存地址（通常只占用4或8个字节在大多数现代系统上）。 您需要跟踪自己分配的实际字节数。

C语言只会跟踪静态分配的缓冲区（堆栈数组）的缓冲区大小，因为数组大小在编译时可用。

int test[100]; printf("Sizeof %d\n", sizeof(test))

这将打印一个等于100 * sizeof(int) （在大多数现代机器上通常为400）

The sizeof operator is returning the size of an int*, which is just a pointer. In C, when you allocate memory dynamically using malloc, the call to malloc returns a pointer to the newly allocated block of memory (usually on the heap.) The pointer itself is just a memory address (which usually only takes up 4 or 8 bytes on most modern systems). You'll need to keep track of the actual number of bytes allocated yourself.

The C language will only keep track of buffer sizes for statically allocated buffers (stack arrays) because the size of the array is available at compile time.

int test[100]; printf("Sizeof %d\n", sizeof(test))

This will print a value equal to 100 * sizeof(int) (usually 400 on most modern machines)