C语言：动态分配含非动态元素的结构体的内存位置及释放问题

Great question—let's walk through this line by line to clear up your confusion.

Absolutely yes. Here's why:

• When you call malloc(sizeof(A)), you're requesting a block of memory on the heap that's exactly the size of your A struct. In this case, since A only contains an int member c, the block is the size of a single int.
• The pointer a stores the address of this heap-allocated block. When you access a->c, you're directly referencing the portion of that heap block reserved for the c member.

To put it simply: the entire struct (and all its members) lives in the heap memory you allocated with malloc. If you had declared a struct on the stack like A a_stack;, then a_stack.c would live on the stack—but that's not what's happening here.

When you call free(a), you're telling the operating system's memory manager that the heap block pointed to by a is no longer needed and can be reused for future allocations. Here's what that means for c:

• The memory that held c is marked as available by the memory manager, but it's not automatically erased. The value 2 might still be present in that memory location for a while, but this is just "garbage" data—your program has no right to access it anymore.
• Accessing a->c after free(a) is undefined behavior. Your program could crash, return a random value, or behave unpredictably—never do this.
• After free(a), the pointer a itself still holds the old address of the freed block (this is called a "dangling pointer"). It's a good practice to set a = NULL immediately after freeing to avoid accidental access to freed memory.

内容的提问来源于stack exchange，提问作者giudaballerino