“被释放的指针未分配”是什么意思？

Question

I have trouble with a program where im trying to copy a string to a structs string variable in c. After copying the string im trying to free the temporary string variable which is copied to the structs string variable. But when i try to free the string the program turns on me and says that "pointer being freed was not allocated". I don't understand exactly what is going on.

char str[]="       ";           //temporary string to copy to structs string
str[3]=s;                       //putting a char s in middle
strcpy(matrix[i-1][j].c, str);  //copying the string
free(str);                      //freeing str that now is useless when copied

Answer 1

Only pointers returned by calls to malloc() , realloc() or calloc() can be passed to free() (dynamically allocated memory on the heap). From section 7.20.3.2 The free function of C99 standard:

The free function causes the space pointed to by ptr to be deallocated, that is, made available for further allocation. If ptr is a null pointer, no action occurs. Otherwise, if the argument does not match a pointer earlier returned by the calloc, malloc, or realloc function, or if the space has been deallocated by a call to free or realloc, the behavior is undefined.

In the posted code, str is not dynamically allocated but is allocated on the stack and is automatically released when it goes out of scope and does not need to be free() d.

Answer 2

Be careful. This code shows confusion about two things:

1. The difference between stack and heap memory
2. The operation of strcpy

Point 1

This has already been answered in a way, but I'll expand a little:

The heap is where dynamic memory is given to your process. When you call malloc (and related functions), memory is returned on the heap. You must free this memory when you are done with it.

The stack is part of your process's running state. It is where ordinary variables are stored. When you call a function, its variables are pushed onto the stack and popped back off automatically when the function exits. Your str variable is an example of something that is on the stack.

Point 2

I'd like to know what that c member of your matrix array is. If it is a pointer, then you might be confused about what strcpy does. The function only copies bytes of a string from one part of memory to another. So the memory has to be available.

If c is a char array (with sufficient number of elements to hold the string), this is okay. But if c is a pointer, you must have allocated memory for it already if you want to use strcpy . There is an alternative function strdup which allocates enough memory for a string, copies it, and returns a pointer. You are responsible for free ing that pointer when you no longer need it.

Answer 3

It used free without using malloc . You can't release memory that hasn't been allocated.

Answer 4

Your code doesn't allocate any memory for the string. It simply copies a string from one string to the memory used by another (a string of spaces).

Memory is allocated using functions like malloc() , realloc() , etc. and then freed with free() . Since this memory was not allocated that way, passing the address to free() results in the error.

Answer 5

You did not use malloc() to allocate space in the heap for str . Therefore, str is allocated on the stack and cannot be deallocated with free() , but will be deallocated when str goes out of scope, ie after the function containing the declaration returns.