“释放的指针未被分配”内存块由于某种原因不会解除分配

我很难找到为什么我不能释放内存块。 指针必定有问题。 用于结构的存储器块在函数中进行，并且所使用的指针存储在数组中。 稍后，从数组中取出指针以释放内存。

我已经弄清楚它是免费的。 我把“//这一个”放在旁边。

#include  typedef enum { INT = 0, FLOAT = 1, STRING = 2, NONE = 3, BOOL = 4 } TYPE; typedef struct { TYPE type; void * data; } Variable; typedef Variable ** var; //Convenience typedef for pointer to variable pointer typedef struct { size_t size; Variable *** pool; //Pointer to the array of pointers to the Variable structure pointers. Pointer required for memory allocation and array consists of pointers to pointers so the pointers to the Variable structures can be changed throughtout different functions. } VarPool; //Variable pools will just be used to deallocate all variables at the end of a scope VarPool * global_pool; //Pool for global scope VarPool ** pool_stack; //Keeps track of pools in stack unsigned int pool_stack_size; void init_pool_stack(){ pool_stack = malloc(sizeof(VarPool *)); //Start with global_pool pool_stack_size = 1; global_pool = malloc(sizeof(VarPool)); pool_stack[0] = global_pool; global_pool->pool = NULL; global_pool->size = 0; } Variable ** new_var(){ //Makes new variable Variable ** return_variable; Variable * new_variable = malloc(sizeof(Variable)); VarPool * var_pool = pool_stack[pool_stack_size-1]; //Current variable pool on top of stack var_pool->size++; var_pool->pool = realloc(var_pool->pool,var_pool->size*sizeof(Variable **)); return_variable = &new_variable; var_pool->pool[var_pool->size - 1] = return_variable; return return_variable; //Return pointer to new pointer so pointer can be changed to NULL when deleted } void empty_pool(){ //Frees all data from variable pool VarPool * var_pool = pool_stack[pool_stack_size-1]; //Current pool on top of stack for (int x = 0; x size; x++) { free(*var_pool->pool[x]); //Free variable data } free(var_pool->pool); //Free pool variable array free(var_pool); //This one pool_stack_size--; pool_stack = realloc(pool_stack, pool_stack_size*sizeof(VarPool *)); } int main (int argc, const char * argv[]) { init_pool_stack(); new_var(); empty_pool(); //Finally empty globals pool which will deallocate pool_stack return 0; } 

感谢您的任何帮助。

Answer 1

在new_var()你有（简化）

 Variable ** new_var(){ //Makes new variable Variable ** return_variable; Variable * new_variable = malloc(sizeof(Variable)); return_variable = &new_variable; return return_variable; } 

一旦函数结束，该函数返回的值将变为无效。 局部变量的地址仅在该变量存在时才有意义。