c - "pointer being freed was not allocated"内存块由于某种原因不会释放

Question

我正在努力寻找无法释放内存块的原因。指针一定有问题。结构的内存块在函数中创建，使用的指针存储在数组中。稍后从数组中获取指针以用于释放内存。

我已经弄清楚它是免费的了。我在它旁边放了“//这个”。

#include <stdlib.h>

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 < var_pool->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;
}

感谢您的帮助。

Answer 1

在 new_var() 中你有(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，从该函数返回的值将失效。局部变量的地址仅在该变量存在时才有意义。