Question

我很难找到为什么我无法释放内存块。指针必定有问题。用于结构的存储块在函数中进行，并且所使用的指针存储在数组中。之后，指针从数组中取出，用于释放内存。

我已经知道它是免费的。我把“//这一个”放在旁边。

#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，此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。

“没有分配被释放的指针”由于某种原因，内存块不会解除分配

1 个答案: