如何正确地分配和打印c中的结构指针?

I have the following two structures.

我有下面两个结构。

struct string_counter {
  char *string;
  int count;
};


struct container{
  struct string_counter *counters;
  int size;
};

And the following functions.

和下面的函数。

void add_string(struct *container, char *string){
    struct string_counter *counter_ptr;
    counter_ptr = new_count(string);

    container->size++;
    container->counters = realloc(container->counters, sizeof(struct string_counter)*container->size);
    container->counters[container->size-1] = *counter_ptr;
    print_string_counter(&container->counters[container->size-1]);
}

struct counter *new_count(char *string){
    struct string_counter *counter_ptr;

    counter_ptr = malloc(sizeof(struct string_counter));
    counter_ptr->string = string;
    counter_ptr->count = 1;

    return counter_ptr;
}

In my main function I am looping over several times callingadd_string(container_ptr, string) with the same container pointer and a different string each time, attempting to add the new string to the end of the string_counter *counters. Later on in my code when I try printing the contents of the counters using the following two functions:

在我的main函数中，我将多次使用相同的容器指针和一个不同的字符串来循环多次callingadd_string(container_ptr, string)，试图将新字符串添加到string_counter *计数器的末尾。稍后在我的代码中，当我尝试使用以下两个函数打印计数器的内容时:

void print_container(struct container *container_ptr){
    int i;
    for(i = 0; i < container_ptr->size; i++){
         print_string_counter(&container_ptr->counters[i]);
    }
}

void print_string_counter(struct string_counter *counter_ptr){
    printf("%s : %d\n", counter_ptr->string, counter_ptr->count);
}

I get extremely incorrect results such as my output below. But earlier when I printed with :

我得到了非常不正确的结果，如下面的输出。但是在我打印的时候:

print_string_counter(&container->counters[container->size-1])

just below :

下面:

container->counters[container->size-1] = *counter_ptr;

in my add_string() function, I got the correct results.

在add_string()函数中，我得到了正确的结果。

Initially my container values are

最初我的容器值是。

struct_counter = NULL;
int size = 0;

Why is my output using the loop later on different from originally when I printed it?

为什么在我打印的时候，我的输出与原来的循环不同?

1 个解决方案

#1

The main problem is likely that your string_counter struct does nat actually store the strings -- it just stores pointers, which will be point at whatever you pass in. In new_count, if that string you pass in is in a temporary buffer (say the buffer you use to read input in main), then when that buffer goes away, the pointer in the string_counter will be left dangling and when you later go to print, you'll get whatever garbage happens to be in memory where your original buffer was.

主要的问题很可能是，您的string_counter结构实际上是nat存储了字符串——它只是存储指针，它将指向您传入的任何东西。new_count,如果你传入的字符串是一个临时缓冲区中(比如缓冲主要用于读取输入),当缓冲区消失,指针的string_counter稍后将左晃来晃去的,当你去打印,你会得到任何垃圾恰好是在内存中原始缓冲区在哪里。

Related (but indpendent) your container struct maintains an array of string_counter objects, not pointers, so when you call add_string, it extends the array AND calls new_count to allocate a new string_counter. It then copies that newly allocated object and leaks it.

Related(但indpendent)您的容器struct维护一个string_counter对象数组，而不是指针，所以当您调用add_string时，它扩展了数组，并调用new_count来分配一个新的string_counter。然后复制新分配的对象并将其泄漏。

So what you likely need is in new_counter to copy the string as well. You can also fix the second problem by having new_count return an object rather than a pointer:

因此，您可能需要在new_counter中复制该字符串。您还可以通过让new_count返回一个对象而不是一个指针来修复第二个问题:

struct counter new_count(char *string){
    struct string_counter counter;

    counter.string = strdup(string);
    counter.count = 1;

    return counter;
}