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如何从函数中的数组返回结构? (指针)

[英]How do you return a struct from an array within a function? (pointers)

Consider this code: 考虑以下代码:

typedef struct fruits_s
{
    char* key;
    char value;
} fruits_t;

static fruits_t fruit_array[] = {
{ "Apple", 1 },
{ "Banana", 2 },
{ "Grape", 3 },
{ "Orange", 4 } };

static fruits_t* getFruitFromValue(char value)
{
    int i;
    for (i = 0; i < sizeof(fruit_array)/sizeof(fruit_array[0]); i++){
        if (value == fruit_array[i].value){
            return fruit_array[i];
        }
    }
}

I am new to C and am still learning when pointers are necessary/used. 我是C的新手,并且在需要/使用指针时仍在学习。 I come spoiled from a Java background. 我从Java背景中被宠坏了。 So, in the above code, what I'm confused of is should the function return a pointer fruits_t* ? 那么,在上面的代码中,我感到困惑的是函数应该返回指针 fruits_t* Or something else? 或者是其他东西? When I do fruit_array[i] is that a pointer to my struct, or the struct itself? 当我做fruit_array[i]是指向我的结构或结构本身的指针?

That being said, later in my code when I want to use the function, is it this: 话虽如此,后来在我的代码中,当我想使用该函数时,是这样的:

 fruits_t* temp = getFruitFromValue(1);

or 要么

 fruits_t temp = getFruitFromValue(1);

or 要么

 fruits_t temp = &getFruitFromValue(1);

The function could return either — your choice. 该功能可以返回 - 您的选择。 You've said you'll return a pointer; 你说你会回指针; that's OK as long as you do. 只要你这样做就没问题。

When you write: 当你写:

static fruits_t *getFruitFromValue(char value)
{
    int i;
    for (i = 0; i < sizeof(fruit_array)/sizeof(fruit_array[0]); i++){
        if (value == fruit_array[i].value){
            return fruit_array[i];
        }
    }
}

There are several problems: 有几个问题:

  1. fruit_array[i] is a structure, not a pointer. fruit_array[i]是一个结构,而不是指针。 Use return &fruit_array[i]; 使用return &fruit_array[i]; .
  2. If the loop exits, you don't return a value from the function at all. 如果循环退出,则根本不从函数返回值。

Fixing those leads to: 修复这些导致:

static fruits_t *getFruitFromValue(char value)
{
    int i;
    for (i = 0; i < sizeof(fruit_array)/sizeof(fruit_array[0]); i++)
    {
        if (value == fruit_array[i].value)
            return &fruit_array[i];
    }
    return NULL;
}

This is OK because the pointer you return is to static data that will outlive the function. 这是可以的,因为您返回的指针是静态数据,该数据将比该函数更长。 If you tried to return a pointer to non-static data, you would (probably) have a bug on your hands, unless you used dynamic memory allocation (via malloc() et al). 如果您尝试返回指向非静态数据的指针,除非您使用动态内存分配(通过malloc()等),否则您可能(可能)手上有一个错误。

You could also return the structure; 你也可以归还结构; handling the error return becomes harder. 处理错误返回变得更难。 If you've got C99, you can use a 'compound literal': 如果你有C99,你可以使用'复合文字':

static fruits_t getFruitFromValue(char value)
{
    int i;
    for (i = 0; i < sizeof(fruit_array)/sizeof(fruit_array[0]); i++)
    {
        if (value == fruit_array[i].value)
            return fruit_array[i];
    }
    return (fruits_t){ .key = "", .value = 0 };
}

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