函数参数一个指针数组 (C)

Question

在我的代码中，我有一个指针数组，其中的指针指向我的结构

struct Container *bucket[sizeOfHashMap];

我有一个函数将返回这些数组指针中的 1 个（例如，它可能返回数组索引 6 处的指针）。作为参数，它需要一个指向此指针的指针。该功能可以在这里看到：

struct Container* getWhichBucket(char word[], struct Container **bucket[10]){
    int value = 0;
    int i = 0;
    int size = strlen(word);
    int hashIndex = 0;

    for(i =0; i < size; i++){
      value += (int)word[i];
    }

    //size of array is worked out by getting memory that array takes up / a slot
    hashIndex =  value % sizeOfHashMap;
    return *bucket[hashIndex];
}

我这样调用函数（其中 test 是一个字符数组）

addToBucket(test, getWhichBucket(test, &bucket));

添加到存储桶如下所示：

void addToBucket(char word[], container **bucket){
    container *temp = (struct Container*)malloc (sizeof(struct Container));
    strcpy(temp->key, word);
    temp->value = 9001;
    temp->next = *bucket;
    *bucket = temp;

    return;
}

但是，当我 compile 代码和 run 它时，编译器会发出 warnings segmentation error。有谁知道为什么？可以在此处查看警告：

cw.c: In function ‘main’:
cw.c:86:2: warning: passing argument 2 of ‘getWhichBucket’ from incompatible pointer type [enabled by default]
cw.c:37:19: note: expected ‘struct Container ***’ but argument is of type ‘struct Container * (*)[(long unsigned int)(sizeOfHashMap)]’
cw.c:86:2: warning: passing argument 2 of ‘addToBucket’ from incompatible pointer type [enabled by default]
cw.c:56:6: note: expected ‘struct container **’ but argument is of type ‘struct Container *’

Answer 1

addToBucket(test, getWhichBucket(test, &bucket));

正在传递一个

struct Container *(*)[10]

到getWhichBucket。正如编译器所说，这是错误的类型。

您可以修复原型和实现

struct Container* getWhichBucket(char word[], struct Container *(*bucket)[10]){
    int value = 0;
    int i = 0;
    int size = strlen(word);
    int hashIndex = 0;

    for(i =0; i < size; i++){
      value += (int)word[i];
    }

    //size of array is worked out by getting memory that array takes up / a slot
    hashIndex =  value % sizeOfHashMap;
    return (*bucket)[hashIndex];
}

或更改调用，但没有简单的方法从struct Container *bucket[10] 获取struct Container **bucket[10]，因此您可能仍想更改getWhichBucket 的类型和实现。

由于这里没有修改bucket参数，所以不需要传递地址，你可以直接传递struct Container *bucket[10]，

struct Container* getWhichBucket(char word[], struct Container *bucket[]){
    int value = 0;
    int i = 0;
    int size = strlen(word);
    int hashIndex = 0;

    for(i =0; i < size; i++){
      value += (int)word[i];
    }

    //size of array is worked out by getting memory that array takes up / a slot
    hashIndex =  value % sizeOfHashMap;
    return bucket[hashIndex];
}

然后打电话

addToBucket(test, getWhichBucket(test, bucket));

Answer 2

您需要将 addToBucket 的声明从

void addToBucket(char word[], container *bucket)
{ 
    container *temp = (struct Container)malloc (sizeof(struct Container)); 
    strcpy(temp->key, word); 
    temp->value = 9001; 
    temp->next = *bucket; 
    *bucket = temp; 
    return; 
} 

到

void addToBucket(char word[], Container *bucket)
{ 
    Container *temp = malloc (sizeof(struct Container)); 
    strcpy(temp->key, word); 
    temp->value = 9001; 
    temp->next = *bucket; 
    *bucket = temp; 
    return; 
} 

注意Container 的大小写变化——C 语言中大小写很重要...container 与Container 不同。

另外...注意...you should not cast malloc in C.