分配和访问结构体中字符串的指针
我试图将字符串存储在结构中包含的数组中，并访问它，但我遇到了困难。该结构如下所示：
typedef struct {
void **storage;
int numStorage;
} Box;
Box 初始化如下：
b->numStorage = 1000000; // Or set more intelligently
Box *b = malloc(sizeof(Box));
// Create an array of pointers
b->storage = calloc(b->numStorage,sizeof(void *));
为了设置字符串，我使用此函数：
void SetString(Box *b, int offset, const char * key)
{
// This may seem redundant but is necessary
// I know I could do strcpy, but made the following alternate
// this isn't the issue
char * keyValue = malloc(strlen(key) + 1);
memcpy(keyValue, key, strlen(key) + 1);
// Assign keyValue to the offset pointer
b->storage[offset*sizeof(void *)] = &keyValue;
// Check if it works
char ** ptr = b->storage[offset*sizeof(void *)];
// It does
printf("Hashcode %d, data contained %s\n", offset, *ptr);
}
问题在于，当我尝试使用完全相同的偏移量再次检索它时：
// Return pointer to string
void *GetString(const Box *b, int offset, const char *key)
char ** ptr = b->storage[offset*sizeof(void *)];
if (ptr != NULL) {
printf("Data should be %s\n", *ptr);
return *ptr;
} else {
return NULL;
}
返回的指针是乱码。可能出什么问题了？
I'm trying to store a string in an array contained within a struct, and access it, but I'm having a hard time. The struct looks like this:
typedef struct {
void **storage;
int numStorage;
} Box;
Box is initialized as such:
b->numStorage = 1000000; // Or set more intelligently
Box *b = malloc(sizeof(Box));
// Create an array of pointers
b->storage = calloc(b->numStorage,sizeof(void *));
In order to set the string, I use this function:
void SetString(Box *b, int offset, const char * key)
{
// This may seem redundant but is necessary
// I know I could do strcpy, but made the following alternate
// this isn't the issue
char * keyValue = malloc(strlen(key) + 1);
memcpy(keyValue, key, strlen(key) + 1);
// Assign keyValue to the offset pointer
b->storage[offset*sizeof(void *)] = &keyValue;
// Check if it works
char ** ptr = b->storage[offset*sizeof(void *)];
// It does
printf("Hashcode %d, data contained %s\n", offset, *ptr);
}
The problem lies when I try to retrieve it again, with the exact same offset:
// Return pointer to string
void *GetString(const Box *b, int offset, const char *key)
char ** ptr = b->storage[offset*sizeof(void *)];
if (ptr != NULL) {
printf("Data should be %s\n", *ptr);
return *ptr;
} else {
return NULL;
}
The returned pointer is gibberish. What could be amiss?
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访问数组时不必指定实际的内存偏移量。只需给它索引，您就会得到正确的元素。
那么，在你的第三个代码块中：
在你的第四个代码块中：
另外，在第二个代码块中，是否已设置
b->numStorage？You don't have to specify the actual memory offset when accessing arrays. Simply give it the index and you will get the correct element.
So, in your third code block:
And in your fourth:
Also, in the second block of code, has
b->numStoragealready been set?这将局部变量
keyValue的地址存储在数组中。一旦函数完成，该地址就变得无效。我想你想要：然后在检索时进行相应的更改。
This stores the address of the local variable
keyValuein the array. Once the function completes, this address becomes invalid. I think you want:and then make the corresponding change while retrieving.
这不就是：
设置storage[offset*sizeof(void*)]指向局部变量keyValue的地址吗？即函数返回后不再有效
Doesn't this:
set storage[offset*sizeof(void*)] to point to the address of the local variable keyValue? i.e. no longer valid after function returns