C中这些访问有什么区别？

Question

i started learning C today, and i have some questions about accessing a pointer data.

I have this function in C:

typedef struct
{
    size_t size;
    size_t usedSize;
    char *array;
} charList;

void addToCharList(charList *list, char *data)
{
    if(list->usedSize == list->size)
    {
        list->size *= 2;
        list->array = realloc(list->array, list->size * sizeof(int));
    }
    list->array[list->usedSize++] = *data;
    printf("1: %d\n", *data);
    printf("2: %s\n", data);
    printf("3: %p\n", &data);
}

I use it to create a "auto growing" array of chars, it works, but i'm not understanding why i need to attribute the value "*data" to my array. I did some tests, printing the different ways i tried to access the variable "data", and i had this outputs (I tested it with the string "test"):

1: 116
2: test
3: 0x7fff0e0baac0

1: Accessing the pointer (i think it's the pointer) gives me a number, that i don't know what is.

2: Just accessing the variable gives me the actual value of the string.

3: Accessing it using the "&" gets the memory location/address.

When i'm attributing the value of my array, i can only pass the pointer, why is that? Shouldn't i be attributing the actual value? Like in the second access.

And what is this number that gives me when i access the pointer? (First access)

Answer 1

So, in the first printf, you're not actually accessing the pointer. If you have a pointer named myPointer , writing *myPointer will actually give you access to the thing that the pointer is pointing to. It is understandable to be confused about this, as you do use the * operator when declaring a variable to specify that it is a pointer.

char* myCharPtr; // Here, the '*' means that myCharPtr is a pointer.

// Here, the '*' means that you are accessing the value that myCharPtr points to.
printf("%c\n", *myCharPtr); 

In the second printf, you're accessing the pointer itself. And in the third printf, you're accessing a pointer to a char pointer. The & operator, when put before a variable, will return a pointer to that variable. So...

char myChar = 'c';

// pointerToMyChar points to myChar.
char* pointerToMyChar = &myChar;

// pointerToPointerToMyChar points to a pointer that is pointing at myChar.
char** pointerToPointerToMyChar = &pointerToMyChar;

When you're trying to store the value in the array, it's forcing you to do *data because the array stores chars, and data is a pointer to a char. So you need to access the char value that the pointer is pointing to. And you do that via *data .

Lastly: the reason that printf("1: %d\n", *data);prints a number is that chars are secretly (or not secretly) numbers. Every character has a corresponding numeric value, behind the scenes. You can see which numerical value corresponds to which character by looking at an ascii table.