为什么动态分配的int数组在scanf，c语言中使用&？

Question

I learned char array don't use & in scanf because char store point, not string itself.

char s1[10];
scanf("%s", s1);

But, why dynamic allocated int array use & in scanf although it use pointer?

int *arr = (int *)malloc(sizeof(int)*3);
scanf("%d", &arr[1]);

Answer 1

This is completely wrong:

int arr[3] = (int *)malloc(sizeof(int)*3);

because you are trying to allocate memory to an array which is not allowed.
It should be:

int * arr = malloc(sizeof(int)*3);

When you do this, the in-memory view would be something like this (assuming malloc is success):

    +-----------+
    |   |   |   |
    +-----------+
    ^
    |
   arr

arr is a pointer pointing to the base address of the memory.
arr[1] is the first element of array (a single element of integer array arr ) and
&arr[1] is the address of first element of array arr .

    +-----------+
    |   |   |   |
    +-----------+
        ^
        |
       &arr[1]

When you take input in first element of array arr you have to pass the address of first element of array to scanf() which is &arr[1] .

& used with scanf() arguments has nothing to do with dynamic allocation. If you want to take input in an element of fixed length array (size determined at compile time), the syntax will be same:

int arr[3];

scanf("%d", &arr[1]);

Please let me know if you have any further questions.

Answer 2

s1 is a char array. The name of the array (s1) is a pointer to char, and so you don't need & with scanf - because %s expects to read the character array.

char s1[10];
scanf("%s", s1);

Note that in this case you can also specify the address of s1 in scanf as well because in terms of address, s1, &s1[0], &s1 all point to the same address.

char s1[10];
scanf("%s", &s1); // work as well

---

arr is array of integer. It's correct that the name of the array (ie arr ) is also a pointer (to int). But for scanf %d, which expect to read an integer, you have to provide the address of the integer you are reading. It's not an array, it's individual integer inside the array that scanf tries to read to - and you have to specify the address of it accordingly: &arr[1]

int arr[3] = (int *)malloc(sizeof(int)*3);
scanf("%d", &arr[1]);

Answer 3

When you scanf the array of chars (string), in your case s1, you don't need & because s1 stores a pointer to the first element (the element with index 0) of this array. With regard to dynamic allocation, malloc return value is a pointer to someplace in memory where your array will be stored. So your declaration should look like this:

int* arr = malloc(sizeof(int) * 3); 
scanf("%d", &add[1]); 
/* here you scanf only the second element of your massive,
 it is int so you have to put & there to pass a pointer */

If you want to initialize the whole massive you can write easy cycle for it. Iwould also recommend you to read some documentation about malloc and an article "calloc vs malloc".

Good luck :)