I doesn't understand the 1st line. What output is to come?

Question

int*arr=(int*) malloc (3*size of(int));
arr[0]=1;
arr[1]=3;
arr[2]=2;
printf ("%d", [arr[0]]);

Answer 1

allocates requested size of bytes and returns a void pointer pointing to the first byte of the allocated space
please read this pos you can learn pointers concepts
https://www.design-reuse.com/articles/25090/dynamic-memory-allocation-fragmentation-c.html
Since the size of int is 4 bytes, this statement will allocate 3*4 bytes of memory. And, the pointer *arr point the first byte in the allocated memory

Answer 2

Suparna Podder Here arr[arr[0]] means here arr[0] is 1 so arr[0] will give 1 so arr[1] and its value is 3 .becoz(arr[1]=3)

{

int*arr=(int*) malloc (3*sizeof(int));
arr[0]=1;
arr[1]=3;
arr[2]=2;
printf ("%d",arr[arr[0]]);
return 0;
}

Answer 3

int* arr=(int*) malloc(3*sizeof(int)); //type errors removed. Check..

malloc is dynamic memory allocation method.
Here allocating 3 locations of size int compatible of int* (Integer pointer type.)
Then initializing values...

arr[0]=1;
arr[1]=3;
arr[2]=2;
printf("%d", arr[0]);

Answer 4

😅🐍🐍😆 thank you . But how it's output to come 3?

Answer 5

Suparna Podder output comes 1 only for arr[0]
If you getting 3,then I guessing you are given full statement as it giving error. So I guessing, you may missing there like arr[arr[0]]. This return 3 as output..
Inner arr[0] return 1, then outer arr[1] return 3.