You notice that I use 3[a] instead of a[3], however, this code works fine but I don't know how. It will be grateful if someone explians to me why and how this code works fine. Thanks.
Total Answers and Comments: 5
Last Update: February 15, 2007 Asked by: Anouar LACHHEB
int a[5]; // This line is creating a unintialized array of 5 integers. a [3] = 1111; // Here Ur intializing the 4th element to 1111printf ("3[a] = %dn", 3[a]); // By saing 3[a] compiler takes it as a[3] and as it is intialized to 1111 u will see 1111 printed on screen.In 3[a].... 3 is displacement/index in the array : a. try out this to understand: int a[5] = {1,1,1,1,1};a[3] = 1111;ptintf("1[a] = %d t 3[a] = %d",1[a],3[a]);Note : diplacement/index starts from 0.
regarding that printf statement.......... a[3] is internally calculated by compiler as (a+3). so 3[a] will be calculated
as (3+a) . and these two statements calculates the same value and this is the reason why that code works. (Name of array itself acts as starting address of the array.)
whenever u declare a array, the name of the array( i.e the array variable) is actually a pointer to the starting address of the memory area that was reserved for the array..for ex:int arr[10];here "arr" is actually a pointer pointing to the starting position of the contiguous memory that was allocated for the arrayi. if the following memory was allocated for the array000 001 002 003 004 005 006 007 008 009then, arr is pointing to 000when u use the [] to access an array, the value within the array is added to the value of the base pointer(arr)i.e arr[3] = *(arr+3) = *(000+3) + *(003)so arr[3] = 3[arr] = *(arr+3) = *(3+arr)hope that solves ur doubt
the above code works fine. This is so because when the program runs, an array is internally converted to a pointer notation,i.e., *(a + 3) for a[3]. The same notation,i.e. *(a+3) is achieved for 3[a]. Hence, the above code works perfectly well. This is valid only in printf statement and not in the scanf statement.
