Suneet Jain

• Nov 20th, 2006

 

you have taken 3[a] as a string. Thats why iot is working. Remove string quotes and try, it won't work.

Vamshi

• Nov 22nd, 2006

 

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.

Bhushan

• Jan 21st, 2007

 

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.)

mytest1

 Profile Answers by mytest1 

• Jan 30th, 2007

 

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

winny gupta

• Feb 15th, 2007

 

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.