Concepts - Take this example:#include int main (){ int a [5]; a [3] = 1111; printf ("3[a] = %dn", 3[a]); return

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Take this example:

#include

int main ()
{
int a [5];
a [3] = 1111;

printf ("3[a] = %dn", 3[a]);

return 0;
}

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

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Best Rated Answer

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Submitted by: winny gupta

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.

Above answer was rated as good by the following members:
sandesh k

November 20, 2006 12:35:06

Suneet Jain

RE: Take this example:#include

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

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November 22, 2006 01:30:17

Vamshi

RE: Take this example:#include

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.

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Overall Rating: -1

January 21, 2007 23:52:10

Bhushan

RE: Take this example:#include

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.)

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January 30, 2007 12:38:11

mytest1

Member Since: January 2007 Contribution: 2

RE: Take this example:#include

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

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February 15, 2007 07:50:49

winny gupta

RE: Take this example:#include

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.

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1 0

Overall Rating: +1

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