homework

Question

Given an array A[-1:6,-2:10]. The base address of array is 1000. If every elements takes 4 bytes for storage then compute the address of element A[5,7]

answer given is 1348

Comments

how to solve ? please help

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rows:  -1 to 6

columns: -2 to 10 --> total columns==> 13

A[5,7]=> to reach 5 rows we need to cross -1 0 1 2 3 4 =>6 row 

each row has 13 columns and size of each element is 4=> 6*13*4=>312

Now we are at starting of 5 , from here we need to get A[5,7]

i.e. to reach 7 we need to cross over -2 -1 0 1 2 3 4 5 6 => 9 columns

size of each element is 4 => 9 * 4==> 36

so get A[5,7]we  need to cross 312+36=> 348

Given base address is 1000=> starting address of A[5,7]= 1000+348==> 1348

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@Hemanth_13 

Your answer is correct and method is also very good.

Answer 1

$A[-1:6;-2:10]$

This is $8\times13$ array.

We know if indexing starting from $'0'$ so we can easily apply the formula.

Now we can convert this array into $'0'$ indexing. Add $+1$ and $+2$

$A[-1+1:6+1;-2+2:10+2]$

$A[0:7;0:12]$

Here $8\times13$ array so we can say $m\times n = 8\times13$

Now we want to find $A[5,7]$,here remember we add $+1$ and $+2$

So $A[5+1,7+2]=A[6,9]=A[i,j]$

Now,Row Major order:

$A[i,j]=$Base address$+((n*i)+j)*$Size of each element

$A[6,9]=1000+((13*6)+9)*4$

$A[6,9]=1000+(78+9)*4$

$A[6,9]=1000+(87*4)$

$A[6,9]=1000+348$

$A[6,9]=1348$

and Column Major order$:$

$A[i,j]=$Base address$+((m*j)+i)*$Size of each element

$A[6,9]=1000+((8*9)+6)*4$

$A[6,9]=1000+(72+6)*4$

$A[6,9]=1000+(78*4)$

$A[6,9]=1000+312$

$A[6,9]=1312$

If nothing is given go for Row major order.