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Array confusion
Hi,

This is from K&B book.



which of the following codes could be inserted at line 7, and still allow the code to compile? (choose four).

a) b2 [0][1] = b;
b) b [0][0] = b3;
c) b2 [1][1][0] = b [0][0];
d) b2 [1][2][0] = b;
e) b2 [0][1][0][0] = b [0][0];
f) b2 [0][1] = big;

I couldn't understand how to work it out the dimensions.
Can anyone please help.

Thanks.
You just need to make sure that the reference types are the same in the LHS and RHS:


And to elaborate:

byte[][][][] b;
Now, b is a 4-dimensional byte array.
This means:
b[0] will be a 3-dimensional byte array, or a reference of type byte[][][]
b[0][0] will be a 2-dimensional byte array, or a reference of type byte[][]
b[0][0][0] will be a 1-dimensional byte array, or a reference of type byte[]
b[0][0][0][0] will be a reference of type byte (you can think of it as a 0-dimensional byte array)
Thanks Ruben for your very clear explanation.

Regards
alph
No problem!
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