array

[help1267]

Hi,

Instead of:
char let[26] = {'A','B','C','D','E',...., 'Z''};

What would be another (shorter) way of writing this ?

Thx,

[rgb]

this is dependent on your character set (ASCII will work), you can declare a char data type initialized at first letter of the aplphabet and then increment it (ex. i++) for the next letter on a loop.

[help1267]

Hi rgb,
thanks for your reply. I'm new at this. Would you be able to elaborate ?

Thx,

[D_C]

Banfa mentioned in an earlier post, that only 0-9 are guaranteed to be consecutive, although as is the case for ASCII, a-z are consecutive, as well as A-Z.

With your array, let[0] = 'a', let[25] = 'z'. For 0 <= N <= 25, you could replace let[N] with ('a'+N)

[rgb]

Banfa mentioned in an earlier post, that only 0-9 are guaranteed to be consecutive, although as is the case for ASCII, a-z are consecutive, as well as A-Z.

as mentioned, the approach is dependent on the charater set. i am not familiar with other sets, most of the time i use ASCII character set.

[rgb]

Hi,

Instead of:
char let[26] = {'A','B','C','D','E',...., 'Z''};

What would be another (shorter) way of writing this ?

Thx,

you can also use this approach:

char let[ ] = "ABCDEF...XYZ";

the array let[ ] will be indexed the same as yours.

let[0] will be equal to 'A'
let[1] will be equal to 'B'
etc
etc
etc
let[25] will be equal to 'Z'

[Talon4196]

char let[26];
for (int x = 65; x < 91; x++)
let[x-65] = char(x);

[Talon4196]

well.. instead of 65, and 91, just use int('A') and int('Z')... 65 and 91 is what the values are for my keyboard...