Dumb Question

[diod]

I'm an extreme newbie, but im trying to make this app thats like a multiplication table. user has to input a letter then a number to get output which might be two numbers or a number with letter. i'm certain i'll be dealing with arrays and maybe pointers?
basically I want to make this into an app that way i can input D and 7 to get 6E as my output. i want the most simplified coding (not to be confused with optimised) dont want small, short coding but the most basics so as i can understand this better..

   0   1   2   3   4   5   6   7   8   9   10  11  12  13  14  15
A  6F  6D  65  20  6F  66  20  74  68  65  0A  66  61  63  69  6C 

B  69  74  69  65  73  20  61  76  61  69  6C  61  62  6C  65  20
                                  
C  6F  6E  20  79  6F  75  72  20  73  63  72  65  65  6E  2E  0A                                      
 
D  43  6C  69  63  68  20  6F  6E  20  74  68  65  20  66  6C  61                                      
 
E  73  68  69  6E  67  20  48  41  52  44  57  41  52  45  20  62                                     
  
F  75  74  74  6F  6E  2E  0A  00  46  69  72  73  74  20  74  68                                   
 
G  69  6E  67  73  20  66  69  72  73  74  20  2D  20  79  6F  75                                      
 
H  20  63  61  6E  20  6D  6F  76  65  20  74  68  69  73  20  74                                      
  
I  75  74  6F  72  69  61  6C  0A  6F  6E  20  74  68  65  20  73                                     
  
J  63  72  65  65  6E  20  62  79  20  6C  65  66  74  2D  63  6C                                     
 
K  69  63  6B  69  6E  67  20  6F  6E  20  74  68  65  20  74  69                                     
 
L  74  6C  65  20  62  61  72  2E  0A  52  69  67  68  74  20  63                                      
  
M  6C  69  63  6B  20  74  6F  20  70  75  74  20  69  74  20  64                                    
  
N  6F  77  6E  20  61  67  61  69  6E  2E  00  00  57  65  6C  63                                      
  
O  6F  6D  65  20  74  6F  20  74  68  65  20  55  70  6C  69  6E  
                                   
P  6B  20  54  75  74  6F  72  69  61  6C  2E  0A  54  68  69  73 
 
Q  20  70  72  6F  67  72  61  6D  20  73  68  6F  75  6C  64  20 
 
R  68  65  6C  70  20  79  6F  75  20  74  6F  20  67  65  74  20 
  
S  73  74  61  72  74  65  64  2E  0A  43  6C  69  63  6B  20  6E 
 
T  65  78  74  20  74  6F  20  63  6F  6E  74  69  6E  75  65  2E 

U  00  00  00  00  50  E5  48  00  80  F8  4C  00  A0  F8  4C  00 
    
V  B0  F8  4C  00  C0  F8  4C  00  D0  F8  4C  00  E0  F8  4C  00                                    
           
W  00  E6  48  00  30  E6  48  00  57  41  52  4E  49  4E  47  3A                            
         
X  20  55  70  6C  69  6E  6B  54  61  73  6B  3A  3A  53  65  74 
           
Y  54  61  72  67  65  74  20  63  61  6C  6C  65  64  20  28  62                                  
           
Z  61  73  65  20  63  6C  61  73  73  29  0A  00  E0  E6  48  C0     

[firedraco]

Sounds like all your need to do is multiply a number by the letter's "value" and then convert it to whatever base you need to output.

[diod]

Not sure that'll get me what i need. the above is a keycard for a game i own which i must refer to everytime i start a new game. I'm trying to figure out how to input a letter as the row and a number as the column to get the necessary "key".

[andywestken]

The simplest way I can think of would be to convert your data to an array (as you suspected)

const int data[26][16] = {
/*        0   1   2   3   4   5   6   7   8   9   10  11  12  13  14  15 */
/* A  */ {6F, 6D, 65, 20, 6F, 66, 20, 74, 68, 65, 0A, 66, 61, 63, 69, 6C}, 
/* B  */ {69, 74, 69, 65, 73, 20, 61, 76, 61, 69, 6C, 61, 62, 6C, 65, 20},
                                  
// SNIP                               
           
/* Z  */ {61, 73, 65, 20, 63, 6C,  61, 73, 73, 29, 0A, 00, E0, E6, 48  C0} };

Then, given a letter and a number, you can access the element using data[row][col], exploting the fact that (char) ch - 'a' = 0 - 25 for 'a' - 'z' (and similarly for the capital letters)

P.S. I was too lazy to add the necessary 0x- prefixes to all the numbers, so this won't compile "out of the box"...

[diod]

Here's my test code so far, and as you can tell its not good. but it gives me what i need. Now is there a way to optimize the "if()" functions to smaller code that initializes the input a - z
to 0 - 25?

#include <iostream>
using namespace std;

char *key_card[5][16] ={
{ "6F", "6D",  "65",  "20",  "6F",  "66",  "20",  "74",  "68",  "65",  "0A",  "66",  "61",  "63",  "69", "6C",},  
 {"69",  "74",  "69",  "65",  "73",  "20", "61",  "76",  "61",  "69",  "6C",  "61",  "62",  "6C",  "65",  "20",},
 {"6F",  "6E",  "20",  "79",  "6F",  "75",  "72",  "20",  "73",  "63",  "72",  "65",  "65",  "6E",  "2E",  "0A",},
 {"43",  "6C",  "69",  "63",  "68",  "20",  "6F",  "6E",  "20",  "74",  "68",  "65",  "20",  "66",  "6C",  "61",},
{"73",  "68",  "69",  "6E",  "67",  "20",  "48",  "41",  "52",  "44",  "57",  "41",  "52",  "45",  "20",  "62",} }; 
                                     
int main() {
	char let;
	int num;
	
	cout << "enter letter: " <<endl;
	cin >> let;
	if('a')
		let = 0;
	if('b')
		let = 1;
	if('c')
		let = 2;
	if('d')
		let = 3;
	if('e')
		let = 4;
	cout << "enter num: " <<endl;
	cin >> num;
	num = num -1;

	cout << key_card[let][num] << endl;

return 0;
}

[DSLq5Di1]

What andy was trying to point out is, that you can subtract 'a' (or 'A' in the case of capitals) from your input to get your row index.

	if('a')
		let = 0;
        ...

	cout << key_card[let-'a'][num] << endl;

It would be prudent to check your input is actually a letter with isalpha(), and perhaps use tolower() to avoid problems with capital letters. #include <cctype> for both these functions.

[diod]

Wow. That seems to have worked.. What can I read to understand why subtracting 'a' makes it work? Thanks guys, This forum is way cool. got answers in less than a day. and reaffirmed that even though I'm still crawling, i'm heading in the right direction..


Edit:nevermind. just did the math after viewing a ANSCII codes table.

[andywestken]

The ascii code for the character 'a' = 97, 'b' = 98, 'c' = 99, ...

So subtracting 'c' - 'a' is the same as 99 - 97 = 2.

http://en.wikipedia.org/wiki/Ascii