Converting string array characters to numbers - c

Question

I am new to C. I am experienced in GWBASIC. In an effort to learn, I am attempting to write a program that will convert the individual chars in a string to a numerical value as so:
1 2 3 4 5 6 7 8 9
a b c d e f g h i
j k l m n o p q r
s t u v w x y z
For example, user input for string A could be 'dog',
said program would then store [d][o][g] as [4][6][7] in string B.
The below code works for a string w/up to four chars, but there must be a more efficient way of doing this.
int main()
{
char a[0];
char b[0];
scanf("%s",a);
if (a[0] == 'a' || a[0] == 'j' || a[0] == 's') b[0] = '1';
if (a[0] == 'b' || a[0] == 'k' || a[0] == 't') b[0] = '2';
if (a[0] == 'c' || a[0] == 'l' || a[0] == 'u') b[0] = '3';
if (a[0] == 'd' || a[0] == 'm' || a[0] == 'v') b[0] = '4';
if (a[0] == 'e' || a[0] == 'n' || a[0] == 'w') b[0] = '5';
if (a[0] == 'f' || a[0] == 'o' || a[0] == 'x') b[0] = '6';
if (a[0] == 'g' || a[0] == 'p' || a[0] == 'y') b[0] = '7';
if (a[0] == 'h' || a[0] == 'q' || a[0] == 'z') b[0] = '8';
if (a[0] == 'i' || a[0] == 'r') b[0] = '9';
if (a[1] == 'a' || a[1] == 'j' || a[1] == 's') b[1] = '1';
if (a[1] == 'b' || a[1] == 'k' || a[1] == 't') b[1] = '2';
if (a[1] == 'c' || a[1] == 'l' || a[1] == 'u') b[1] = '3';
if (a[1] == 'd' || a[1] == 'm' || a[1] == 'v') b[1] = '4';
if (a[1] == 'e' || a[1] == 'n' || a[1] == 'w') b[1] = '5';
if (a[1] == 'f' || a[1] == 'o' || a[1] == 'x') b[1] = '6';
if (a[1] == 'g' || a[1] == 'p' || a[1] == 'y') b[1] = '7';
if (a[1] == 'h' || a[1] == 'q' || a[1] == 'z') b[1] = '8';
if (a[1] == 'i' || a[1] == 'r') b[1] = '9';
if (a[2] == 'a' || a[2] == 'j' || a[2] == 's') b[2] = '1';
if (a[2] == 'b' || a[2] == 'k' || a[2] == 't') b[2] = '2';
if (a[2] == 'c' || a[2] == 'l' || a[2] == 'u') b[2] = '3';
if (a[2] == 'd' || a[2] == 'm' || a[2] == 'v') b[2] = '4';
if (a[2] == 'e' || a[2] == 'n' || a[2] == 'w') b[2] = '5';
if (a[2] == 'f' || a[2] == 'o' || a[2] == 'x') b[2] = '6';
if (a[2] == 'g' || a[2] == 'p' || a[2] == 'y') b[2] = '7';
if (a[2] == 'h' || a[2] == 'q' || a[2] == 'z') b[2] = '8';
if (a[2] == 'i' || a[2] == 'r') b[2] = '9';
if (a[3] == 'a' || a[3] == 'j' || a[3] == 's') b[3] = '1';
if (a[3] == 'b' || a[3] == 'k' || a[3] == 't') b[3] = '2';
if (a[3] == 'c' || a[3] == 'l' || a[3] == 'u') b[3] = '3';
if (a[3] == 'd' || a[3] == 'm' || a[3] == 'v') b[3] = '4';
if (a[3] == 'e' || a[3] == 'n' || a[3] == 'w') b[3] = '5';
if (a[3] == 'f' || a[3] == 'o' || a[3] == 'x') b[3] = '6';
if (a[3] == 'g' || a[3] == 'p' || a[3] == 'y') b[3] = '7';
if (a[3] == 'h' || a[3] == 'q' || a[3] == 'z') b[3] = '8';
if (a[3] == 'i' || a[3] == 'r') b[3] = '9';
printf("%s\n",b);
return 0;
}

Answer 1

Assuming that your compiler uses an ASCII encoding then you can use the following simple arithmetic to get your answer:
1 + (strA[i] - 'a') % 9
You really don't want to implement this with a long list of if statements or indeed a switch statement.
Naturally you will have input validation issues if you have non alphabetical characters, numeric characters, upper-case characters and so on. I presume you can simply ignore those for a learning exercise.

Answer 2

To correct your original approach, you need to do two things:
Use single quotes around character constants;
Use == to check for equality;
Terminate statements with ;.
... so your snippet becomes:
if (strA[0] == 'a')
strB[0] = '1';
if (strA[0] == 'b')
strB[0] = '2';
if (strA[0] == 'c')
strB[0] = '3';

Answer 3

You can type the following exactly into a gwbasic editor and it will solve your problem
10 INPUT A$
12 L = LEN(A$)
15 FOR T = 1 TO L
20 M$ = MID$(A$,T,1)
25 GOSUB 70
30 B$ = B$ + V$
35 NEXT T
40 PRINT B$
50 END
55 REM -----------------
70 REM - Subroutine to convert m$ into v$
72 X = ASC(M$) : REM this is the ascii value of m$ (eg. "a" = 97)
74 X = X - 96 : REM so that 97 becomes "1"
80 IF X > 9 THEN X = X - 9 : GOTO 80
90 V$ = STR$(X) : REM just converting to a string type variable
95 RETURN : REM takes you back to line 30 where this value is added to the
96 REM final resulting B$ - when I say added I mean a char added to a string
97 REM such that "APPL" + "E" = "APPLE"
98 REM ------------------------------------------ DONE

Answer 4

For ASCII, it would go somewhat like:
... make sure strB has enough space ...
for (i = 0; i < ... length of strA ... ; i++) {
/* you should always check your input is valid */
if (strA[i] >= 'a' && strB[i] <= 'z')
strB[i] = (strA[i] - 'a') % 9 + 1;
else
strB[i] = ... some default value ...
}
For EBCDIC:
for (i = 0; i < ... length of strA ... ; i++) {
/* you should always check your input is valid */
if (strA[i] >= 'a' && strB[i] <= 'r')
strB[i] = strA[i] & 0xF;
else if (strA[i] >= 's' && strB <= 'z')
strB[i] = (strA[i] & 0xF) - 1;
else
strB[i] = ... some default value ...
}

Answer 5

Have a closer look at an ASCII table. You'll see that all letters are encoded with some integer value. As a start, if you only have lower case letters, it's enough to substract the character code of 'a' from any letter to get what you want
int nr = strA[0] - 'a' + 1;
//now you'd need to convert back to a string; better: strB should be an array of integer
Also, = is the assignment operator; you need to use == to check for equality.

Answer 6

try this :)
int strB[MAX_LEN] = {0};
char *strA = malloc (MAX_LEN * sizeof(char));
int i,c = 0,x;
scanf("%s",strA);
for(i = 0 ; i<strlen(strA) ; i++){
x = strA[i] - 'a' + 1;
if(x >= 1 && x <= 9)
strB[c] = x;
else if(x <= 18){
strB[c] = x - 10;
else if(x <= 26){
strB[c] = x - 19;
if(x <= 26)
c++;
}
or you can use ninjalj approach in the for loop, if you aleady checked the input:
for(i=0 ; i<strlen(strA) ; i++){
strB[i] = (strA[i] - 'a') % 9 + 1;
}
or this :
for(i=0 ; i<strlen(strA) ; i++){
if(strA[i] >= 'a' && strA[i] <= 'z'){
strB[c] = (strA[i] - 'a') % 9 + 1;
c++;
}
}