About Rot13-Implementation - c

Question

I am trying to implement the rot13-algorithm in C.
But since I am not very familiar with that language, I have some problems with my code right here.
Basically, I want to rotate every letter in args[] to 13 positions up.
But this code seems to be pretty sluggish:
#include <stdio.h>
char[] rotate(char c[]) {
char single;
int i;
int alen = sizeof(c)/sizeof(c[0]);
char out[alen];
for(i=0;i<=alen;i+=1) {
if(c[i]>='a' && (c[i]+13)<='z'){
out[i] = c[i]+13;
}
}
return out;
}
int main(int argc, char *argv[]) {
printf("The given args will be rotated\n");
int i;
char rotated[sizeof(argv)/sizeof(argv[0])];
rotated = rotate(argv);
/* printing rotated[] later on */
return 0;
}
I know there a lot of holes here - could you show me how to fix this?

Answer 1

Thanks a lot guys, I solved the problem with this code
#include <stdio.h>
int rot13(int c){
if('a' <= c && c <= 'z'){
return rot13b(c,'a');
} else if ('A' <= c && c <= 'Z') {
return rot13b(c, 'A');
} else {
return c;
}
}
int rot13b(int c, int basis){
c = (((c-basis)+13)%26)+basis;
return c;
}
int main() {
printf("The given args will be rotated");
int c;
while((c = getchar()) != EOF){
c = rot13(c);
putchar(c);
}
return 0;
}

Answer 2

How #Michael said this char out[alen] is not accepted by the compiler because you can't declare an array size with a non constant value. Another problem of your code is the for loop for( i = 0; i < = alen; i+=1 ) the arrays start on 0 so if you do the for until the lenght's position you will be out of the array.
About the code:
You must use a pointer to the start of the string as argument of the function, because You can't return arrays in C (But you can return pointers ).
Your if( str[i] >= 'a' && (str[i]+13) <='z') is incorrect because you will convert some letters into symbols take a look.
________
--------------------------!
void rotate( char * str )
{
int i = 0;
/* You do this until you find a '\0' */
for( i = 0; str[ i ] != '\0' ; i++ ){
/* Use the pointer notation if you passed a pointer. */
/* If the letter is between a and m you can simply sum it. */
if( *( str + i ) >= 'a' && *( str + i ) < 'n')
*( str + i ) += 13;
/* If the letter is between the n and z you have to do the opposite.*/
else if( *( str + i ) >= 'n' && *( str + i ) <= 'z')
*( str + i ) -= 13;
}
}

Answer 3

Size of arrays in C must be set at compile time, so you can't use non constant expression for array size.
Consider the below implementation:
// in place rotate
void rotate(char *str)
// str must be a zero-terminated string
{
int i =0;
// loop until str itself is not NULL and str[i] is not zero
for(i=0;str && str[i]; ++i) // ++i is a pre-increment
{
if(str[i] >= 'a' && (str[i]+13) <='z')
{
str[i] = str[i]+13; // modifying str in place
}
}
}
Then your main() can look like this:
int main(int argc, char *argv[])
{
printf("The given args will be rotated: %s\n", argv[1]);
rotate(argv[1]);
printf("Rotated: %s\n", argv[1]);
return 0;
}
Update More advanced version of the transform that takes care of case when str[i] + 13 > 'z'
for(i=0;str && str[i]; ++i) // ++i is a pre-increment
{
// ignore out of range chars
if (str[i] < 'a' || str[i] > 'z') continue;
// rotate
for (off = 13; off > ('z' - str[i]); )
{
off-= (1 + 'z' - str[i]);
str[i] = 'a';
}
str[i]+=off;
}

Answer 4

This function can encode/decode to/from rot13 string. It's compatible with VIM's g? rot13 encoder.
void rot13 (char *s) {
if (s == NULL)
return;
int i;
for (i = 0; s[i]; i++) {
if (s[i] >= 'a' && s[i] <= 'm') { s[i] += 13; continue; }
if (s[i] >= 'A' && s[i] <= 'M') { s[i] += 13; continue; }
if (s[i] >= 'n' && s[i] <= 'z') { s[i] -= 13; continue; }
if (s[i] >= 'N' && s[i] <= 'Z') { s[i] -= 13; continue; }
}
}