String Character Changing - Printable Version

String Character Changing - Printable Version

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String Character Changing - Aerotactics - 03.06.2014

Alright, I'm having an issue with string character replacement:

Код:

if(!strcmp(input[i],"A") && input[i] != '\0'){input[i] = 'C'; continue;}

Picture 26 of these lines total, and they're in a for loop. Each line represents a letter of the alphabet, and a letter to replace it with.

Issue: When I enter a string such as "STRING", only the last letter is changed, instead of all of the characters.

Re: String Character Changing - Vince - 03.06.2014

When you use input[i] it returns the entire string from that point onwards. For example:

pawn Код:

new input[6] = "hello";
printf("%s", input[1]);
//prints: ello

The proper way to compare a single character would be:

pawn Код:

if(input[i] == 'A')

Note: single quotes.
But you may want to look at an implementation of rot13 or similar; 26 similar lines does not seem efficient at all.

Re: String Character Changing - Aerotactics - 03.06.2014

Quote:

Originally Posted by Vince

When you use input[i] it returns the entire string from that point onwards. For example:

pawn Код:

new input[6] = "hello";
printf("%s", input[1]);
//prints: ello

The proper way to compare a single character would be:

pawn Код:

if(input[i] == 'A')

Note: single quotes.
But you may want to look at an implementation of rot13 or similar; 26 similar lines does not seem efficient at all.

While it might not be efficient, It's easy to read (but not easy on the eyes). Thanks for that correction, and I did some research on rot13. The difference is, you could change the character set in those 26 lines, with rot13, it's all the same key (it seems, from a short research period).

EDIT: I wrote this script as a sort of proof of concept to myself and to a friend, I might turn this into a stock to use for a wider range of jumbling