09.06.2018, 20:11
Quote:
|
Consider this I have not tested it but pretty sure I got it right the rest is up to you to test and make sure it works right.
Код:
// Replacement rules (these can have any number of characters)
static LetterA[3] = { 'g', 'h', 'd' };
static LetterE[3] = { 'g', 'h', 'd' };
static LetterI[3] = { 'g', 'h', 'd' };
static LetterO[3] = { 'g', 'h', 'd' };
// Only has one
static LetterU = 'g';
BabbleText(string[])
{
new output[144], len, currpos;
len = strlen(string);
// Loop through string
for(new i = 0; i < len; i++)
{
// Check for letter rules
switch(string[i])
{
// Found a letter that has an addition
case 'a', 'e', 'i', 'o', 'u':
{
// Letter needs to be added no matter what
output[currpos] = string[i];
// Chance of letter actually being added
if(random(2) == 1)
{
// Only need to increment position if we actually need to replace
currpos++;
switch(string[i])
{
case 'a': { output[currpos] = LetterA[random(strlen(LetterA))]; }
case 'e': { output[currpos] = LetterE[random(strlen(LetterE))]; }
case 'i': { output[currpos] = LetterI[random(strlen(LetterI))]; }
case 'o': { output[currpos] = LetterO[random(strlen(LetterO))]; }
// Only has one no need for random
case 'u': { output[currpos] = LetterU; }
}
}
}
// No rule on letter
default: { output[currpos] = string[i]; }
}
// Always increment current position
currpos++;
}
return output;
}
|
Please assist me some more, thanks so far.


