[S4t3K]

Hi.

I simply used "format" to...format a string, but when compiling, I got these errors :

Код:

date_kilou.pwn(19) : error 012: invalid function call, not a valid address
date_kilou.pwn(19) : error 029: invalid expression, assumed zero
date_kilou.pwn(19) : warning 215: expression has no effect
date_kilou.pwn(19) : error 001: expected token: ";", but found "sizeof"
date_kilou.pwn(19) : fatal error 107: too many error messages on one line

My code is the following :

PHP код:


new string[500] = "";
    new cdate[3], time[3];
    gettime(time[0], time[1], time[2]);
    getdate(cdate[0], cdate[1], cdate[2]);
    for(new i = 0; i < strlen(format); i++)
    {
        
        switch(format[i])
        {
            /*************************
                    Jour
            *************************/
            
            case 'd': format(string, sizeof(string), "%s%02d", string, cdate[2]); 


(I try to make a date function, with almost the same features as the php version but in pawn).

I absolutely don't know the source of the errors, so if someone was able to help me, thanks to him in advance.

[Golf]

mybe

PHP код:

  case 'd': format(string, sizeof(string), "  '%s' '%02d'  ", string, cdate[2]); 


[S4t3K]

Already tried, already failed.
This isn't designed to be assigned to a SQL query, so I don't understand the meaning of the simple quotes here.
You'd give me an explaination ?

[NaClchemistryK]

Just adding simple " ' " s inside a string won't make any difference.
OT:
It's actually just a simple format. I don't understand the problem of the pawn compilers, they are especially designated to kill someone
And I don't understand why he is expecting a semi colon when you still need several more parameters inside a format.
Maybe try doing simple thing: like putting it in braces, or try simply formatting it with something simple. Like that we might find a way to a solution

[S4t3K]

I added the "format" into brackets (which gives me now

PHP код:

case 'd': // line 19
{ // line 20
      format(string, sizeof(string), "%s%02d", string, cdate[2]);  // line 21
} // line 22 


And now, the errors are more numerous :

Код:

date_kilou.pwn(21) : error 012: invalid function call, not a valid address
date_kilou.pwn(21) : warning 215: expression has no effect
date_kilou.pwn(21) : warning 215: expression has no effect
date_kilou.pwn(21) : warning 215: expression has no effect
date_kilou.pwn(21) : warning 215: expression has no effect
date_kilou.pwn(21) : warning 215: expression has no effect
date_kilou.pwn(21) : error 001: expected token: ";", but found ")"
date_kilou.pwn(21) : error 029: invalid expression, assumed zero
date_kilou.pwn(21) : fatal error 107: too many error messages on one line

So I tried adding a simple "format" just before, and I get exactly the same errors as above but on the line of the format I added (still line 21) :

PHP код:

case 'd': // line 19
            { // line 20
                format(string, sizeof(string), "%s%s", string, "123"); // line 21
                format(string, sizeof(string), "%s%02d", string, cdate[2]); // line 22
            } // line 23 


[Vince]

Doesn't it seem obvious that you can't use a function's name as a variable's name?

[S4t3K]

@Vince : I renamed "format" to "formatt" and it works absolutely fine.
Thanks.