Re: Can't understand how to get needed pointer.

[Tony Clarke <sam@palamon.ie>]

Try changing the lines

char inbuf[MAX_SIZE];
char *inptr = inbuf;

to

const char inbuf[MAX_SIZE];
const char * inptr = inbuf;

The warning should go away then, i think...

Tony

>Hello.
>I've wrote function that convert in_str from charset to UTF-8.
>It uses libiconv.
>int convert_str(char *in_str,const char *charset,char *out_str)
>{
>        char *wrptr = (char *) out_str;
>        char inbuf[MAX_SIZE];
>        char *inptr = inbuf;
>        size_t insize = strlen(in_str)+1;
>        size_t avail = MAX_SIZE;
>        size_t nconv;
>
>        iconv_t cd;
>        sprintf (inbuf,"%s",in_str);
>        cd=iconv_open ("UTF-8",charset);
>        nconv = iconv (cd, &inptr, &insize, &wrptr, &avail);
>        iconv (cd, NULL, NULL, &wrptr, &avail);
>        iconv_close(cd);
>        return 0;
>}
>But I'v not understood how to get "const char* * inbuf" without
>manipulations that I've done in code above. This code works fine, but
>when I was compiling it gcc prints warning:
>       gd_ft_test.c: In function `convert_str':
>       gd_ft_test.c:83: warning: passing arg 2 of `iconv' from incompatible pointer type
>I don't understand what I have to do with "char *in_str" to pass it to
>iconv?
>Here is declaration of function iconv:
>size_t iconv (iconv_t cd,
>                     const char* * inbuf, size_t * inbytesleft,
>                     char* * outbuf, size_t * outbytesleft);
>
>Thanks beforehead.
>_________________________________________________________________________
>Sorry for my English.
>
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