output nummero without zero .

left justified without zero !

input > 200.123
output > with zero -0000400.246
without zero > -400.246


gruss

Code:


include "cpc6128.ccz80";

array byte text[12];
array byte a[5];
array byte b[5];
byte n,i,z,d;

input(text);
atof(a,text);

mul(a,atof(b,"-2.0"));

ftoa(text,a);
prints("\r\n");
prints(text);

prints("\r\n");

d=0;
if (*(text)=='-')
{
   prints("-");
   d=1;
}

for (i =d; i <= 6; ++i)
{
 n=*(text+i);
 if (n>48) goto weiter;
}
weiter:
for (z =i; z <= 11; ++z)
{
 n=*(text+z);
 printc(n);
}

return;


Well work, but you can write this program so:

Code:

include "cpc6128.ccz80";

string text[12];
array byte a[5], b[5];

input(text);
atof(a,text);

mul(a,atof(b,"-2.0"));

ftoa(text,a);
prints("\r\n");
prints(text);
prints("\r\n");

prints(strltrm(text + (*text == '-'), '0'));
return;

Remarks:
- All relational expressions (with comparision operators, return 1 if true or 0 if false). We can use this for get 0 or 1 easily.
- The function strltrm remove, from the string as 1st parameter, the character specified as 2nd parameter until another character is found. Example: the expression strltrm("aaaaabbccaaa", 'a') remove the first 5 letters and returns the address from string "bbccaaa".
- For any expression that use a string, if we specify the string name, we use the string from begin, if we specify string + n, we use the string that starts in address of string + n characters.

This is as an hieroglyphic game for know as work an instruction .

mul(a,atof(b,"-2.0"));

input "123.123" not result print "-246.246" the print "246,246", it is not ok.


gruss

Sorry , lack this instruction before prints(strltrm ...):

if (*text == '-') printc('-');

It's much like your idea, which is the same as strltmb ago.

whery got.
thanks.

why can in the asm-routine ?

Code:


if (*text == '-') printc('-');
prints(strltrm(text + (*text == '-'), '0'));