Jean-Paul a écrit :
> Le Mardi 8 Juillet 2008 11:41, rufus shadowolf a écrit :
>
>> for(i = 0; i>3; i++)
>> {
>> lcdOut(str[1]);
>> }
>> show me 3 strange identical char whitch change when i turn off the pic
>> alimentation.
>>
>
> With lcdOut(str[1]) as you wrote it, you will get three times the same
> character.
>
> You want to try lcdOut(str[i])
>
>
>> i try to change the stack size/position without any changement.
>>
>>
>> i'm sorry for my poor english,
>>
> (Tu peux être désolé ;-)
>
> Jean-Paul
>
>
hum hum, of curse i don't paste what i mean to.
yes, as you say:
for(i = 0; i>3; i++)
{
lcdOut(str[i]);//with lcdOut(str[1]); it work fine
}
after more and more test like it:
int main (void)
{
char i;
char* chaine = "salut";
for (i = 0; i < 3; i++)
{
DATA = chaine[3];
enable=1;
delay10tcy(90);
enable=0;
delay10tcy(90);
}
for (i = 0; i < 3; i++)
{
DATA = chaine[i];
enable=1;
delay10tcy(90);
enable=0;
delay10tcy(90);
}
for (i = 0; i < 4; i++)
{
DATA = chaine[3];
enable=1;
delay10tcy(90);
enable=0;
delay10tcy(90);
}
while(1);
}
i really don't understand, it show: "uuu???uu" (why two "u" at the end and not
4? i don't know) and as i say "??? " are tree time the same char and change on
"reboot".
i think, my routinework, cause whit a "static" var i see what i send, but when
i use arrays...
thanks a lot for time you spend for me.
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