I need to send the variable (double) into the function as a char*
tried this:Code:void changecolor(char* word, int color){ textcolor(color); //changes text to the color given through the argument cprintf("%s", word); //outputs the string in the given color textcolor(LIGHTGRAY); //change color back to the default light gray } //end void changecolor
monthly_interest is a double?? why? what did I go wrong? thanks!Code:char* mon; mon = (char)monthly_interest; changecolor(mon, color);
i dont know why you would want to cast a double to a char *, but heres how you do it...
(char *)&monthly_interest
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>> (char)monthly_interest;
Typecasting your double to a char will do no good. You need to convert it to a string first.
>> (char *)&monthly_interest
That will not do it either. In the function he passed it to, he's printing it as a string. So he'll need to convert the float to a string.
EDIT - if you're still having trouble, I suggest you look at the sprintf function
Last edited by Cshot; 10-05-2002 at 06:21 PM.
Try not.
Do or do not.
There is no try.
- Master Yoda
I tried and it s garbled if I use that way.Originally posted by *ClownPimp*
i dont know why you would want to cast a double to a char *, but heres how you do it...
(char *)&monthly_interest
Basically, I just need a way to input a variable, and send it to that function and have it output in the color I choose..
Thanks for response
void changecolor(char* word, int color){
textcolor(color); //changes text to the color given through the argument
cprintf("%s", word); //outputs the string in the given color
textcolor(LIGHTGRAY); //change color back to the default light gray
} //end void changecolor --------------------------------------------------------------------------------
tried this:
code:------------------------------------------------------------------------ char* mon;
mon = (char)monthly_interest;
changecolor(mon, color);
++++++++++++++++++++
I believe the person is using that function to output both text and variables if sent in as arguments, like if he went changecolor("Word", 1); or changecolor(variable, 1), it'd still do the same.
BTW how WOULD you do that?
>> BTW how WOULD you do that?
Actually the changecolor function's first parameter is char* which points to a string of characters. So you just convert all your variables to a string before passing it to the function.
Try not.
Do or do not.
There is no try.
- Master Yoda
How would you change a given datatype into a string before passing it?
Thats one I want to know! lol..Originally posted by Nakeerb
How would you change a given datatype into a string before passing it?
You shoud've read my suggestions then and tried them out. But since you can't seem to do that here's a sample:
Now that wasn't too hardCode:#include <stdio.h> void printString(char *blah) { printf("%s\n", blah); } int main() { double x = 6.925; char bleh[20]; sprintf(bleh, "%f", x); printString(bleh); return 0; }
Try not.
Do or do not.
There is no try.
- Master Yoda
Nm this post
Last edited by Nakeerb; 10-05-2002 at 07:34 PM.
I should say, it does work, but
something that that won't work on itCode:cout.setf(ios::showpoint); cout.setf(ios::fixed); cout.precision(2);
Actually i think it is impossible to do that because it is a string, you can't "round a string", lol.
Not sure what you're saying. But you can't mix those statements with printf. Use cout instead. You now know how to convert a variable to a string so I don't see what the problem is.
Try not.
Do or do not.
There is no try.
- Master Yoda
with cout you can use precision, for instance if you had a double n = 4.553535 and used the cout.precision(2), outputting that variable would make it 4.55, but you CANNOT do this with a string
There's no functions to set precision for a string. However you CAN still do it manually. Instead of printing the string, print character by character until:
1) you've reached the end of the string '\0'
2) or until you've reached a '.' then print 2 more characters after that if you want a precision of 2
Try not.
Do or do not.
There is no try.
- Master Yoda
