casting int to char

i've been looking for this for a while, but unfortunatelly not too many descriptive insights.....

well,.... correct me if i am wrong ......

i need to cast from integer to character...

i did............

Code:

int Selection;
char Choice;
Choice = (char) Selection;

considering it is possible, so far so good... at least the compiler isn't complaining...

the trick is, after converting that "int" to "char" i need to check whether that converted "Choice" is a numeric character......

so i write:

Code:

#include <ctype.h>

if(isdigit(Choice))
    cout << "is digit";

.......all good
after doing all of the above the compiler is all happy however the output is nowhere near... nothing works properly.....
maybe i am doing this wrong ....



any help would be appreciated
Regards,
mathe917

i got this to compile, but i don't know if its what you want.

Code:

#include <iostream.h>
#include <iomanip.h>

#define tab '\t'

main()
{
	int a=113;
	char b;
	b=static_cast<char>(a);
	cout<<b;
	return(0);
}

thanx for input.....
however the casting is done fine...yours and mine.......

the casting isn't the problem

seams like the comparison isn't doing the job


Regards,
matheo917

You're passing a char to isdigit. The param for isdigit is int.

From MSDN:

//
int isdigit( int c );

Parameter

c

Integer to test

//

i guess i don't understand what you mean by "numeric character" or what your trying to do with the comparison. i'd like to help. . .

Does selection have an integer value such as 5, or a character value '5'? If it's the former then the test will fail unless you force a character conversion, not simply casting. This works peachy.

Code:

#include <iostream>
#include <cctype>

int main ( void )
{
  int selection = 5;
  char choice = selection + '0';
  if ( isdigit ( choice ) )
    std::cout<< choice <<"\nThis works\n";
  return 0;
}

-Prelude

think of a simple do ... while loop ...
which keeps looping around if the input is a number smaller than 0 or grater than 5, as well as if you would enter a letter "w" in place of an integer the program would crash, kind of loop forever, the reason why i needed that isdigit is to determine if the input is a letter "w" or a simple so called number value "4"......

maybe i was using a wrong function???

what you think??

Regards,
matheo917

i played around with it from a different direction and still nothing...


here's the code of this perticular loop:

Code:

int Selection;
  char Choice[2];
  do
   {
       cout << "1.  HELP MENU \n\n";
       cout << "2.  INVENTORY DIRECTORY \n\n";
       cout << "3.  DELIVERY INFORMATION \n\n";
       cout << "4.  ORDERING DIRECTORY \n\n";
       cout << "5.  WAITING LIST \n\n";
       cout << "0.  EXIT \n\n\n\n";

        cout << "Enter a number (#) corresponding to your selection - ";

        cin >> Selection;

        itoa(Selection, Choice, 2);

        clrscr();                               // clears the screen...

   }while((isalpha(Choice[0])) || (Selection < 0 || Selection > 5));

.......
for some reasong i can't get it to work.... maybe i am using a wrong algorithm in a while loop delimeter...... i am sure there's gotta be a way, i am sure people have run into this problem before when asked for int someone entered a char and the loop crashed...??

?? any ideas?

Regards,
matheo917

anyone care to give another perhaps a different solution??

hmmm..... no answers...

anyone please....

thank you..

Regards,
matheo917

Is there any reason you can't do it like this?

Code:

  char Choice;
  do
   {
       cout << "1.  HELP MENU \n\n";
       cout << "2.  INVENTORY DIRECTORY \n\n";
       cout << "3.  DELIVERY INFORMATION \n\n";
       cout << "4.  ORDERING DIRECTORY \n\n";
       cout << "5.  WAITING LIST \n\n";
       cout << "0.  EXIT \n\n\n\n";

        cout << "Enter a number (#) corresponding to your selection - ";

        cin >> Choice;

        clrscr();                               // clears the screen...

   }while(Choice < '0' || Choice > '5');

in this particular case the compiler will not detect whether a character is less than the other based on the less than or greater than principal.......at least that's what i think and compiler too i guess....

i am working on something similar now, although i am making my loop to be infinite ....... here it is:

Code:

do
   {
       cout << "1.  HELP MENU \n\n";
       cout << "2.  INVENTORY DIRECTORY \n\n";
       cout << "3.  DELIVERY INFORMATION \n\n";
       cout << "4.  ORDERING DIRECTORY \n\n";
       cout << "5.  WAITING LIST \n\n";
       cout << "0.  EXIT \n\n\n\n";

        cout << "Enter a number (#) corresponding to your selection: ";

        cin >> Selection;
        Temp2 = Selection;
        Temp1 = (int)Temp2;
        Choice = Temp1;

        clrscr();                               // clears the screen...

   }while ((Selection < '0' || Selection > '5') || (Choice < 0 || Choice > 5));

i am thinking of a ASCII representation of these number when it comes to characters i.e. char(32) maybe that will work......

what do you think?

Regards,
matheo917

got it to work, however it's not too roboust and too innovative, but i kind of got tired of trying to do it the hard way and so here it goes......

Code:

string Selection;

  do
   {
       cout << "1.  HELP MENU \n\n";
       cout << "2.  INVENTORY DIRECTORY \n\n";
       cout << "3.  DELIVERY INFORMATION \n\n";
       cout << "4.  ORDERING DIRECTORY \n\n";
       cout << "5.  WAITING LIST \n\n";
       cout << "0.  EXIT \n\n\n\n";

        cout << "Enter a number (#) corresponding to your selection: ";

        cin >> Selection;

    clrscr();

   }while ((Selection != "0" && Selection != "1" &&
            Selection != "2" && Selection != "3" &&
            Selection != "4" && Selection != "5"));

thanks for the effort...


matheo917

in this particular case the compiler will not detect whether a character is less than the other based on the less than or greater than principal.......at least that's what i think and compiler too i guess....

It does on my computer (Windows NT). But I guess if you're not comfortable with it, go with what feels right.