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how to avoid to printing the value of charecter in C?

Discussion in 'C++' started by gopi, Feb 7, 2008.

  1. gopi

    gopi New Member

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    Hi Friends,
    I wrote a program in c to print a table for ant in teger as shown below

    /**----------------------------------------**/
    int a,b,c;
    printf("Enter the value of a\n");
    scanf("%d",&a);

    for(b=1; b<= 20; b++)
    {
    c=a*b;
    printf("%d X %d = %d\n",a, b,c );
    }
    /**-----------------------------------------**/

    This program is working fine,when i give input as a number.

    However,

    [COLOR=[B]DarkRed]in case,the given input is a charecter[/B],[/COLOR]
    Then it is showing some different output.
     
  2. shabbir

    shabbir Administrator Staff Member

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    Thats because the character when converted into ASCII and multiplied will be numbers not as per expectation and so its giving unexpected output.
     
  3. gopi

    gopi New Member

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    Hi Shabbir,
    Thankyou very much for your reply.
    Now what is the solution?
    how can i use the function "isalpha()"
    to avoid the above problem
     
  4. shabbir

    shabbir Administrator Staff Member

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    Code:
       char c;
       scanf( "%c", &c );
       if( isalpha(c) )
         printf( "You entered a letter of the alphabet\n" ); 
    
     
  5. logical_brain

    logical_brain New Member

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    Occupation:
    using my mind
    Location:
    India
    Check This code ( I only use cout and cin instead of scanf() and printf() )

    Code:
    #include <iostream>
    #include <cstdlib>
    #include <cctype>
    using namespace std;
    
    int valid_data(const char *data)
    {
        int i=0;
        //TO Check -ve Number
        if(data[0] == '-')
            i++;
        //To check if only '-' sign is their
        if(data[i] == '\0' )
            return 1;
        while(data[i] != '\0')
        {
            if(!isdigit(data[i]))
                return 1;
            i++;
        }
        return 0;
    }
    
    int main()
    {
    	string n;
    	cout<<"Enter Value of n : ";
    	cin>>n;
    	if(!valid_data(n.c_str()))
    	{
    	    int a = atoi(n.c_str());
            for(int i=1;i<=20;i++)
                cout<<a<<" X "<<i<<" = "<<a*i<<endl;
    	}
        else
            cout<<"Invalid Value";
    	return 0;
    }
    
    If you want a simpler code you read this about aoti() function
    You can check whether returned value is zero or not. If it is zero then you can display message. "Invalid Value". But here we cannot check if user has entered zero (0) as value. If user has entered zero (0) as value then also your program will display "Invalid Value". Hence i do not suggest this.


    Regards
     
  6. gopi

    gopi New Member

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    Thanks allot!!!!
     

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