help with a printf statement

This is a discussion on help with a printf statement within the C Programming forums, part of the General Programming Boards category; Code: #include<stdio.h> #include<conio.h> void main() { char var; scanf("%c",&var); printf("%c\n%d\n%d ",var-'0',(int)var); getch(); } what does var-'0' means . Please help ...

  1. #1
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    Exclamation help with a printf statement

    Code:
    #include<stdio.h>
    #include<conio.h>
    
    void main()
    {
         char var;
       
         scanf("%c",&var);
         printf("%c\n%d\n%d ",var-'0',(int)var);
         getch();
       
    }
    what does var-'0' means . Please help .

  2. #2
    C++ Witch laserlight's Avatar
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    Assuming that var is in the range ['0', '9'] (i.e., digit characters), it gives you the corresponding value in the range [0, 9].
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
    Look up a C++ Reference and learn How To Ask Questions The Smart Way

  3. #3
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    The format string expects three parameters.

  4. #4
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    First, you have one too many arguments in the quotes. Remove the last %d.

    Now. If you compile this:

    Code:
    	char var = 0;
    	// scanf("%c",&var);
    	printf("%c\n%d\n%d", (var-'0') ,(int)var, '0');
    Output:
    \320
    0
    48

    You'll understand that '0' integer representation is 48.
    So, I set var to 0 for simplicity. It would be the character representation of 0 - 48. Which gives the character /320 (on this Mac).

    Also, I do not believe you needed to type cast var. But it's good practice I suppose.

  5. #5
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    Quote Originally Posted by Dr.Xperience View Post
    Code:
    #include<stdio.h>
    #include<conio.h>
    
    void main()
    {
         char var;
       
         scanf("%c",&var);
         printf("%c\n%d\n%d ",var-'0',(int)var);
         getch();
       
    }
    what does var-'0' means . Please help .
    You are calling the printf function incorrectly... It wants 3 parameters not 2...

    Try this...
    ]
    Code:
    #include<stdio.h>
    #include<conio.h>
    
    void main()
    {
         char var;
       
         scanf("%c",&var);
         printf("%c\n%d\n%d ",var,var-'0',(int)var);
         getch();
       
    }
    Maybe it will make more sense to you.

  6. #6
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    Thanks for reply and sorry for extra %d in printf statement.

    But tilll the extend I have got the idea here '0' represent zero and we are reducing var ie. character variable var with '0'. So the ASCII code for 0 is 48 and if i give input for var as q.
    So, ASCII code of q i.e.113 this means 113-48=65and we get ans as A which have ASCII code 65.
    So, indeed this just an arithmetic expression in which ASCII code of character stored in var is reduced by ASCII code of 0 which is 48.

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