Printing a variable's name.

This is a discussion on Printing a variable's name. within the C Programming forums, part of the General Programming Boards category; Hello everyone. I'm new to C and I'm having trouble printing a variable's name. Such as if I have int ...

  1. #1
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    Printing a variable's name.

    Hello everyone. I'm new to C and I'm having trouble printing a variable's name.
    Such as if I have int a =2, is there a way to print "a".

    I would like my print statement to look like:

    printf("variable's name %d \n",a);

    and "a 2" to be printed to the screen.

    I've been trying to find an answer on google but haven't had any luck.

  2. #2
    Woof, woof! zacs7's Avatar
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    > Such as if I have int a =2, is there a way to print "a".
    No.

  3. #3
    Registered User NeonBlack's Avatar
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    We only have variable names so humans don't get confused (like it really helps).
    After your program gets turned into assembly and eventually machine code, the computer doesn't give a damn what you name your variables.
    I copied it from the last program in which I passed a parameter, which would have been pre-1989 I guess. - esbo

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    Unhappy

    Quote Originally Posted by NeonBlack View Post
    We only have variable names so humans don't get confused (like it really helps).
    After your program gets turned into assembly and eventually machine code, the computer doesn't give a damn what you name your variables.
    That makes sense. Too bad it doesn't help me. Thanks anyway.

  5. #5
    Jack of many languages Dino's Avatar
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    Code:
    #include <stdio.h>
    
    #define print_int(x) printf("Variable " #x "= &#37;d\n", x)  
    #define print_char(x) printf("Variable " #x "= %c\n", x)  
    #define print_string(x) printf("Variable " #x "= %s\n", x)  
    
    int main (int argc, const char * argv[]) {
    	int a = 2 ; 
    	char letter = 'Z' ; 
    	char phrase[] = "Look, I can do this!!" ; 
    	
    	print_int(a) ;
    	print_char(letter) ; 
    	print_string(phrase) ; 
    	return 0;
    }
    Output is
    Code:
    [Session started at 2008-03-11 23:01:17 -0500.]
    Variable a= 2
    Variable letter= Z
    Variable phrase= Look, I can do this!!
    Todd
    Mac and Windows cross platform programmer. Ruby lover.

    Quote of the Day
    12/20: Mario F.:I never was, am not, and never will be, one to shut up in the face of something I think is fundamentally wrong.

    Amen brother!

  6. #6
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    Code:
    #include<stdio.h>
    #define display(n) printf(#n);
    int main()
    {
        int n;
        display(n);
        return 0;
    }
    The preprocessor directive # when used with the variable in the macro outputs the name of the variable that was used.
    Code:
    >+++++++++[<++++++++>-]<.>+++++++[<++++>-]<+.+++++++..+++.[-]>++++++++[<++++>-] <.>+++++++++++[<++++++++>-]<-.--------.+++.------.--------.[-]>++++++++[<++++>- ]<+.[-]++++++++++.

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    Awesome, I really appreciate the help. Thanks.

  8. #8
    Captain Crash brewbuck's Avatar
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    Ah yes, the good old "stringizing" operator. Also useful is the token pasting operator '##'. Don't ask what I've used it for, it's wrong. Very, very wrong.

  9. #9
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    One more question if you don't mind.

    I'm using the #define print_int(x) printf("Variable " #x "= %d\n", x) method; but mine looks like this:
    #define print_int(x) printf(#x "\t%d\t %d %% \n", x-1, ((x-1)*100)/total)

    Is there a way to right justify the #x? I've tried putting %5c in quotes in front of it with a space as the char and a bunch of other stuff like this but no luck.

  10. #10
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    Code:
    #define print_int(x) printf("%5s\t%d\t %d %% \n", #x, x-1, ((x-1)*100)/total)
    --
    Mats
    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

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