strings of ASCII

This is a discussion on strings of ASCII within the C Programming forums, part of the General Programming Boards category; As a newbie I was trying to declare an ASCII string ending with a "0", but got a compiler error. ...

  1. #1
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    strings of ASCII

    As a newbie I was trying to declare an ASCII string ending with a "0", but got a compiler error.

    Code:
    constant char STRING[7] = {"string",0};
    What is wrong?

  2. #2
    Devil's Advocate SlyMaelstrom's Avatar
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    Because when you initialize a string, you need to do it one character at a time.

    Like this:
    Code:
    const char STRING[] = {'s', 't', 'r', 'i', 'n', 'g', '0'};
    or you can do this...
    Code:
    //Also, include <string.h>  
      char STRING[10];
      strcpy(STRING,"string");
      strcat(STRING,"0");
    Last edited by SlyMaelstrom; 10-30-2005 at 01:09 AM.
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  3. #3
    ATH0 quzah's Avatar
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    Quote Originally Posted by SlyMaelstrom
    Because when you initialize a string, you need to do it one character at a time.
    Code:
    const char astringy[] = "You do?";

    Quzah.
    Hope is the first step on the road to disappointment.

  4. #4
    Devil's Advocate SlyMaelstrom's Avatar
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    Ok you don't... but you could.
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  5. #5
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    Quote Originally Posted by Andy_P
    As a newbie I was trying to declare an ASCII string ending with a "0", but got a compiler error.

    Code:
    constant char STRING[7] = {"string",0};
    What is wrong?
    you don't have to explicitly put the terminating 0 there, the compiler will do that for you. You also don't need the braces, nor specify the string size unless you want the character array to be larger than the initialization string.

    Code:
    constant char STRING[] = "string";

  6. #6
    cwr
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    Quote Originally Posted by Ancient Dragon
    Code:
    constant char STRING[] = "string";
    And it's const, not constant.

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