2 simple C program questions.

This is a discussion on 2 simple C program questions. within the C Programming forums, part of the General Programming Boards category; 1. How do you declare a variable s containing the string hello? My attempt: char s = "hello"; 2. Why ...

  1. #1
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    2 simple C program questions.

    1. How do you declare a variable s containing the string hello?

    My attempt:
    char s = "hello";

    2. Why is it that when the following code executes:

    int i = -1;
    int k = i++;

    k is -1, isn't k 0?

    Thank you!

  2. #2
    C++ Witch laserlight's Avatar
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    Both of these questions should be easy to answer after having read some introductory material on C, so re-read your book/notes on strings and on post-increment.
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    Okay I found out the first one is chars[] = "hello";

    But for the second question I cannot find it in my notes :S.

    I was thinking that i++ only works when it is on it's own, ex:

    int i = -1;
    i++;
    int k = i;

  4. #4
    C++ Witch laserlight's Avatar
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    Quote Originally Posted by justiliang
    Okay I found out the first one is chars[] = "hello";
    With the space you had before, yes

    Quote Originally Posted by justiliang
    But for the second question I cannot find it in my notes :S.

    I was thinking that i++ only works when it is on it's own, ex:
    If you cannot find "post-increment", then just try "increment", or look back to where the operator was introduced to you. You might also recall pre-incement, e.g., ++i. The crux of this question is how does i++ differ from ++i
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    OOOO, yeah I found increment.

    So i++ is a postfix position and it looks like

    int k = i++;

    is the same as

    int k = i;
    i = i + 1;

    ?

    Thanks for the help

  6. #6
    C++ Witch laserlight's Avatar
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    Quote Originally Posted by justiliang
    So i++ is a postfix position and it looks like

    int k = i++;

    is the same as

    int k = i;
    i = i + 1;

    ?
    For the purposes of this discussion, yes, you have the right idea.
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