Assign this to a variable

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  1. #1
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    Assign this to a variable

    Hi I'm new in C, from this url, how do I assign the value to a variable from this printf result.

    Code:
    printf("The substring is: %.*s\n", j - i, &ch[i]);
    Thanks!

  2. #2
    DESTINY BEN10's Avatar
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    Maybe like this:
    Code:
    int j;
    j=printf("The substring is: %.*s\n", j - i, &ch[i]);
    HOPE YOU UNDERSTAND.......

    By associating with wise people you will become wise yourself
    It's fine to celebrate success but it is more important to heed the lessons of failure
    We've got to put a lot of money into changing behavior


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  3. #3
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    printf() returns the number of bytes it prints, unless there is an error. I doubt that's what you want to assign.

    Looks like ch[i] is the base address of the string you want. Why not strcpy() it into some char array that you want it to go into? Lots of ways to do this.

  4. #4
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    Use sprintf() instead:
    Code:
    int i, j;
    char str[100];
    sprintf( str, "The substring is: %.*s\n", j - i, &ch[i] );
    Although, you're missing a "%d" for the j-i part.
    Last edited by cpjust; 11-25-2009 at 07:11 AM.
    "I am probably the laziest programmer on the planet, a fact with which anyone who has ever seen my code will agree." - esbo, 11/15/2008

    "the internet is a scary place to be thats why i dont use it much." - billet, 03/17/2010

  5. #5
    DESTINY BEN10's Avatar
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    Quote Originally Posted by cpjust View Post
    Use sprintf() instead:
    Code:
    int i, j;
    char str[100];
    sprintf( str, "The substring is: %.*s\n", j - i, &ch[i] );
    Although, you're missing a "%d" for the j-i part.
    The '*' will be replaced by the 'j-i' expression. For eg. j=5 and i=2, then the printf will be treated as %.3s.
    So, there's no need of the %d.
    Another eg.
    Code:
    float i=9.76,j=5,k=2;
    printf("%*.*f",j,k,i); // %5.2f
    HOPE YOU UNDERSTAND.......

    By associating with wise people you will become wise yourself
    It's fine to celebrate success but it is more important to heed the lessons of failure
    We've got to put a lot of money into changing behavior


    PC specifications- 512MB RAM, Windows XP sp3, 2.79 GHz pentium D.
    IDE- Microsoft Visual Studio 2008 Express Edition

  6. #6
    and the hat of sweating
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    Quote Originally Posted by BEN10 View Post
    The '*' will be replaced by the 'j-i' expression. For eg. j=5 and i=2, then the printf will be treated as %.3s.
    So, there's no need of the %d.
    Another eg.
    Code:
    float i=9.76,j=5,k=2;
    printf("%*.*f",j,k,i); // %5.2f
    Ah OK, good to know. I never use those fancy format specifiers...
    "I am probably the laziest programmer on the planet, a fact with which anyone who has ever seen my code will agree." - esbo, 11/15/2008

    "the internet is a scary place to be thats why i dont use it much." - billet, 03/17/2010

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