int wrong value being printed?

This is a discussion on int wrong value being printed? within the C Programming forums, part of the General Programming Boards category; Code: int test = 012452; printf("\n%d", test); why is the printed result: 5418 nb: it happens to ints that have ...

  1. #1
    Learner Axel's Avatar
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    Unhappy int wrong value being printed?

    Code:
    int test = 012452;
    printf("\n%d", test);
    why is the printed result:

    5418

    nb: it happens to ints that have a 0 as the first digit ??

  2. #2
    Its hard... But im here swgh's Avatar
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    Zero is false, so in some terms it is skipped, I use C++, so I could be wrong

  3. #3
    Reverse Engineer maxorator's Avatar
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    I think 0 in the front means the number is counted as octal.
    "The Internet treats censorship as damage and routes around it." - John Gilmore

  4. #4
    Learner Axel's Avatar
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    is the only way of getting around it is to make test a string? or can i keep as an int and somehow manipulate it?

  5. #5
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    Quote Originally Posted by Axel
    nb: it happens to ints that have a 0 as the first digit ??
    Yes, the 0 prefix denotes that the number is represented in the octal number system (See the first line on pg 193, K&R 2nd Edition).

    In printf, %d means decimal integer, so it's printing it in the decimal format.

    Try printing it in octal form using
    Code:
    printf("\n%o", test);

  6. #6
    Learner Axel's Avatar
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    hmm this time it omits the 0..
    Code:
    int test =  012345;
    printf("\n%o", test);
    12345

  7. #7
    Reverse Engineer maxorator's Avatar
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    See the first line on pg 193, K&R 2nd Edition
    WT*?
    "The Internet treats censorship as damage and routes around it." - John Gilmore

  8. #8
    Registered User
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    Quote Originally Posted by Axel
    hmm this time it omits the 0..
    Code:
    int test =  012345;
    printf("\n%o", test);
    12345
    It would have to be this (if I remmember correctly):
    Code:
    int test =  012345;
    printf("\n%06o", test);
    But I'm not sure if you want to limmit test to just having digits 0-7.
    Last edited by King Mir; 10-13-2006 at 09:21 AM.
    It is too clear and so it is hard to see.
    A dunce once searched for fire with a lighted lantern.
    Had he known what fire was,
    He could have cooked his rice much sooner.

  9. #9
    Registered User SKeane's Avatar
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    Or
    Code:
    int main()
    {
        int test =  012345;
        printf("\n%#o", test);
        return(0);
    }

  10. #10
    Frequently Quite Prolix dwks's Avatar
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    Or:
    Code:
    #include <stdio.h>
    
    int main(void) {
        const char *test = "0123456";
        puts(test);
        return 0;
    }
    If you don't know how many zeros precede the number then you're stuck with using a string representation like my sample above.
    dwk

    Seek and ye shall find. quaere et invenies.

    "Simplicity does not precede complexity, but follows it." -- Alan Perlis
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  11. #11
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    Axel: you know what octal is?

    King Mir: it is just an integer. If it's in octal, then its digits are [0-7]. If decimal, then [0-9]. If hex then [0-9a-f]. It's the same integer in either case. Digital representation has nothing to do with the values an integer can take on.
    Code:
    #include <stdio.h>
    
    void J(char*a){int f,i=0,c='1';for(;a[i]!='0';++i)if(i==81){
    puts(a);return;}for(;c<='9';++c){for(f=0;f<9;++f)if(a[i-i%27+i%9
    /3*3+f/3*9+f%3]==c||a[i%9+f*9]==c||a[i-i%9+f]==c)goto e;a[i]=c;J(a);a[i]
    ='0';e:;}}int main(int c,char**v){int t=0;if(c>1){for(;v[1][
    t];++t);if(t==81){J(v[1]);return 0;}}puts("sudoku [0-9]{81}");return 1;}

  12. #12
    The C eater *munch*
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    that's saying, hey, here, i have a number in octal and on the printf, please represent it in decimal
    Code:
    printf("%d", test);
    and by any chance, 5418 is decimal for 12452 in octal


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