unsigned long long division print

This is a discussion on unsigned long long division print within the C Programming forums, part of the General Programming Boards category; I have the following code: Code: ....... typedef unsigned long long CONTADOR; ....... for ( i = 0; i < ...

  1. #1
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    unsigned long long division print

    I have the following code:

    Code:
    .......
    typedef unsigned long long CONTADOR;
    .......
    	for ( i = 0; i < g_i_hwm ; ++i ) {
    		l_prof_time += g_tag[i].l_total;
    		l_time_tipos[g_tag[i].tipo] += g_tag[i].l_total;
    	}
    .....
    
    	for (int m = 0; m < 4; m++) {
    	  switch (m) {
    	    case P_A:
            fprintf ( salida , "A: % 12u (%4.1f%%)\n",l_time_tipos[m],(double) (l_time_tipos[m]*100/l_prof_time));
            break;
    	    case P_B:
            fprintf ( salida , "B: % 12u (%4.1f%%)\n",l_time_tipos[m],(double)(l_time_tipos[m]*100/l_prof_time));
            break;
    	    case P_C:
            fprintf ( salida , "C: % 12u (%4.1f%%)\n",l_time_tipos[m],(double)(l_time_tipos[m]*100/l_prof_time));
            break;
    	    case P_D:
            fprintf ( salida , "D: % 12u (%4.1f%%)\n",l_time_tipos[m],(double) (l_time_tipos[m]*100/l_prof_time));
            break;
          default:
            break;
    	  }
    	}
      fprintf ( salida , "Total: %12u\n", l_prof_time);
    And I get the following output:

    Code:
    A:     31011584 ( 0.0%)
    B:     38044808 ( 0.0%)
    C:    155035960 ( 0.0%)
    D:    337083256 ( 0.0%)
    Total:    561175608
    My code is not showing the percent. I have trying replacing "double" with "float" without success.
    How can I print % for unsigned long long?
    thanks in advance

  2. #2
    CSharpener vart's Avatar
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    l_time_tipos[m]*100/l_prof_tim

    this is integer division and as such result cannot contain decimal part

    if you want it - convert to double before deviding
    for example
    Code:
    l_time_tipos[m]*100.0/l_prof_time
    %u is for unsigned int
    it could be not suitable for unsigned long and definitely is wrong for unsigned long long - check your compiler specification to find out correct printf format
    Last edited by vart; 01-30-2009 at 06:15 AM.
    The first 90% of a project takes 90% of the time,
    the last 10% takes the other 90% of the time.

  3. #3
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    Quote Originally Posted by vart View Post
    l_time_tipos[m]*100/l_prof_tim

    this is integer division and as such result cannot contain decimal part

    if you want it - convert to double before deviding
    for example
    Code:
    l_time_tipos[m]*100.0/l_prof_time
    %u is for unsigned int
    it could be not suitable for unsigned long and definitely is wrong for unsigned long long - check your compiler specification to find out correct printf format
    This have not solved the problem. This is the output
    Code:
    A:     32582144 (-19668521817147667000000000000000000000000000000000
    00000000000000000000000000000000000000000000000000000000000000000000000000000000
    000000000000000000000000000000000000000000000000000000000000000000000.0%)
    B:     37161824 (-0.0%)
    C:    147305256 (-0.0%)
    D:    317159280 (46548150217338782000.0%)
    Total:    534208504
    About %u it is true not for this case, I will look for it, but currently, my problem is not in the %u printf but in the percent output.

  4. #4
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    If what you are printing with %u is a long long, then you will need to format it as "%I64u" or "%llu" depending on the OS (or maybe "%lu" if you have a 64-bit linux machine).

    --
    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.

  5. #5
    Registered User Codeplug's Avatar
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    More specifically, it depends on the CRT implementation.
    MinGW and VC 6.0 (which use the older MS-CRT) use "I64" prefix.
    Most everything else can use C99's "ll" prefix.

    gg

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