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

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

Quote:

---

Originally Posted by vart

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.

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

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