CODEWITHSUNDEEP
c
Liam Frutchey
Liam Frutchey

Reputation: 11

C Printf won't print certain 0s

A program I am working on for a school project ignores 0s when printing. For example, this function would not print the first arm (Arm 0) so I added an extra printf before the for loop but whenever my armArray[i].count is equal to 0 it ignores it like this:

Total Payoff: 75.67
Arm   Count   Avg. Payoff   Value
  0       3         -0.02    0.59
  1      74          0.94    0.86
  2      19          0.25    0.18
  3       4          0.33    0.89
  4                  0.00    0.25

void PrintStatistics(const BanditArm armArray[], int numArms) {
    double tPayoff = 0.0;
    for(int i =0; i<numArms;i++){
        tPayoff+=armArray[i].totalPayoff;
    }
    printf("Total Payoff:%6.2f\n",tPayoff);
    printf("Arm   Count   Avg. Payoff   Value\n");
    printf("  0%8.0d%14.2f%8.2lf\n",armArray[0].count,armArray[0].avgPayoff,armArray[0].payoffMean);
    for(int i =1; i<numArms;i++){
        if(armArray[i].count==0)
        printf("%3d%8.0d%14.2f%8.2lf\n",i,0,armArray[i].avgPayoff,armArray[i].payoffMean);
        else
        printf("%3d%8.0d%14.2f%8.2lf\n",i,armArray[i].count,armArray[i].avgPayoff,armArray[i].payoffMean);
    }
}

If more of the code is needed I can add the rest.

Upvotes: 1

Views: 75

Answers (1)

kaylum
kaylum

Reputation: 14046

printf("%3d%8.0d%14.2f%8.2lf\n",i,0,armArray[i].avgPayoff,armArray[i].payoffMean);

The second format specifier %8.0d provides a precision of 0. For %d the precision gives the minimum number of digits to appear. Since you gave 0 precision it means nothing will be printed for 0 values. There is no need for the precision in this case. So either remove it completely or give a precision of 1.

That is, %8d or %8.1d.

As a side note, you don't really need to have that last if/else condition. The printf in the else case is sufficient for both the zero and non-zero cases.

Upvotes: 2

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