Define constant integer with if condition in c++

Newbie
1

To use a constant integer with the if condition I am using this logic in c++ code which compile with the g++ but not with root compiler.

#include <iostream>
using namespace std;

int main() {
     int wafer = 26;
    int nof1;

    if (wafer == 26)
    {
        nof1 = 25;
    }
    if (wafer == 10)
    {
        nof1 = 40;
    }
     const int nof = nof1;

     int array[nof] = {0};
     return 0;
}

In the root compilation I am getting this error, /home/sawan/FoCal_India/position_scan_pi/check.cxx:18:16: error: variable-sized object may not be initialized
int array[nof] = {0};
^~~

2

may be:

   int array[nof];
   for (int i=0; i<nof; i++) array[i] = 0;
3

Hi,

instead of

int array[nof] = {0};

do

int array[nof];
memset(array, 0, sizeof(array));
4

I have many arrays, histograms and graphs in to code whose size will be given by nof. So does this method works for histograms also? Like TH1F *histograms[nof];

5

You mean you have arrays of histograms and arrays of graphs? Why would you need to nullify the elements of such arrays?

6

You do not need to initialize the array to 0. Example:

void sawan() {
   int wafer = 26;
   int nof1;

   if (wafer == 26) nof1 = 25;
   if (wafer == 10) nof1 = 40;

   const int nof = nof1;

   int array[nof];
   memset(array, 0, sizeof(array));

   TH1F *histograms[nof];

   auto c = new TCanvas("c","c",1000,1000);
   c->Divide(5,5);
   for (int i=1; i<=nof; i++) {
      histograms[i-1] = new TH1F(Form("h%d",i),Form("h%d",i),100, -4,4);
      histograms[i-1]->FillRandom("gaus",20000);
      c->cd(i);
      histograms[i-1]->Draw();
   }
}
1 Like
7

What about std::array?

#include <array>

void f()
{
    constexpr int a = 20;
    constexpr int b = (a == 20 ? 30 : 40);

    std::array<int, b> ar;
    ar.fill(0);
}
1 Like
8

For arrays, I am only filling the array at certain position and rest of the position I need to fill it with zero. This is why I was initializing them with 0;