No, nothing will be printed on the first loop only. If you wanted to be picky you could reduce the amount of loops required -
Both i and i_loops are set to 0, and after i_loops is incremented the first time, it's already larger than i, so nothing will print. i is never incremented, so when is i_loops ever smaller than i, allowing printing?
const int max =5;
for(int i=1; i< max;i++)
i_loops = 1;
while(i_loops <= i) std::cout << i_loops++ << std::endl;
if(i != 0) std::cout << std::endl;