Looking for solution to array and cout interaction in C++ - c++

Question

I want a function or loop to run through the array and print each element out until it has printed 10 elements. In which case, a new line is started and the printing continues. eg. 1 2 3 4 5
6 7 8 9 10
This is for a program that works an array like a 50's housewife performing many calculations and alterations to said array.
This is my current attack on the logic behind my issue.
int main()
{
test = new int[100];
for (int i = 0; i < 100; i++){
//Set array to rand
test[i] = rand() % 44 + 55;
}
printf("original List\n");
for (int i = 0; i < 100; i++){
// print original order
printf("%d\n", test[i]);
}
sortArr;
// function call sort array ascend
printf("\Sorted List\n");
for (int i = 0;i < 100;i++) {
// print sorted order
printf("%d , ", test[i]);
int temp;
//temp counter for width of printout
for (temp = 0;temp < 10;temp++) cout<< "\n" << endl;
sum += test[i];
}
Expected is a block of output consisting of 100 array elements in a grid with a width of 10 elements per line.
Actual result is a mess of new line loops and further headache to me.

Answer 1

Pretty common issue just use a modulus based on the index i:
for (int i = 0;i < 100;i++) {
printf("%d , ", test[i]);
if ((i + 1) % 10 == 0) {
printf("\n");
}
}
If you want nicely formatted output however you'll need:
#include <iomanip>
and
std::cout << std::setw(/*max_len*/) << test[i];

Answer 2

The simplest solution would be to print the delimiter (i%10 == 0) ? "\n" : ", ". You correctly recognized that taking the remainder on every iteration of the loop is inefficient, and wanted to write one that would print ten elements followed by a newline.
The trick there is to write an inner loop that increments a second counter, j, do all your output within the inner loop, and then update i at the bottom of the outer loop. A simplified example:
#include <array>
#include <iostream>
#include <stdlib.h>
#include <time.h>
using std::cout;
int main()
{
constexpr size_t ARRAY_LEN = 100;
std::array<int, ARRAY_LEN> test;
{
// Quick and dirty initialization of the random seed to the lowest 30
// or so bits of the system clock, which probably does not really have
// nanosecond precision. It’ll do for this purpose.
timespec current_time;
timespec_get( &current_time, TIME_UTC );
srand(current_time.tv_nsec);
}
for (int i = 0; i < test.size(); i++){
//Set array to rand
test[i] = rand() % 44 + 55;
}
for ( int i = 0, j = 0;
i < test.size();
i += j ) {
for ( j = 0; j < 10 && i + j < test.size(); ++j ) {
cout << test[i + j] << ' ';
}
cout << '\n';
}
return EXIT_SUCCESS;
}
Note that the version you wrote does not initialize the standard library’s random seed, so you get the same (poorly-distributed) random numbers. You could write a version that uses the much superior STL random-number generators and perhaps <chrono>, instead of the C versions, but that’s a bit outside the scope of your question.