Print an array in C++ - c++

Question

I'm trying to do some of my C++ homework, but I seem to have run into an issue. I need to make it so that the user inputs 8 numbers, and those said 8 get stored in an array. Then, if one of the numbers is greater than 21, to output said number. The code is below, and it's kind of sloppy. Yes, first year C++ learner here :p
#include <iostream>
using namespace std;
int main() {
const int NUM_ELEMENTS = 8; // Number of elements
int userVals[NUM_ELEMENTS]; // User numbers
int i = 0; // Loop index
int sumVal = 0; // For computing sum
int prntSel = 0; // For printing greater than 21
// Prompt user to populate array
cout << "Enter " << NUM_ELEMENTS << " integer values..." << endl;
for (i = 0; i < NUM_ELEMENTS; ++i) {
cin >> userVals[i];
}
for (int i = NUM_ELEMENTS - 1; i > 21; i--)
cout << "Value: " << sumVal << endl;
// Determine sum
sumVal = 0;
for (i = 0; i < NUM_ELEMENTS; ++i) {
sumVal = sumVal + userVals[i];
}
cout << "Sum: " << sumVal << endl;
return 0;
}

Answer 1

Don't reinvent the wheel, use standard algorithms:
std::copy_if(std::begin(userVals), std::end(userVals),
std::ostream_iterator<int>(std::cout, "\n"),
[] (auto x) { return x > 21; });
I improved the rest of your program as well:
#include <algorithm>
#include <iostream>
#include <iterator>
#include <numeric>
#include <vector>
auto constexpr count = 8;
int main() {
std::vector<int> numbers(count);
std::cout << "Enter " << count << " integer values...\n";
std::copy_n(std::istream_iterator<int>(std::cin), numbers.size(), numbers.begin());
std::copy_if(numbers.begin(), numbers.end(),
std::ostream_iterator<int>(std::cout, "\n"),
[] (auto x) { return x > 21; });
auto sum = std::accumulate(numbers.begin(), numbers.end(), 0);
std::cout << "Sum: " << sum << '\n';
return 0;
}
See it live on Coliru!

Answer 2

Ok, I'm going to explain this to you and keep it simple. This loop
`for (int i = NUM_ELEMENTS - 1; i > 21; i--)`
will never execute because in your first iteration you are checking if (NUM_ELEMENTS-1=7)>21. You are then decrementing i so this will take the series (6,5,4,...) and nothing would ever happen here.
If you have to sum the numbers greater than 21, which I presume is what you need then you will have to remove the above loop and modify your second loop to:
for (i = 0; i < NUM_ELEMENTS; i++) {
if(userVals[i]>21)
sumVal = sumVal + userVals[i];
}
This way, you add the numbers in the array that are only greater than 21. The index of userVals is determined by the i variable which also acts as a counter.

Answer 3

You're on the right track. There's just a few things wrong with your approach.
#include <iostream>
#include <stdlib.h>
using namespace std;
int main() {
const int NUM_ELEMENTS = 8;
int userVals[NUM_ELEMENTS];
int i = 0;
int sumVal = 0;
int prntSel = 0;
int size = sizeof(userVals) / sizeof(int); // Get size of your array
// 32/4 = 8 (ints are 4 bytes)
cout << "Enter " << NUM_ELEMENTS << " integer values..." << endl;
for (i = 0; i < NUM_ELEMENTS; ++i) {
cin >> userVals[i];
}
for(int i = 0; i < size; i++) {
if(userVals[i] > 21) { // Is number > 21?
cout << userVals[i] << endl; // If so, print said number
exit(0); // And exit
}
else
sumVal += userVals[i]; // Else sum your values
}
cout << "Sum: " << sumVal << endl;
return 0;
}

Answer 4

#include <iostream>
using namespace std;
int main() {
const int NUM_ELEMENTS = 8; // Number of elements
int userVals[NUM_ELEMENTS]; // User numbers
int i = 0; // Loop index
int sumVal = 0; // For computing sum
int prntSel = 0; // For printing greater than 21
// Prompt user to populate array
cout << "Enter " << NUM_ELEMENTS << " integer values..." << endl;
for (i = 0; i < NUM_ELEMENTS; ++i) {
cin >> userVals[i];
}
// for (int i = NUM_ELEMENTS - 1; i > 21; i--)
// cout << "Value: " << sumVal << endl;
for( i = 0; i < NUM_ELEMENTS; ++i )
{
if( userVals[ i ] > 21 )
{
cout << "Value: " << i << " is " << userVals[ i ] << endl;
}
}
for (i = 0; i < NUM_ELEMENTS; ++i) {
sumVal = sumVal + userVals[i];
}
cout << "Sum: " << sumVal << endl;
return 0;
}

Answer 5

Try
for (int i = NUM_ELEMENTS - 1; i > 21; i--)
cout << "Value: " << sumVal << endl;
to
for (i = 0; i < NUM_ELEMENTS; ++i) {
if(userVals[i] > 21)
cout << "Value: " << userVals[i] << endl;
}
This line isnt needed as well, as you arent using it.
int prntSel = 0; // For printing greater than 21

Answer 6

for (int i = NUM_ELEMENTS - 1; i > 21; i--)
cout << "Value: " << sumVal << endl;
Here you are printing the value of sumVal, not the value of the array in the position i. The line should be:
cout << "Value: " << usersVals[i] << endl;
Also that that your for is not doing what you think it does. for doesn't use the condition you gave to decide if will execute the current iteration or not, it uses the condition to decide if the loop should continue or not. So when you put i > 21, means that it will continue running while i is bigger than 21. To achieve your goal, you should make a test (if statement) inside the loop.
The final result it would be:
for (i = 0; i < NUM_ELEMENTS; ++i) {
if (usersVals[i] > 21) {
cout << "Value: " << usersVals[i] << endl;
}
}