While cicle and array

[Zhylian]

Hi guys, this program asks to input numbers between 0 and 5 and print the time a number occurred; and it works. BUT: I cannot understand why, if I inizialize n=0, the outcomes go crazy. theoretically any n>=0 get me into the while and then, the first thing the program does is to put the imput into n. I don't see any problems with that(maybe is something else).
I really need your help. Thank you all for your time.

  int main() {
	const int p=6;
	int n=6;
	int v[n]={};  

	while(n>=0){
		cout << "insert an integer number between 0 and 5(negative if you want to end the program) \n";
		cin >> n;
		if(n>5) {
			cout << "try agan \n";
			continue;
		}
		v[n]++;	
	}
	for(int i=0; i<p; i++) {
		if(v[i]>0) cout << "recurrences of " << i << ": \t" << v[i] << endl;
	}
	return 0;
}

[lastchance]

You are not diagnosing the problem correctly. It is perfectly legitimate to initialise n as 0.

The problem is that if you do input a negative number ... you will still carry out the line
v[n]++;
and v[negative number] isn't defined.

You won't test for negative n until the next cycle through the loop.

Either refactor, or exit the loop immediately on getting a negative number.

[Zhylian]

Ok, thanks! anyway even now the same problem remain. If I inizialize n to a number lower than 6, the values of the array seems to be random. here's the code:

int main() {
	const int p=6;
	int n=6;
	int v[n]={};  

	while(true){
		cout << "insert an integer number between 0 and 5(negative if you want to end the program) \n";
		cin >> n;
		if (n<0) break;
		if(n>5) {
			cout << "try agan \n";
			continue;
		}
		if(n>=0) v[n]++;	
	}
	for(int i=0; i<p; i++) {
		if(v[i]>0) cout << "recurrences of " << i << ": \t" << v[i] << endl;
	}
	return 0;
}

[kbw]

int v[n]={};
This isn't valid C++.

https://www.cplusplus.com/forum/beginner/232570/

[Zhylian]

Thanks man, now it works plus I understood the problem.

[seeplus]

#include <iostream>

int main() {
	const size_t p {6};
	int v[p] {};

	for (int n {}; n >= 0; ) {
		std::cout << "insert an integer number between 0 and " << p - 1 << " (negative if you want to end the program) \n";
		std::cin >> n;

		if (n >= 0)
			if (n > p - 1)
				std::cout << "try again \n";
			else
				++v[n];
	}

	for (size_t i = 0; i < p; ++i)
		if (v[i] > 0)
			std::cout << "recurrences of " << i << ": \t" << v[i] << '\n';
}

[Zhylian]

Wow, I'll investigate on that beautiful for! thanks man

[deleted account xyzzy]

int v[n]={};

If'n you want to initialize your entire array when created:

#include <iostream>

int main()
{
   // if you don't initialize the array it holds "garbage" values
   int arr1[5];

   // use a range-based for loop to "walk through" the array elements
   for (const auto& itr : arr1)
   {
      std::cout << itr << ' ';
   }
   std::cout << "\n\n";

   // zero initialize the array when created
   int arr2[5] { };

   // whitespace can be your friend
   for (const auto& itr : arr2) { std::cout << itr << ' '; }
   std::cout << "\n\n";

   // trying to use 'uniform initialization' on an array is tricky
   int arr3[5] { 10 };

   for (const auto& itr : arr3) { std::cout << itr << ' '; }
   std::cout << "\n\n";

   // proper uniform initialization of an array, specify each element
   int arr4[] { 5, 10, 15, 20, 25 };

   for (const auto& itr : arr4) { std::cout << itr << ' '; }
   std::cout << '\n';
}

0 9711856 0 9705188 1998279686

0 0 0 0 0

10 0 0 0 0

5 10 15 20 25

[seeplus]

Braced array initialisation will only initialise with the number of specified elements. If there are fewer specified than the number of elements, the rest are default initialised.

To fill an array with the same value:

std::fill_n(arr1, 5, 10);

[Zhylian]

thanks guys, I really appreciate it!