I am writing a simple code to calculate Fabonacci numbers as an exercise. The code works, but i don't get why. I have some special cases for n=1 and n=2 which is the place of the number in the sequence (the numbers are 0 and 1). However after those, the number is calculated in this loop.
while(n>LoopCount)
{
Fib_n=Fib_1+Fib_2;
Fib_2=Fib_1;
Fib_1=Fib_n;
LoopCount++;
}
Going in to the loop, Fib_1=0, Fib_2=0, and Fib_n=1. Why does not this loop just spit out 0 no matter what? The whole code is below.
#include <iostream>
using namespace std;
int main()
{
cout <<"Which number of the Fibonacci sequence do you want to calculate?" <<endl;
int n;
cin >>n;
cout <<endl;
int Fib_n;
int Fib_1;
int Fib_2;
int LoopCount=1;
if(n>1)
{
Fib_n=1;
LoopCount++;
while(n>LoopCount)
{
Fib_n=Fib_1+Fib_2;
Fib_2=Fib_1;
Fib_1=Fib_n;
LoopCount++;
}
}
cout <<Fib_n;
return 0;
}
int Fib_1;
int Fib_2;
were never initialized. Therefore, the first time you calculate Fib_n=Fib_1+Fib_2;, Fib_n will get the sum of two uninitialized variables.
I have modified your code so it would work.
#include <iostream>
using namespace std;
int main()
{
cout <<"Which number of the Fibonacci sequence do you want to calculate?" <<endl;
int n;
cin >> n;
cout << endl;
int Fib_1 = 1;
int Fib_2 = 1;
int count = 0;
while(n > count)
{
Fib_1 = Fib_1 + Fib_2;
Fib_2 = Fib_1 - Fib_2;
count++;
}
cout << Fib_1;
return 0;
}
Fib_1
You have that as an uninitalized variable, so you may get a garbage value for output.
Fib_2 = Fib_1
Next, you initialize Fib_2 with Fib_1, meaning they both share the same (random) value.
In debug mode, these are both initialized to 0, and adding them:
Fib_n=Fib_1+Fib_2;
makes the sum equal 0. In release mode, you can expect random values from the compiler. Here is more info on Uninitialized Variables.
Related
I have been given some integers and I have to count the frequency of a specific digit in the number.
example input:
5
447474
228
6664
40
81
The first number says number of integers in the list. I am finding frequency of 4 in this case. I tried to change the integer to an array, but it is not working.
#include<iostream>
#include<cmath>
#include<vector>
using namespace std;
int main() {
int n;
cin>>n;
for (int i=0; i<n; i++)
{
int x;
cin>>x;
int frequency=0;
int t=log10(x);
int arr[t];
for (i=t; i>0; i--)
{
arr[i]=x%10;
x=x/10;
}
for(int i=0; i<t; i++)
{
if(arr[i]==4)
{
frequency++;
}
}
std::cout << frequency << std::endl;
}
return 0;
}
No need to create an array, or to determine the number of digits. Just loop until the number reaches zero.
int digitCount(int n, int d) {
if(n < 0) n = -n;
int count = 0;
for(; n != 0; n /= 10)
if(n % 10 == d) count++;
return count;
}
Test:
cout << digitCount(447474, 4) << endl;
cout << digitCount(-447474, 4) << endl;
Output:
4
4
Your code uses VLAs which are not standard C++. See Why aren't variable-length arrays part of the C++ standard?.
log10(x) is not the number of digits. For example log10(1234) == 3.09131516 but it is 4 digits. Also you are accessing the array out of bounds in the first iteration of the loop: arr[t]. Valid indices in an array of size t are 0,1,2,...,t-1. Trying to access arr[t] is undefined behavior.
Actually you dont need any array. Instead of storing the digits in an array you can immediately check whether it is a 4 and count.
Even simpler would be to read the user input as a std::string:
#include <string>
#include <algorithm>
#include <iostream>
int main() {
std::string input;
std::cin >> input;
std::cout << std::count(input.begin(),input.end(),'4');
}
Perhaps you should add some checks to verify that the user input is actually a valid number. However, also when reading an int you should validate the input.
Program in C++ that takes 3 numbers and send them to a function and then calculate the average function of these 3 numbers.
I know how to do that without using a function ,for example for any n numbers I have the following program:
#include<stdio.h>
int main()
{
int n, i;
float sum = 0, x;
printf("Enter number of elements: ");
scanf("%d", &n);
printf("\n\n\nEnter %d elements\n\n", n);
for(i = 0; i < n; i++)
{
scanf("%f", &x);
sum += x;
}
printf("\n\n\nAverage of the entered numbers is = %f", (sum/n));
return 0;
}
Or this one which do that using arrays:
#include <iostream>
using namespace std;
int main()
{
int n, i;
float num[100], sum=0.0, average;
cout << "Enter the numbers of data: ";
cin >> n;
while (n > 100 || n <= 0)
{
cout << "Error! number should in range of (1 to 100)." << endl;
cout << "Enter the number again: ";
cin >> n;
}
for(i = 0; i < n; ++i)
{
cout << i + 1 << ". Enter number: ";
cin >> num[i];
sum += num[i];
}
average = sum / n;
cout << "Average = " << average;
return 0;
}
But is it possible to use functions?if yes then how? thank you so much for helping.
As an alternative to using fundamental types to store your values C++ provides std::vector to handle numeric storage (with automatic memory management) instead of plain old arrays, and it provides many tools, like std::accumulate. Using what C++ provides can substantially reduce your function to:
double avg (std::vector<int>& i)
{
/* return sum of elements divided by the number of elements */
return std::accumulate (i.begin(), i.end(), 0) / static_cast<double>(i.size());
}
In fact a complete example can require only a dozen or so additional lines, e.g.
#include <iostream>
#include <vector>
#include <numeric>
double avg (std::vector<int>& i)
{
/* return sum of elements divided by the number of elements */
return std::accumulate (i.begin(), i.end(), 0) / static_cast<double>(i.size());
}
int main (void) {
int n; /* temporary integer */
std::vector<int> v {}; /* vector of int */
while (std::cin >> n) /* while good integer read */
v.push_back(n); /* add to vector */
std::cout << "\naverage: " << avg(v) << '\n'; /* output result */
}
Above, input is taken from stdin and it will handle as many integers as you would like to enter (or redirect from a file as input). The std::accumulate simply sums the stored integers in the vector and then to complete the average, you simply divide by the number of elements (with a cast to double to prevent integer-division).
Example Use/Output
$ ./bin/accumulate_vect
10
20
34
done
average: 21.3333
(note: you can enter any non-integer (or manual EOF) to end input of values, "done" was simply used above, but it could just as well be 'q' or "gorilla" -- any non-integer)
It is good to work both with plain-old array (because there is a lot of legacy code out there that uses them), but equally good to know that new code written can take advantage of the nice containers and numeric routines C++ now provides (and has for a decade or so).
So, I created two options for you, one use vector and that's really comfortable because you can find out the size with a function-member and the other with array
#include <iostream>
#include <vector>
float average(std::vector<int> vec)
{
float sum = 0;
for (int i = 0; i < vec.size(); ++i)
{
sum += vec[i];
}
sum /= vec.size();
return sum;
}
float average(int arr[],const int n)
{
float sum = 0;
for (int i = 0; i < n; ++i)
{
sum += arr[i];
}
sum /= n;
return sum;
}
int main() {
std::vector<int> vec = { 1,2,3,4,5,6,99};
int arr[7] = { 1,2,3,4,5,6,99 };
std::cout << average(vec) << " " << average(arr, 7);
}
This is an example meant to give you an idea about what needs to be done. You can do this the following way:
// we pass an array "a" that has N elements
double average(int a[], const int N)
{
int sum = 0;
// we go through each element and we sum them up
for(int i = 0; i < N; ++i)
{
sum+=a[i];
}
// we divide the sum by the number of elements
// but we first have to multiply the number of elements by 1.0
// in order to prevent integer division from happening
return sum/(N*1.0);
}
int main()
{
const int N = 3;
int a[N];
cin >> a[0] >> a[1] >> a[2];
cout << average(a, N) << endl;
return 0;
}
how to do that without using a function
Quite simple. Just put your code in a function, let's call it calculateAverage and return the average value from it. What should this function take as input?
The list of numbers (array of numbers)
Total numbers (n)
So let's first get the input from the user and put it into the array, you have already done it:
for(int i = 0; i < n; ++i)
{
cout << i + 1 << ". Enter number: ";
cin >> num[i];
}
Now, lets make a small function i.e., calculateAverage():
int calculateAverage(int numbers[], int total)
{
int sum = 0; // always initialize your variables
for(int i = 0; i < total; ++i)
{
sum += numbers[i];
}
const int average = sum / total; // it is constant and should never change
// so we qualify it as 'const'
//return this value
return average
}
There are a few important points to note here.
When you pass an array into a function, you will loose size information i.e, how many elements it contains or it can contain. This is because it decays into a pointer. So how do we fix this? There are a couple of ways,
pass the size information in the function, like we passed total
Use an std::vector (when you don't know how many elements the user will enter). std::vector is a dynamic array, it will grow as required. If you know the number of elements beforehand, you can use std::array
A few problems with your code:
using namespace std;
Don't do this. Instead if you want something out of std, for e.g., cout you can do:
using std::cout
using std::cin
...
or you can just write std::cout everytime.
int n, i;
float num[100], sum=0.0, average;
Always initialize your variables before you use them. If you don't know the value they should be initialized to, just default initialize using {};
int n{}, i{};
float num[100]{}, sum=0.0, average{};
It is not mandatory, but good practice to declare variables on separate lines. This makes your code more readable.
There is a task. It is necessary in a one-dimensional array of N real numbers to calculate the number of the maximum modulo element among unpaired numbers.
I wrote the code, but it does not work. I can’t understand what’s wrong with him.
#include <iostream>
#include <math.h>
using namespace std;
int main() {
setlocale(0, "");
const int KolEl = 5;
int mas[KolEl];
int max = abs(mas[0]);
int result;
for (int i = 0; i < KolEl; i++)
{
cout << " Введите елемент[" << i << "] = ";
cin >> mas[i];
if (mas[i] % 2 == 1) {
if (abs(mas[i]) > max) {
result = i;
cout << result << endl;
}
}
}
system("pause");
}
You initialize max as:
int mas[KolEl];
int max = abs(mas[0]);
However, the values in mas[] are garbage values (read: undefined behavior). So now the value in max is also UB.
You then go on to use that value to compare to the input you take:
if (abs(mas[i]) > max) {
So the result of that comparison is undefined.
You probably meant to declare max as something like:
int max = INT_MIN;
So that the first comparison will always be true (every int except INT_MIN will be greater than it).
I'm trying to write a program that creates and fills a vector with int values, then searches through it and returns the minimum value, recursively. I have the code written out and building, but it returns a weirdly large value for minimum every time- I have a feeling it's not properly assigning the smallest value to int minimum, but I'm not sure. Any thoughts?
#include <iostream>
#include <conio.h>
#include <vector>
using namespace std;
int vectorSize;
int minimum;
int result = -1;
int start;
int ending;
int answer;
int test;
int recursiveMinimum(vector<int>, int, int);
void main() {
cout << "How many values do you want your vector to be? ";
cin >> vectorSize;
cout << endl;
vector<int> searchVector(vectorSize);
start = 0;
ending = searchVector.size() - 1;
for (int i = 0; i < vectorSize; i++) {
cout << "Enter value for position " << i << " " << endl;
cin >> searchVector[i];
}
for (int x = 0; x < vectorSize; x++) {
cout << searchVector[x] << " ";
}
int answer = recursiveMinimum(searchVector, start, ending);
cout << "The smallest value in the vector is: " << answer;
_getch();
}
int recursiveMinimum(vector<int> searchVector, int start, int end) {
if (start < end) {
if (searchVector[start] < minimum) {
minimum = searchVector[start]; //this part seems to not work
}
start++;
recursiveMinimum(searchVector, start, end);
}
else {
return minimum;
}
}
`
Your minimum variable is not initialised, which leads to undefined behaviour. It should be set to the first value in the vector:
minimum = searchVector[0];
int answer = recursiveMinimum(searchVector, start, ending);
Additionally, ending is off by one, which makes it pick 6 as the smallest value out of [6, 9, 8, 4].
So, ultimately, your code should look like this:
minimum = searchVector[0];
int answer = recursiveMinimum(searchVector, start, ending + 1); // note the + 1
While irrelevant to the question, I advise you to use a tail call in recursiveMinimum, as explained here:
start++;
return recursiveMinimum(searchVector, start, end);
The main issue is that you do not initialise minimum. Hence, comparison searchVector[start] < minimum might never become true, and minimum remains uninitialized.
As a quick fix, write int minimum = MAX_INT; instead of int minimum;. MAX_INT is the maximum positive integer value (defined in limits.h). So the values in your array will never be greater that this value, and your minimum search loop will work (unless there are other issues; but for that, please consult the debugger :-) )
Here is my code for printing the divisors and then number of divisors of a given number.
Now suppose I take 2 test cases: 5 and 8; this code gives count of 5 as 2 and 8 as 6 (i.e it adds the previous count).
Even if I declare it as int count = 0; it returns the same output.
The other problem arises when I declare int count = 0 inside function factors.
The code gives count as 0 for all cases.
#include<iostream>
using namespace std;
int count;
long long factors(long n, long f=1)
{
if(n%f==0) {
cout << f << endl;
count++;
}
if(f==n) {
return 0;
}
factors(n,f+1);
return count;
}
int main()
{
int n;
int t;
cin >> t;
while(t--)
{
cin >> n;
cout << factors(n) << endl;
}
return 0;
}
Using globals is not usually a good idea. It is especially bad in recursive functions, which should preferably be re-entrant. Of course you can fix your function by resetting the count in the loop, like this:
while(t--)
{
cin>>n;
count = 0; // Reset count before the recursive call
cout << factors(n) << endl;
}
You could also make factors "wrapper" that resets the count to free the callers from the need to reset count before calling factors, like this:
long long factors(long n) {
count = 0;
return factors(n, 1);
}
long long factors(long n,long f /* Remove the default */) {
... // the rest of your code
}
you can achieve this by passing count as reference -
#include<iostream>
using namespace std;
long long factors(long n, int& count, long f=1)
{
if(n%f==0)
{
cout<<f<<endl;
count = count + 1;
}
if(f==n)
return 0;
factors(n, count, f+1);
return 0;
}
int main()
{
int n,t;
cin>>t;
while(t--)
{
cin>>n;
int count = 0;
factors(n, count);
cout << count << endl;
}
return 0;
}
-Gaurav
First, why are you declaring the count variable in the global space?
Second, you can not perform arithmetic operations to an undeclared variable (the int "count" in this case is never declared).
Third, why do you create an infinite loop by doing while(t--)?
You said the function gives you count as 0 for all input,
Can this be due to count never being declared?