int main()
{
int numbers[30];
int i;
// making array of 30 random numbers under 100
for(i=0;i<30;i++)
{
numbers[i] = rand() % 100;
}
// finding the greatest number in the array
int greatest = 0;
srand(1);
for(i=0;i<30;i++)
{
if ( numbers[i] > greatest)
greatest = numbers[i];
}
How do I then tell the program to display the max value of the array??
Thank you
To display it in the basic console output:
#include <iostream>
...
std::cout << "Max value is: " << greatest << "\n";
#include <iostream>
std::cout << greatest << '\n';
On a sidenote, you might want to call srand() before your call rand() (and might want to supply a more meaningful parameter).
If you are not doing this for home work I would suggest using std::max_element (available in <algorithm>).
std::cout << "Max Value: " << *(std::max_element(number, numbers+30)) << std::endl;
Otherwise, in your program all thats left to do is to print the value. You could use std::cout (available in <iostream>). After you've computed the great in the for loop.
// Dump the value greatest to standard output
std::cout << "Max value: " << greatest << std::endl;
Is this what you're referring to?
printf("%d", greatest);
Make sure to include "cstdio".
std::cout << greatest <<endl;
Related
This code is outputting 1 correct answer - which is always the one associated with 'random_number', so the first cout statement is always true. But who wants this kind of a quiz?
srand((int)time(0));
int random_number = rand() % max_event_number;
std::cout <<"\n" << final_years[random_number] << std::endl;
std::cout << "\n" << final_years[1 + random_number] << std::endl;
std::cout << "\n" << final_years[2 + random_number] << std::endl;
std::cout << "\n" << final_years[3 + random_number] << std::endl;
std::cout << "\n" << "Please type the correct year : " << std::endl;
Yes..., I can generate some random answers from the entire array, but they won't necessarily include the correct answer.
I don't want to change the way the correct answer is generated by the first 'random_number', because it takes only one line of code to check if the answer is true or not...
If only I could shuffle every time those 4 cout statement...
How would you do it?
Maybe not the best solution, but it works:
#include <iostream>
#include <string>
#include <cstdlib>
#include <algorithm>
#include <vector>
std::vector<std::string> possible_answers = { final_years[random_number], final_years[1 + random_number], final_years[2 + random_number], final_years[3 + random_number] };
// I initialized a vector like an array with all the 4 answers, including the correct one (since C++11)
std::vector<std::string>::iterator it;
it = possible_answers.begin();
std::random_shuffle(possible_answers.begin(), possible_answers.end()); //shuffled its contents
for (it = possible_answers.begin(); it < possible_answers.end(); it++) //some tinkering to output vector's content
std::cout <<"\n" << *it;
Now the randomly generated numbers are displayed in a random order on the screen, but the correct answer is always the first generated number.
Doing an exercise to find the mean and mode of a list of numbers input by a user. I have written the program and it works, but I'm wondering if my function 'calcMode' is too large for this program. I've just started looking into functions which is a first attempt. Would it be better to write smaller functions? and if so what parts can I split? Im pretty new to C++ and also looking if I can improve this code. Is there any changes I can make to make this run more efficient?
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int calcMean(vector<int> numberList)
{
int originNumber = numberList[0];
int nextNumber;
int count = 0;
int highestCount = 0;
int mean = 0;
for (unsigned int i = 0; i <= numberList.size() - 1; i++)
{
nextNumber = numberList[i];
if (nextNumber == originNumber)
count++;
else
{
cout << "The Number " << originNumber << " appears " << count << " times." << endl;
count = 1;
originNumber = nextNumber;
}
}
if (count > highestCount)
{
highestCount = count;
mean = originNumber;
}
cout << "The Number " << originNumber << " appears " << count << " times." << endl;
return mean;
}
int main()
{
vector<int> v;
int userNumber;
cout << "Please type a list of numbers so we can arrange them and find the mean: "<<endl;
while (cin >> userNumber) v.push_back(userNumber);
sort(v.begin(), v.end());
for (int x : v) cout << x << " | ";
cout << endl;
cout<<calcMean(v)<<" is the mean"<<endl;
return 0;
}
One thing to watch out for is copying vectors when you don't need to.
The function signature
int calcMode(vector<int> numberList)
means the numberList will get copied.
int calcMode(const & vector<int> numberList)
will avoid the copy. Scott Meyer's Effective C++ talks about this.
As an aside, calling is a numberList is misleading - it isn't a list.
There are a couple of points that are worth being aware of in the for loop:
for (unsigned int i = 0; i <= numberList.size()-1; i++)
First, this might calculate the size() every time. An optimiser might get rid of this for you, but some people will write
for (unsigned int i = 0, size=numberList.size(); i <= size-1; i++)
The size is found once this way, instead of potentially each time.
They might even change the i++ to ++i. There used to a potential overhead here, since the post-increment might involve an extra temporary value
One question - are you *sure this gives the right answer?
The comparison nextNumber == originNumber is looking at the first number to begin with.
Try it with 1, 2, 2.
One final point. If this is general purpose, what happens if the list is empty?
Would it be better to write smaller functions?
Yes, you can make do the same job using std::map<>; which could be
a much appropriate way to count the repetition of the array elements.
Secondly, it would be much safer to know, what is the size of the
array. Therefore I suggest the following:
std::cout << "Enter the size of the array: " << std::endl;
std::cin >> arraySize;
In the calcMode(), you can easily const reference, so that array
will not be copied to the function.
Here is the updated code with above mentioned manner which you can refer:
#include <iostream>
#include <algorithm>
#include <map>
int calcMode(const std::map<int,int>& Map)
{
int currentRepetition = 0;
int mode = 0;
for(const auto& number: Map)
{
std::cout << "The Number " << number.first << " appears " << number.second << " times." << std::endl;
if(currentRepetition < number.second )
{
mode = number.first; // the number
currentRepetition = number.second; // the repetition of the that number
}
}
return mode;
}
int main()
{
int arraySize;
int userNumber;
std::map<int,int> Map;
std::cout << "Enter the size of the array: " << std::endl;
std::cin >> arraySize;
std::cout << "Please type a list of numbers so we can arrange them and find the mean: " << std::endl;
while (arraySize--)
{
std::cin >> userNumber;
Map[userNumber]++;
}
std::cout << calcMode(Map)<<" is the mode" << std::endl;
return 0;
}
Update: After posting this answer, I have found that you have edited your function with mean instead of mode. I really didn't get it.
Regarding mean & mode: I recommend you to read more. Because in general, a data set can have multiple modes and only one mean.
I personally wouldn't split this code up in smaller blocks, only if i'd want to reuse some code in other methods. But just for this method it's more readable like this.
The order of excecution is aroun O(n) for calc which is quite oke if you ask me
What is the correct way to determine if a number (in my case it is a value of power of two calculated by pow(2,n)) is within the limits of values that one variable type can take? I'm doing it like this: if(pow (2,128)>std::numeric_limits<float>::max()), but this is evaluated as true although it is expected that float's maximum value is 2^128 or something more. Is there any better way to do this comparison?
For these kinds of limit checking, you can move the terms around to stay within the limits of the type.
In this case, pow(2,n) == exp(ln(2)*n) mathematically, so, rearranging terms, you can use n > ln(maxval)/ln(2)
You can take the base 2 logarithm of the maximum limit for the type of variable and compare it to n. For example: if(n > std::log2(std::numeric_limits<float>::max()). You probably don't want n to be exactly on the limit though, since I think stuff like floating point error might cause some problems.
First of all can you answer what is the result of pow(2, 128)?
The real question is what is the type for this expression?
The second question is do you know how floating point numbers work?
Take a look on this code to give you a hints:
#include <cmath>
#include <iostream>
#include <limits>
template<class T>
void printInfo(const std::string& desc, T x)
{
std::cout << desc << ' ' << typeid(x).name() << ' ' << x << std::endl;
}
int main()
{
printInfo("A", std::pow(2, 128));
printInfo("B", std::pow(2.0f, 128));
printInfo("A", std::pow(2, 128.0f));
auto c = std::pow(2.0f, 128.0f);
printInfo("C", c);
std::cout << (c > std::numeric_limits<float>::max()) << std::endl;
std::cout << (c == std::numeric_limits<float>::infinity()) << std::endl;
return 0;
}
https://wandbox.org/permlink/bHdKqToDKdC0hSvW
I recommend review documentation of numeric_limits.
And analyze this code:
#include <cmath>
#include <iostream>
#include <limits>
template<class T>
void print2exp()
{
std::cout << typeid(T).name() << '\n';
std::cout << "Radix = " << std::numeric_limits<T>::radix << '\n';
auto maxExp = std::numeric_limits<T>::max_exponent;
std::cout << "Max exp = " << maxExp << '\n';
std::cout << "2^maxExp = " << std::pow(static_cast<T>(2), static_cast<T>(maxExp)) << '\n';
std::cout << "2^(maxExp - 1) = " << std::pow(static_cast<T>(2), static_cast<T>(maxExp - 1)) << '\n';
}
int main()
{
print2exp<float>();
print2exp<double>();
print2exp<long double>();
return 0;
}
https://wandbox.org/permlink/J0hACKUKvKlV8lYK
So proper approach to this is (assuming that radix is 2):
if (x < std::numeric_limits<T>::max_exponent) {
return std::pow(static_cast<T>(2), static_cast<T>(x));
} else {
throw invalid_argument("x is to big to be use as 2^x");
}
I was trying to generate some random doubles in C++ (MSVC, though that isn't too important to me—I just didn't have another compiler to test) and I noticed that my quick program never generated negative numbers:
#include <iostream>
#include <random>
#include <ctime>
int main() {
std::mt19937 generator(clock());
std::uniform_real_distribution<double>
rand_dbl(std::numeric_limits<double>::min(),
std::numeric_limits<double>::max());
std::cout << "Double Limits: (" << std::numeric_limits<double>::min()
<< "," << std::numeric_limits<double>::max() << ")"
<< std::endl << std::endl;
int total_neg = 0;
for (int i=0; i<100; i++) {
double d = rand_dbl(generator);
if (d<0) total_neg++;
std::cout << d << " ";
}
std::cout << std::endl << std::endl
<< "Total negative random double is: " << total_neg << std::endl;
return 0;
}
No matter how many numbers I have it generate, it never generates a negative one. I understand why most of the numbers generated are in the 10307 - 10308 range (which isn't exactly what I wanted), but not why the numbers are always positive. I tried a few different generators (default, mt19937, minstd_rand0) without any difference in this aspect.
Can anyone describe why this is the case?
You set it up that way with the limits that you provided. std::numeric_limits<double>::min() gives the smallest positive double, and you used that as the lower bound on the distribution.
std::numeric_limits<double>::min()
Will return DBL_MIN which is the smalles value closest to 0 a double can hold. If you want the largest negative value then you need to use
std::numeric_limits<double>::lowest()
Which will return -DBL_MAX which is the largest negative value a double can hold.
From cppreference:
For floating-point types with denormalization, min returns the minimum positive normalized value.
(emphasis mine)
So it's pretty normal you only get positive values.
Could you tell what is displayed by those lines?
std::cout << "Double Limits: (" << std::numeric_limits<double>::min()
<< "," << std::numeric_limits<double>::max() << ")"
<< std::endl << std::endl;
This question already has answers here:
How to generate different random numbers in a loop in C++?
(13 answers)
Closed 7 years ago.
I'm having trouble with a random generator.
I'm trying to print out random values and I'm getting almost the same value every single time.
This is what I have:
void Deck::shuffle() {
StackNode<Card>* top = stack->top;
for (int i = 0; i < stack->numNodes - 1; i++) {
int x = random(i);
StackNode<Card>* temp = findCard(x);
//cout << "Random index was: " << random(i) << endl;
//cout << "Face value of random was: " << temp->data.getFaceVal() << endl;
cout << "Top: " << top->data.getFaceVal() << endl;
cout << "Temp: " << temp->data.getFaceVal() << endl;
swapX(top,temp);
}
}
Here's my random generator function:
int random(int index) {
int r;
srand(time(NULL));
cout << "Index: " << index << endl;
r = rand() % 50;
cout << "Random value: " << r << endl;
return r;
}
I think you can use std::shuffle here for your problem. Like this:
#include <vector>
#include <algorithm>
void Deck::shuffle() {
StackNode<Card>* top = stack->top;
std::vector<StackNode<Card>*> cards;
for (int i = 0; i < stack->numNodes - 1; i++) {
cards.push_back(findCard(i))
}
std::shuffle(cards.begin(), cards.end());
for (auto card : cards) {
std::cout << card->data.getFaceVal() << std::endl;
}
}
By the way, I would recommend you to call srand only once in your code.
rand() is a pseudo random number generator. The numbers it generates appear to be random, but they are generated by a completely deterministic function. The seed that you give it with sand() determines the starting point for the function. If you give it the same seed it will generate the same sequence of random numbers. You can try this and see for your self by seeding with a literal, like srand(200) and running the program several times, you will get the exact same results.
If you want different results each time you have to seed with something that will be different each time the program runs, so time is often used as a seed. In your case you are in a very tight loop so many of the calls in a row use the same time value.
If you call srand() once, before your loop this problem will go away.