if I have the array N[12] and I have the code:
float N[12];
for (int X = 1; X < 12; X++) {
N[X] = N[X] + 2;
cout << N[X] << endl;
}
what else do I need to get this to display odd numbers starting at 1 and increasing until I have 12?? (so until 23)
Its outputting a really weird slew of numbers
I'm pretty new to c++, I know this is a silly question, sorry...
Your array is not initialized to any values, so depending on how your compiler does it, it most likely will be filled with random values. This would explain why you are seeing weird numbers.
Since you want to start with a initial value of 1, you can initialize your array like this:
float N[12] = { 1 };
This will set the first element of the array to 1 and the rest of the elements will be set to 0. But then the other person is right, in that you want to add 2 to the previous element in your array. So it would make the code into this:
float N[12] = { 1 }
int X;
for(X=1; X<12; X++){
N[X]= N[X-1] + 2;
cout << N[X] << endl;
}
Couple of issues in your code as mentioned below.
Major issue : your array is uninitialized. So when you perform N[x] = N[x] + 2, it is meaningless as N[x] doesn't have any value.
Minor issue : No need to declare array as float as we are going to store integer values.
Below is the code.
#include <iostream>
using namespace std;
int main()
{
int N[12];
for(int x = 0; x < 12; x++)
{
N[x] = (2 * x) + 1;
cout << N[X] << endl;
}
return 0;
}
I think you mean to add two to the previous element, not the current one, since the current one is uninitialized. Also, since odd numbers are integers, you don't need floats.
int n[12];
n[0] = 1; // start off with the first odd
for (int x = 0; x < 12; x ++) {
// we already have a value for n[0] so we shouldn't overwrite it
if (x != 0) {
// n[x - 1] gets the previous value
n[x] = n[x - 1] + 2;
}
std::cout << n[x] << std::endl;
}
Related
So im working on a class assignment where I need to take a base 2 binary number and convert it to its base 10 equivalent. I wanted to store the binary as a string, then scan the string and skip the 0s, and at 1s add 2^i. Im not able to compare the string at index i to '0, and im not sure why if(binaryNumber.at(i) == '0') isnt working. It results in an "out of range memory error". Can someone help me understand why this doesnt work?
#include <iostream>
using namespace std;
void main() {
string binaryNumber;
int adder;
int total = 0;
cout << "Enter a binary number to convert to decimal \n";
cin >> binaryNumber;
reverse(binaryNumber.begin(),binaryNumber.end());
for (int i = 1; i <= binaryNumber.length(); i++) {
if(binaryNumber.at(i) == '0') { //THIS IS THE PROBLEM
//do nothing and skip to next number
}
else {
adder = pow(2, i);
total = adder + total;
}
}
cout << "The binary number " << binaryNumber << " is " << total << " in decimal form.\n";
system("pause");
}
Array indices for C++ and many other languages use zero based index. That means for array of size 5, index ranges from 0 to 4. In your code your are iterating from 1 to array_length. Use:
for (int i = 0; i < binaryNumber.length(); i++)
The problem is not with the if statement but with your loop condition and index.
You have your index begin at one, while the first character of a string will be at index zero. Your out memory range error is caused by the fact that the loop stops when less than or equal, causing the index to increase one too many and leave the memory range of the string.
Simply changing the loop from
for (int i = 1; i <= binaryNumber.length(); i++) {
if(binaryNumber.at(i) == '0') {
}
else {
adder = pow(2, i);
total = adder + total;
}
}
To
for (int i = 0; i < binaryNumber.length(); i++) {
if(binaryNumber.at(i) == '0') {
}
else {
adder = pow(2, i);
total = adder + total;
}
}
Will solve the issue.
Because your started from 1 and not 0
for (int i = 1; i <= binaryNumber.length(); i++)
Try with that
for (int i = 0; i < binaryNumber.length(); i++)
First of all, sorry for the mis-worded title. I couldn't imagine a better way to put it.
The problem I'm facing is as follows: In a part of my program, the program counts occurences of different a-zA-Z letters and then tells how many of each letters can be found in an array. The problem, however, is this:
If I have an array that consists of A;A;F;A;D or anything similar, the output will be this:
A - 3
A - 3
F - 1
A - 3
D - 1
But I am required to make it like this:
A - 3
F - 1
D - 1
I could solve the problem easily, however I can't use an additional array to check what values have been already echoed. I know why it happens, but I don't know a way to solve it without using an additional array.
This is the code snippet (the array simply consists of characters, not worthy of adding it to the snippet):
n is the size of array the user is asked to choose at the start of the program (not included in the snippet).
initburts is the current array member ID that is being compared against all other values.
burts is the counter that is being reset after the loop is done checking a letter and moves onto the next one.
do {
for (i = 0; i < n; i++) {
if (array[initburts] == array[i]) {
burts++;
}
}
cout << "\n\n" << array[initburts] << " - " << burts;
initburts++;
burts = 0;
if (initburts == n) {
isDone = true;
}
}
while (isDone == false);
Do your counting first, then loop over your counts printing the results.
std::map<decltype(array[0]), std::size_t> counts;
std::for_each(std::begin(array), std::end(array), [&counts](auto& item){ ++counts[item]; });
std::for_each(std::begin(counts), std::end(counts), [](auto& pair) { std::cout << "\n\n" << pair.first << " - " pair.second; });
for (i = 0; i < n; i++)
{
// first check if we printed this character already;
// this is the case if the same character occurred
// before the current one:
bool isNew = true;
for (j = 0; j < i; j++)
{
// you find out yourself, do you?
// do not forget to break the loop
// in case of having detected an equal value!
}
if(isNew)
{
// well, now we can count...
unsigned int count = 1;
for(int j = i + 1; j < n; ++j)
count += array[j] == array[i];
// appropriate output...
}
}
That would do the trick and retains the array as is, however is an O(n²) algorithm. More efficient (O(n*log(n))) is sorting the array in advance, then you can just iterate over the array once. Of course, original array sequence gets lost then:
std::sort(array, array + arrayLength);
auto start = array;
for(auto current = array + 1; current != array + arrayLength; ++current)
{
if(*current != *start)
{
auto char = *start;
auto count = current - start;
// output char and count appropriately
}
}
// now we yet lack the final character:
auto char = *start;
auto count = array + arrayLength - start;
// output char and count appropriately
Pointer arithmetic... Quite likely that your teacher gets suspicious if you just copy this code, but it should give you the necessary hints to make up your own variant (use indices instead of pointers...).
I would do it this way.
#include <iostream>
#include <string>
#include <vector>
using namespace std;
int main()
{
string s;
vector<int> capCount(26, 0), smallCount(26, 0);
cout << "Enter the string\n";
cin >> s;
for(int i = 0; i < s.length(); ++i)
{
char c = s.at(i);
if(c >= 'A' && c <= 'Z')
++capCount[(int)c - 65];
if(c >= 'a' && c <= 'z')
++smallCount[(int)c - 97];
}
for(int i = 0; i < 26; ++i)
{
if(capCount[i] > 0)
cout << (char) (i + 65) << ": " << capCount[i] << endl;
if(smallCount[i] > 0)
cout << (char) (i + 97) << ": " << smallCount[i] << endl;
}
}
Note: I have differentiated lower and upper case characters.
Here's is the sample output:
output
I'd like to make non-graphic (text) C++ game (you move your character using n, s, w, e and every location is described). Locations will be an array of an objects (there will be location description and other information in this array). I've started making this game, but I have a question: Is it possible to make arrays with dimensions x - from -100 to 100 and z - from -100 to 100? If it is not possible, are there other ways to do it? (I don't want a [0][0] position in one of 4 corners, but on the middle.)
An Array can have only positive indexes:
Location loc[201][201];
define a Funktion that returns your desired Location:
Location getLocation(int xCoord, int yCoord)
{
if (abs(x)>100 || abs(y)>100)
throw std::invalid_argument( "value out of range");
return loc[xCoord+100][yCoord+100];
}
Then you can get the Location by calling the function getLocation(x,y)
One common (but rather sketchy) method is to do something like this (example for 21 x 21 board):
#include <iostream>
using namespace std;
int main()
{
typedef int (*board_ptr)[21];
int board_data[21][21];
board_ptr board = (board_ptr)&board_data[10][10];
for (int i = -10; i <= 10; ++i)
for (int j = -10; j <= 10; ++j)
board[i][j] = 0;
board[-10][-10] = 1;
board[-10][10] = 2;
board[10][-10] = 3;
board[10][10] = 4;
board[0][0] = 5;
for (int i = -10; i <= 10; ++i)
{
for (int j = -10; j <= 10; ++j)
{
cout << " " << board[i][j];
}
cout << endl;
}
return 0;
}
This creates a normal 21x21 array but then it also creates a pointer to a fake array which is initialised to point at the centre of the real array. You can then use this fake pointer as if it were a real array with indices ranging from -10 to +10 (inclusive).
LIVE DEMO
So, i've made a program which is able to sort arrays, and i'm trying to sort an array containing double FP's, including 2-3 random ones i enter, pos inf, neg inf and a single NaN. so for this purpose i wish to sort the NaN.
So my code works, however when trying to sort the NaN, i'm unable to do so. What i'd like to do is sort it to the end, or have it put at the end of the sorted array. Is there anyway I can actually do this? Thanks in advance!!! code is as follows:
int main()
{
int start_s = clock();
int n, k = 4, j; // k is number of elements
double x = -0.0;
double i = 0;
double swap = 0;//used in the function as a place holder and used for swapping between other variables
double a[100] = { (1/x) + (1/i), 2.3, 1/x *0, 1/i };//array of double elements // 1/i * 0 is NaN
//(1 / i) * 0
for (n = 0; n < (k - 1); n++) // for loop consists of variables and statements in order to arrange contents of array
{
for (j = 0; j < k - n - 1; j++)
{
if (a[j] > a[j + 1])
{
swap = a[j];
a[j] = a[j + 1];
a[j + 1] = swap;
}
}
}
cout << "The list of sorted elements within the array, is: " << endl; /* Output message to user */
for (int i = 0; i < k; i++)// Loop up to number of elements within the array
{
cout << a[i] << " ";/* Output contents of array */
}
cout << endl; //new line
int stop_s = clock();
cout << "The execution time of this sort, is equal to: " << (stop_s - start_s) / double(CLOCKS_PER_SEC) * 1000 << " milliseconds" << endl;
return 0;
Since you're in C++ land anyway, why not use it to the full. First, indeed, move the NaN's and then sort. I've taken out 'noise' from your code and produced this, it compiles and runs (edit: on gcc-4.4.3). The main difference is that the NaN's are at the beginning but they're easily skipped since you will get a pointer to the start of non-NaN's.
#include <iostream>
#include <algorithm>
#include <math.h>
int main()
{
int n, k = 4, j; // k is number of elements
double x = -0.0;
double i = 0;
double a[100] = { (1/x) + (1/i), 2.3, 1/x *0, 1/i };//array of double elements // 1/i * 0 is NaN]
double *ptr; // will point at first non-NaN double
// divide the list into two parts: NaN's and non-NaN's
ptr = std::partition(a, a+k, isnan);
// and sort 'm
// EDIT: of course, start sorting _after_ the NaNs ...
std::sort(ptr, a+k);
cout << "The list of sorted elements within the array, is: " << endl; /* Output message to user */
for (int i = 0; i < k; i++)// Loop up to number of elements within the array
{
cout << a[i] << " ";/* Output contents of array */
}
cout << endl; //new line
return 0;
}
Do a linear scan, find the NaNs, and move them to the end - by swapping.
Then sort the rest.
You can also fix your comparator, and check for NaN there.
For the actual check see: Checking if a double (or float) is NaN in C++
you can use isnan() in cmath to check for NaNs. So, you can just change your comparison line from:
if (a[j] > a[j + 1])
to:
if (!std::isnan(a[j + 1]) && std::isnan(a[j]) || (a[j] > a[j + 1]))
just a reminder, you need to have:
#include <cmath>
at the top of your code.
For this code I created that outputs the ASCII characters corresponding to ints, I need to print out 16 ASCIIs per line. How would I go about doing so? I'm not sure how to approach these? Do I create another for loop inside?
int main()
{
int x = 0;
for (int i = 0; i <= 127; i++)
{
int x = i;
char y = (char) x;
cout << y;
}
return 0;
}
Or should I put the cout outside with 16 separate lines? I am trying to print 17 ASCIIs starting from 1 in a row.
Use another variable that counts up along with i. When it reaches 16, reset it and print a new line. Repeat until the loop terminates.
i.e.(I may be off by one here, I didn't think about it too deeply)
for (int i=0, j=1; i<=127; i++,j++)
{
int x = i;
char y = (char) x;
cout << y;
if (j == 16) {
j = 0;
cout << '\n';
}
}
Alternatively, you could just check if (i % 16 == 0)
You don't need another variable to track it. i is already an int.
so if i modulo 16 equals 0 then print a newline
else print (char)i
EDIT:
Note, using variables like i is ok for simple iteration but its always good practice to name them better.
So think about how changing i to ascii in your program improves the readability. It instantly makes it even more clear what is it that you are trying to do here.
int main()
{
int charsThisLine =0;
for (int currentChar=0; currentChar<128; currentChar++)
{
if(charsThisLine==16)
{
cout<<endl;
charsThisLine = 0;
}
else
{
cout<<(char)currentChar;
charsThisLine++;
}
}
}
How about:
#include <iostream>
int main()
{
for(int i = 0, j = 0; i < 128; ++i, ++j)
{
if(j == 16)
{
j = 0;
std::cout << std::endl;
}
std::cout << static_cast<char>(i);
}
return 0;
}
Every iteration, j increases by 1; after 16 iterations, j is reset to 0, and a newline is printed.
Alternatively, as #Sujoy points out, you could use:
if((i % 16) == 0)
std::cout << std::endl;
But this introduces the problem of printing an extra newline character at the beginning of the output.
Yes, you need a second loop inside the first. (I misunderstood what is being requested.)
You also need to clean up the code. The first x is unused; the second x isn't needed since you could perfectly well use char y = (char)i; (and the cast is optional). You should normally use a loop for (int i = 0; i < 128; i++) with a < condition rather than <=.
You will also need to generate a newline somewhere (cout << endl; or cout << '\n';). Will you be needing to deal with control characters such as '\n' and '\f'?
Finally, I'm not sure that 'asciis' is a term I've seen before; the normal term would be 'ASCII characters'.