I'm wondering whether how you can write something like this recursively or using a different loop system:
std::string a = "00000000";
for (int i = 0; i<8; i++) {
a[i] = '1';
for (int j = 0; j<8; j++) {
if (i!=j) {
a[j] = '1';
... //more for loops with the same structure
std::cout<<a[j]<<"\n";
a[j] = '0';
}
a[i] = '0';
}
I'm trying to print out every possible eight bit combination of 0s and 1s without using any libraries (except bitset if I have to). If I do it this way, I'll end up with 8 for loops, which is a bit much. I'm wondering whether there is a way to condense this using either recursion or a clever trick with using the standard do/while/for loops.
This task can be achieved with a simple for loop and binary operations.
Bitshift i by an amount, then & it by 1 to mask that bit.
#include <iostream>
void printBinary()
{
for(int i = 0; i < 256; i++){
for(int bit = 7; bit >= 0; bit--){
std::cout << (i >> bit & 1);
}
std::cout << std::endl;
}
}
First, your loops are incorrect: they run from 0 to 7, inclusive, while they should be running from 0 to 1, inclusive, because a bit is either zero or one.
As far as going through all 8-bit combinations goes, you can do it with a single loop: use an int counting from 0 to 255, inclusive, and print its binary representation:
for (int i = 0 ; i != 256 ; i++) {
cout << bitset<8>(i).to_string() << endl;
}
Related
Okay so I'm tryna create a program that:
(1) swaps my array
(2) performs caesar cipher substitution on the swapped array
(3) convert the array from (2) that is in decimal form into 8-bit binary form
And so far I've successfully done the first 2 parts but I'm facing problem with converting the array from decimal to binary form.
And this is my coding of what I've tried
#include<iostream>
#include<bits/stdc++.h>
using namespace std;
void swapfrontback(int a[], int n);
int main()
{
int a[10], i, n;
cout << "enter size" << endl;
cin >> n;
if (n == 0)
{
cout << "Array is empty!\n";
}
else
{
cout << "p = " << endl;
for (i = 0; i < n; i++)
{
cin >> a[i];
}
}
swapfrontback(a,n);
//caesar cipher
int shift = 0;
cout << "input shift: ";
cin >> shift;
int modulus = 0;
cout << "input modulus: ";
cin >> modulus;
cout << "p''=" << endl;
for (i = 0; i < n; i++)
{
a[i] = (a[i] + shift) % modulus;
cout << a[i] << endl;
}
// Function that convert Decimal to binary
int b;
b = 8;
cout<< "p'''=" << endl;
for (i = 0; i < n; i++)
{
for(int i=b-1;i>=0;i--)
{
if( a[i] & ( 1 << i ) ) cout<<1;
else cout<<0;
}
}
return 0;
}
void swapfrontback(int a[], int n)
{
int i, temp;
for (i = 0; i < n / 2; i++)
{
temp = a[i];
a[i] = a[n - i-1];
a[n - i-1] = temp;
}
cout << "p' = '" << endl;
for (i = 0; i < n; i++)
{
cout << a[i] << endl;
}
}
the problem is that instead of converting the array of decimal from the 2nd part which is the caesar cipher into its binary form, I'm getting 000000010000000100000001 .
My initial array is
3
18
25
Shift 8 and modulo 26. If anyone knows how to fix this please do help me.
Well, there seems to be something that may be an issue in the future (like the n being larger than 10, but, regarding your question, this nested for sentence is wrong.
for (i = 0; i < n; i++)
{
for(int i=b-1;i>=0;i--) //here you are using the variable 'i' twice
{
if( a[i] & ( 1 << i ) ) cout<<1; //i starts at 7, which binary representation in 4 bits is 0111
else cout<<0;
}
}
When you're using nested for sentences, it is a good idea to not repeat their iterating variables' names since they can affect each other and create nasty things like infinite loops or something like that. Try to use a different variable name instead to avoid confusion and issues:
for(int j=b-1;j>=0;j--) //this is an example
Finally, the idea behind transforming a base 10 number to its binary representation (is to use the & operator with the number 1 to know if a given bit position is a 1 (true) or 0 (false)) for example, imagine that you want to convert 14 to its binary form (00001110), the idea is to start making the & operation with the number 1, an continue with powers of 2 (since them will always be a number with a single 1 and trailing 0s) 1-1 2-10 4-100 8-1000, etc.
So you start with j = 1 and you apply the & operation between it and your number (14 in this case) so: 00000001 & 00001110 is 0 because there is not a given index in which both numbers have a '1' bit in common, so the first bit of 14 is 0, then you either multiply j by two (j*=2), or shift their bits to the left once (j = 1<<j) to move your bit one position to the left, now j = 2 (00000010), and 2 & 14 is 2 because they both have the second bit at '1', so, since the result is not 0, we know that the second bit of 14 is '1', the algorithm is something like:
int j = 128; 128 because this is the number with a '1' in the 8th bit (your 8 bit limit)
int mynumber = 14;
while(j){ // when the j value is 0, it will be the same as false
if(mynumber & j) cout<<1;
else cout<<0;
j=j>>1;
}
Hope you understand, please ensure that your numbers fit in 8 bits (255 max).
I need to find all the prime numbers from 2 to n using the Sieve of Eratosthenes. I looked on Wikipedia(Sieve of Eratosthenes) to find out what the Sieve of Eratosthenes was, and it gave me this pseudocode:
Input: an integer n > 1
Let A be an array of Boolean values, indexed by integers 2 to n,
initially all set to true.
for i = 2, 3, 4, ..., not exceeding √n:
if A[i] is true:
for j = i2, i2+i, i2+2i, i2+3i, ..., not exceeding n :
A[j] := false
Output: all i such that A[i] is true.
So I used this and translated it to C++. It looks fine to me, but I have a couple errors. Firstly, if I input 2 or 3 into n, it says:
terminate called after throwing an instance of 'Range_error'
what(): Range_error: 2
Also, whenever I enter a 100 or anything else (4, 234, 149, 22, anything), it accepts the input for n, and doesn't do anything. Here is my C++ translation:
#include "std_lib_facilities.h"
int main()
{
/* this program will take in an input 'n' as the maximum value. Then it will calculate
all the prime numbers between 2 and n. It follows the Sieve of Eratosthenes with
the algorithms from Wikipedia's pseudocode translated by me into C++*/
int n;
cin >> n;
vector<string>A;
for(int i = 2; i <= n; ++i) // fills the whole table with "true" from 0 to n-2
A.push_back("true");
for(int i = 2; i <= sqrt(n); ++i)
{
i -= 2; // because I built the vector from 0 to n-2, i need to reflect that here.
if(A[i] == "true")
{
for(int j = pow(i, 2); j <= n; j += i)
{
A[j] = "false";
}
}
}
//print the prime numbers
for(int i = 2; i <= n; ++i)
{
if(A[i] == "true")
cout << i << '\n';
}
return 0;
}
The issue is coming from the fact that the indexes are not in line with the value they are representing, i.e., they are moved down by 2. By doing this operation, they no longer have the same mathematical properties.
Basically, the value 3 is at position 1 and the value 4 is at position 2. When you are testing for division, you are using the positions as they were values. So instead of testing if 4%3==0, you are testing that 2%1=0.
In order to make your program works, you have to remove the -2 shifting of the indexes:
int main()
{
int n;
cin >> n;
vector<string>A;
for(int i = 0; i <= n; ++i) // fills the whole table with "true" from 0 to n-2
A.push_back("true");
for(int i = 2; i <= sqrt(n); ++i)
{
if(A[i] == "true")
{
for(int j = pow(i, 2); j <= n; j += i)
{
A[j] = "false";
}
}
}
//print the prime numbers
for(int i = 2; i <= n; ++i)
{
if(A[i] == "true")
cout << i << '\n';
}
return 0;
}
I agree with other comments, you could use a vector of bools. And directly initialize them with the right size and value:
std::vector<bool> A(n, false);
Here you push back n-1 elements
vector<string>A;
for(int i = 2; i <= n; ++i) // fills the whole table with "true" from 0 to n-2
A.push_back("true");
but here you access your vector from A[2] to A[n].
//print the prime numbers
for(int i = 2; i <= n; ++i)
{
if(A[i] == "true")
cout << i << '\n';
}
A has elements at positions A[0] to A[n-2]. You might correct this defect by initializing your vector differently. For example as
vector<string> A(n+1, "true");
This creates a vector A with n+1 strings with default values "true" which can be accessed through A[0] to A[n]. With this your code should run, even if it has more deficits. But I think you learn most if you just try to successfully implement the sieve and then look for (good) alternatives in the internet.
This is painful. Why are you using a string array to store boolean values, and not, let's say, an array of boolean values? Why are you leaving out the first two array elements, forcing you to do some adjustment of all indices? Which you then forget half the time, totally breaking your code? At least you should change this line:
i -= 2; // because I built the vector from 0 to n-2, i need to reflect that here.
to:
i -= 2; // because I left the first two elements out, I that here.
// But only here, doing it everywhere is too annoying.
As a result of that design decision, when you execute this line:
for(int j = pow(i, 2); j <= n; j += i)
i is actually zero which means j will stay zero forever.
while(i < length)
{
pow = 1;
for(int j = 0; j < 8; j++, pow *=2)
{
ch += (str[j] - 48) * pow;
}
str = str.substr(8);
i+=8;
cout << ch;
ch = 0;
}
This seems to be slowing my program down a lot. Is it because of the string functions I'm using in there, or is this approach wrong in general. I know there's the way where you implement long division, but I wanted to see if that was actually more efficient than this method. I can't think of another way that doesn't use the same general algorithm, so maybe it's just my implementation that is the problem.
Perhaps you want might to look into using the standard library functions. They're probably at least as optimised as anything you run through the compiler:
#include <iostream>
#include <iomanip>
#include <cstdlib>
int main (void) {
const char *str = "10100101";
// Use str.c_str() if it's a real C++ string.
long int li = std::strtol (str, 0, 2);
std::cout
<< "binary string = " << str
<< ", decimal = " << li
<< ", hex = " << std::setbase (16) << li
<< '\n';
return 0;
}
The output is:
binary string = 10100101, decimal = 165, hex = a5
You are doing some things unnecessarily, like creating a new substring for each each loop. You could just use str[i + j] instead.
It is also not necessary to multiply 0 or 1 with the power. Just use an if-statement.
while(i < length)
{
pow = 1;
for(int j = 0; j < 8; j++, pow *=2)
{
if (str[i + j] == '1')
ch += pow;
}
i+=8;
cout << ch;
ch = 0;
}
This will at least run a bit faster.
short answer could be:
long int x = strtol(your_binary_c++_string.c_str(),(char **)NULL,2)
Probably you can use int or long int like below:
Just traverse the binary number step by step, starting from 0 to n-1, where n is the most significant bit(MSB) ,
multiply them with 2 with raising powers and add the sum together. E.g to convert 1000(which is binary equivalent of 8), just do the following
1 0 0 0 ==> going from right to left
0 x 2^0 = 0
0 x 2^1 = 0;
0 x 2^2 = 0;
1 x 2^3 = 8;
now add them together i.e 0+0+0+8 = 8; this the decimal equivalent of 1000. Please read the program below to have a better understanding how the concept
work. Note : The program works only for 16-bit binary numbers(non-floating) or less. Leave a comment if anything is not clear. You are bound to receive a reply.
// Program to convert binary to its decimal equivalent
#include <iostream>
#include <math.h>
int main()
{
int x;
int i=0,sum = 0;
// prompts the user to input a 16-bit binary number
std::cout<<" Enter the binary number (16-bit) : ";
std::cin>>x;
while ( i != 16 ) // runs 16 times
{
sum += (x%10) * pow(2,i);
x = x/10;
i++;
}
std::cout<<"\n The decimal equivalent is : "<<sum;
return 0;
}
How about something like:
int binstring_to_int(const std::string &str)
{
// 16 bits are 16 characters, but -1 since bits are numbered 0 to 15
std::string::size_type bitnum = str.length() - 1;
int value = 0;
for (auto ch : str)
{
value |= (ch == '1') << bitnum--;
}
return value;
}
It's the simplest I can think of. Note that this uses the new C++11 for-each loop construct, if your compiler can't handle it you can use
for (std::string::const_iterator i = str.begin(); i != str.end(); i++)
{
char ch = *i;
// ...
}
Minimize the number of operations and don't compute things more than once. Just multiply and move up:
unsigned int result = 0;
for (char * p = str; *p != 0; ++p)
{
result *= 2;
result += (*p - '0'); // this is either 0 or 1
}
The scheme is readily generalized to any base < 10.
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'.
I am making a program that adds two binary numbers (up to 31 digits) together and outputs the sum in binary.
I have every thing working great but I need to remove the leading zeros off the solution.
This is what my output is:
char c[32];
int carry = 0;
if(carry == '1')
{
cout << carry;
}
for(i = 0; i < 32; i++)
{
cout << c[i];
}
I tried this but it didn't work:
char c[32];
int carry = 0;
bool flag = false;
if(carry == '1')
{
cout << carry;
}
for(i=0; i<32; i++)
{
if(c[i] != 0)
{
flag = true;
if(flag)
{
for(i = 0; i < 32; i++)
{
cout << c[i];
}
}
}
}
Any ideas or suggestions would be appreciated.
EDIT: Thank you everyone for your input, I got it to work!
You should not have that inner loop (inside if(flag)). It interferes with the i processing of the outer loop.
All you want to do at that point is to output the character if the flag is set.
And on top of that, the printing of the bits should be outside the detection of the first bit.
The following pseudo-code shows how I'd approach this:
set printing to false
if carry is 1:
output '1:'
for each bit position i:
if c[i] is 1:
set printing to true
if printing:
output c[i]
if not printing:
output 0
The first block of code may need to be changed to accurately output the number with carry. For example, if you ended up with the value 2 and a carry, you would want either of:
1:10 (or some other separator)
100000000000000000000000000000010 (33 digits)
Simply outputting 110 with no indication that the leftmost bit was a carry could either be:
2 with carry; or
6 without carry
The last block ensures you have some output for the value 0 which would otherwise print nothing since there were no 1 bits.
I'll leave it up to you whether you should output a separator between carry and value (and leave that line commented out) or use carry to force printing to true initially. The two options would be respectively:
if carry is 1:
output '1 '
and:
if carry is 1:
output 1
set printing to true
And, since you've done the conversion to C++ in a comment, that should be okay. You state that it doesn't work, but I typed in your code and it worked fine, outputting 10:
#include <iostream>
int main(void)
{
int i;
int carry = 0;
int c[] = {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0};
bool print = false;
// This is the code you gave in the comment, slightly modified.
// vvvvvv
if(carry == 1) {
std::cout << carry << ":";
}
for (i = 0; i < 32; i++) {
if (c[i] == 1) {
print = true;
}
if (print) {
std::cout << c[i];
}
}
// ^^^^^^
std::cout << std::endl;
return 0;
}
const char * begin = std::find(c, c+32, '1');
size_t len = c - begin + 32;
std::cout.write(begin, len);
Use two fors over the same index. The first for iterates while == 0, the second one prints starting from where the first one left off.