For this program, I will input a Binary number and it will convert into a decimal number. At the end I wanted to return the number of digits in the Binary number that I had input. For example, 1001 - 4 Binary digits.The output of the digits of Binary number is always 0. Should I use size_type to do it ?
#include<iostream>
#include<string>
#include<bitset>
#include<limits>
#include<algorithm>
using namespace std;
int multiply(int x);
int multiply(int x)
{
if (x == 0)
{
return 1;
}
if (x == 1)
{
return 2;
}
else
{
return 2 * multiply(x - 1);
}
}
int count_set_bit(int n)
{
int count = 0;
while (n != 0)
{
if (n & 1 == 1)
{
count++;
}
n = n >> 1;
}
return count;
}
int main()
{
string binary;
cout << "\n\tDualzahlen : ";
cin >> binary;
reverse(binary.begin(), binary.end());
int sum = 0;
int size = binary.size();
for (int x = 0; x < size; x++)
{
if (binary[x] == '1')
{
sum = sum + multiply(x);
}
}
cout << "\tDezimal : " << sum << endl;
int n{};
cout << "\tAnzahl der Stelle : " << count_set_bit(n) << endl;
}
It looks like you are on the right track for unsigned integers. Signed integers in a multiply are generally converted to positive with a saved result sign.
You can save some time with a bit data solution, as value testing is not cost free plus 'and' has no carry delay. Of course, some CPUs can ++ faster than += 1, but hopefully the compiler knows how to use that:
int bits( unsigned long num ){
retVal = 0 ;
while ( num ){
retVal += ( num & 1 );
num >>= 1 ;
}
return retVal ;
}
I recall the H-4201 multiplied 2 bits at a time, using maybe shift, maybe add, maybe carry/borrow, so 0 was no add, 1 was add, 2 was shift and add, 3 was carry/borrow add (4 - 1)! That was before IC multiply got buried in circuits and ROMs. :D
Related
I have an assignment to make a program that should convert a number from it's integer value to a binary value. For some reason my array is always filled with zeroes and won't add "1"'s from my if statements. I know there are probably solutions to this assignment on internet but I would like to understand what is problem with my code. Any help is appreciated.
Here is what I tried:
#include <iostream>
/*Write a code that will enable input of one real number in order to write out it's binary equivalent.*/
int main() {
int number;
int binaryNumber[32] = { 0 };
std::cout << "Enter your number: ";
std::cin >> number;
while (number > 0) {
int i = 0;
if ((number / 10) % 2 == 0) {
binaryNumber[i] = 0;
}
if ((number / 10) % 2 != 0) {
binaryNumber[i] = 1;
}
number = number / 10;
i++;
}
for (int i = 31; i >= 0; i--) {
std::cout << binaryNumber[i];
}
return 0;
}
You need to remove number/10 in both the if statements. Instead, just use number. you need the last digit every time to get the ith bit.
Moreover, you need to just half the number in every iteration rather than doing it /10.
// Updated Code
int main() {
int number;
int binaryNumber[32] = { 0 };
std::cout << "Enter your number: ";
std::cin >> number;
int i = 0;
while (number > 0) {
if (number % 2 == 0) {
binaryNumber[i] = 0;
}
if (number % 2 != 0) {
binaryNumber[i] = 1;
}
number = number / 2;
i++;
}
for (int i = 31; i >= 0; i--) {
std::cout << binaryNumber[i];
}
return 0;
}
The first thing is the variable 'i' in the while loop. Consider it more precisely: every time you iterate over it, 'i' is recreated again and assigned the value of zero. It's the basics of the language itself.
The most relevant mistake is logic of your program. Each iteration we must take the remainder of division by 2, and then divide our number by 2.
The correct code is:
#include <iostream>
int main()
{
int x = 8;
bool repr[32]{};
int p = 0;
while(x)
{
repr[p] = x % 2;
++p;
x /= 2;
}
for(int i = 31; i >= 0; --i)
std::cout << repr[i];
return 0;
}
... is always filled with zeroes ... I would like to understand what is problem with my code
int i = 0; must be before the while, having it inside you only set the index 0 of the array in your loop because i always values 0.
But there are several other problems in your code :
using int binaryNumber[32] you suppose your int are on 32bits. Do not use 32 but sizeof(int)*CHAR_BIT, and the same for your last loop in case you want to also write 0 on the left of the first 1
you look at the value of (number / 10) % 2, you must look at the value of number % 2
it is useless to do the test then its reverse, just use else, or better remove the two ifs and just do binaryNumber[i] = number & 1;
number = number / 10; is the right way when you want to produce the value in decimal, in binary you have to divide by 2
in for (int i = 31; i >= 0; i--) { except for numbers needing 32 bits you will write useless 0 on the left, why not using the value of i from the while ?
There are some logical errors in your code.
You have taken (number/10) % 2, instead, you have to take (number %2 ) as you want the remainder.
Instead of taking i = 31, you should use this logic so you can print the following binary in reverse order:
for (int j = i - 1; j >= 0; j--)
{
cout << BinaryNumb[j];
}
Here is the code to convert an integer to its binary equivalent:
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
// function to convert integer to binary
void DecBinary(int n)
{
// Array to store binary number
int BinaryNumb[32];
int i = 0;
while (n > 0)
{
// Storing remainder in array
BinaryNumb[i] = n % 2;
n = n / 2;
i++;
}
// Printing array in reverse order
for (int j = i - 1; j >= 0; j--)
{
cout << BinaryNumb[j];
}
}
// Main Program
int main()
{
int testcase;
//Loop is optional
for(int i = 0; i < testcase; i++)
{
cin >> n;
DecToBinary(n);
}
return 0;
}
I am trying to write a code that takes a binary number input as a string and will only accept 1's or 0's if not there should be an error message displayed. Then it should go through a loop digit by digit to convert the binary number as a string to decimal. I cant seem to get it right I have the fact that it will only accept 1's or 0's correct. But then when it gets into the calculations something messes up and I cant seem to get it correct. Currently this is the closest I believe I have to getting it working. could anyone give me a hint or help me with what i am doing wrong?
#include <iostream>
#include <string>
using namespace std;
string a;
int input();
int main()
{
input();
int decimal, x= 0, length, total = 0;
length = a.length();
// atempting to make it put the digits through a formula backwords.
for (int i = length; i >= 0; i--)
{
// Trying to make it only add the 2^x if the number is 1
if (a[i] = '1')
{
//should make total equal to the old total plus 2^x if a[i] = 1
total = total + pow(x,2);
}
//trying to let the power start at 0 and go up each run of the loop
x++;
}
cout << endl << total;
int stop;
cin >> stop;
return 0;
}
int input()
{
int x, x2, count, repeat = 0;
while (repeat == 0)
{
cout << "Enter a string representing a binary number => ";
cin >> a;
count = a.length();
for (x = 0; x < count; x++)
{
if (a[x] != '0' && a[x] != '1')
{
cout << a << " is not a string representing a binary number>" << endl;
repeat = 0;
break;
}
else
repeat = 1;
}
}
return 0;
}
I don't think that pow suits for integer calculation. In this case, you can use shift operator.
a[i] = '1' sets the value of a[i] to '1' and return '1', which is always true.
You shouldn't access a[length], which should be meaningless.
fixed code:
int main()
{
input();
int decimal, x= 0, length, total = 0;
length = a.length();
// atempting to make it put the digits through a formula backwords.
for (int i = length - 1; i >= 0; i--)
{
// Trying to make it only add the 2^x if the number is 1
if (a[i] == '1')
{
//should make total equal to the old total plus 2^x if a[i] = 1
total = total + (1 << x);
}
//trying to let the power start at 0 and go up each run of the loop
x++;
}
cout << endl << total;
int stop;
cin >> stop;
return 0;
}
I would use this approach...
#include <iostream>
using namespace std;
int main()
{
string str{ "10110011" }; // max length can be sizeof(int) X 8
int dec = 0, mask = 1;
for (int i = str.length() - 1; i >= 0; i--) {
if (str[i] == '1') {
dec |= mask;
}
mask <<= 1;
}
cout << "Decimal number is: " << dec;
// system("pause");
return 0;
}
Works for binary strings up to 32 bits. Swap out integer for long to get 64 bits.
#include <iostream>
#include <stdio.h>
#include <string>
using namespace std;
string getBinaryString(int value, unsigned int length, bool reverse) {
string output = string(length, '0');
if (!reverse) {
for (unsigned int i = 0; i < length; i++) {
if ((value & (1 << i)) != 0) {
output[i] = '1';
}
}
}
else {
for (unsigned int i = 0; i < length; i++) {
if ((value & (1 << (length - i - 1))) != 0) {
output[i] = '1';
}
}
}
return output;
}
unsigned long getInteger(const string& input, size_t lsbindex, size_t msbindex) {
unsigned long val = 0;
unsigned int offset = 0;
if (lsbindex > msbindex) {
size_t length = lsbindex - msbindex;
for (size_t i = msbindex; i <= lsbindex; i++, offset++) {
if (input[i] == '1') {
val |= (1 << (length - offset));
}
}
}
else { //lsbindex < msbindex
for (size_t i = lsbindex; i <= msbindex; i++, offset++) {
if (input[i] == '1') {
val |= (1 << offset);
}
}
}
return val;
}
int main() {
int value = 23;
cout << value << ": " << getBinaryString(value, 5, false) << endl;
string str = "01011";
cout << str << ": " << getInteger(str, 1, 3) << endl;
}
I see multiple misstages in your code.
Your for-loop should start at i = length - 1 instead of i = length.
a[i] = '1' sets a[i] to '1' and does not compare it.
pow(x,2) means and not . pow is also not designed for integer operations. Use 2*2*... or 1<<e instead.
Also there are shorter ways to achieve it. Here is a example how I would do it:
std::size_t fromBinaryString(const std::string &str)
{
std::size_t result = 0;
for (std::size_t i = 0; i < str.size(); ++i)
{
// '0' - '0' == 0 and '1' - '0' == 1.
// If you don't want to assume that, you can use if or switch
result = (result << 1) + str[i] - '0';
}
return result;
}
I think I've almost got it, but I feel like I'm go in circles trying to figure this out.
The challenge to out cout without using strings or arrays. I took the number 56 as an example and 56 should equal 111000 this is not the case as it goes through fine till 7 then the number equals number*2 + number%2 makes it equal to 15 and outputs all 1's. Idk anymore, this is driving me to the moon and back.
#include <iostream>
using namespace std;
int main()
{
int number = 0;
int n = 1;
int x = n;
cin>>number;
cout<<n%2;
while(n <= number)
{
if(n%2 == 0)
{
n = n*2;
cout<<0;
}
else
{
n = n*2 + n%2;
cout<<n%2;
}
}
}
You can use the binary operator & to check if a single bit is 1 or 0.
for (int i=512; i>0; i/=2) {
cout << ( ( number & i ) != 0 ) ;
}
Note that this WILL print leading 0's.
Also, I'm assuming you only want to print positive integers.
Alternative:
for (int i=512; i>0; i/=2) {
if (number >= i) {
cout << 1;
number -= i;
} else {
count << 0;
}
}
You can use recursion
void decimal_to_binary(int decimal)
{
int remainder = decimal % 2;
if (decimal < 1)
return;
decimal_to_binary(decimal / 2);
cout << remainder;
}
This function will take the decimal, get its remainder when divided to 2. Before it the function call itself again, it checks if the decimal is less than 1(probably 0) and return to execute the printing of 1's and 0's
I had this type of problem assigned to me recently. This code example work up to a maximum of 10 binary digits (per the problem guidelines) and keep prompting for input until 0 is entered (sentinel value). This can certainly be improved but the math is correct:
#include <iostream>
#include <cmath>
using namespace std;
int main ()
{
//Declare Variables
int inputValue = 0;
int workingValue = 0;
int conversionSum = 0;
//Begin Loop
do{
//Prompt for input
cout << "Enter a binary integer (0 to quit): ";
cin >> inputValue;
//Reset Variables
workingValue = inputValue;
conversionSum = 0;
//Begin processing input
//10 digits max, so 10 iterations
for (int i=0; i<10; i++) {
//Check for non-binary entry
if ((workingValue % 10) != 1 && (workingValue % 10 != 0)){
cout << "Invalid!\n";
workingValue = 0;
conversionSum = 0;
break;
}
//check to see if 2^i should be added to sum
if (workingValue%2 == 1){
conversionSum += pow(2,i);
workingValue--;
}
//divide by 10 and continue loop
workingValue= workingValue / 10;
}
//output results
cout << "converted to decimal is: " << conversionSum << endl;
}while (inputValue != 0);
}
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
cout << "enter a number";
int number, n, a=0;
cin >> number;
n = number;
do
{
n=n/2;
a=a+1;
}
while (n>=1);
cout << "a is" << a;
int c = a;
int b = a;
cout << "binary is";
for(int i=0; i<=c; i++)
{
int k = number / pow(2,b);
cout << k;
number = number - k * pow(2,b);
b = b-1;
}
return 0;
}
Although asked in C I have used C++. I have used the logic that if you have to convert decimal to binary we have to find the maximum power of 2 contained in the number which when added by 1 becomes the number of digit of required binary .. leftmost digit is the number of highest available power of 2 (ex in 8 highest power of 2 is 3 and 1 such is available)...then subtract this from the number and (ex 8-8=0)and search for number of next highest available power of 2 and so on.
If we have classical decimal format in int long, we can do something like that:
const long int NUMBER = 4577;
const long int DIGIT_TO_FIND = 5;
long int thisNumber = NUMBER >= 0 ? NUMBER : -NUMBER;
long int thisDigit;
while (thisNumber != 0)
{
thisDigit = thisNumber % 10;
thisNumber = thisNumber / 10;
if (thisDigit == DIGIT_TO_FIND)
{
printf("%d contains digit %d", NUMBER, DIGIT_TO_FIND);
break;
}
}
But what about binary representing or octal representing in int long?
We have:
const long int NUMBER = 01011111; // octal
const long int DIGIT_TO_FIND1 = 0;
const long int DIGIT_TO_FIND2 = 1;
Correct input:
01010101
11111111
Bad input:
02010101 (because two)
00000009 (because nine)
We need to check if int long contains only 0 or 1.
What is the simpliest way to check correct input for that? Maybe just easy question, but no idea, thank you.
To check whether a digit is a valid binary digit, compare it with 1:
if (thisDigit > 1)
{
printf("%d contains digit %d", NUMBER, thisDigit);
break;
}
As noted by #Pubby, you have an octal number instead of a decimal number, so use 8 instead of 10 to calculate digits:
thisDigit = thisNumber % 8;
thisNumber = thisNumber / 8;
If I understand your question something like this -
#include <iostream>
bool check(int number, int search_1, int search_2)
{
while(number > 0)
{
int digit = number % 10;
number = number / 10;
if (digit != search_1 && digit != search_2)
return false;
}
return true;
}
int main()
{
const int DIGIT_TO_FIND1 = 0;
const int DIGIT_TO_FIND2 = 1;
int number1 = 1001001;
std::cout << "Checking " << number1 << " : " << check(number1, DIGIT_TO_FIND1, DIGIT_TO_FIND2) <<" \n";
int number2 = 12110101;
std::cout << "Checking " << number2 << " : " << check(number2, DIGIT_TO_FIND1, DIGIT_TO_FIND2) <<" \n";
}
My solution. It's really simple.
bool check(int long number)
{
return ((number % 10 != 0) && (number % 10) != 1);
}
I have an integer:
int iNums = 12476;
And now I want to get each digit from iNums as integer. Something like:
foreach(iNum in iNums){
printf("%i-", iNum);
}
So the output would be: "1-2-4-7-6-".
But i actually need each digit as int not as char.
Thanks for help.
void print_each_digit(int x)
{
if(x >= 10)
print_each_digit(x / 10);
int digit = x % 10;
std::cout << digit << '\n';
}
Convert it to string, then iterate over the characters. For the conversion you may use std::ostringstream, e.g.:
int iNums = 12476;
std::ostringstream os;
os << iNums;
std::string digits = os.str();
Btw the generally used term (for what you call "number") is "digit" - please use it, as it makes the title of your post much more understandable :-)
Here is a more generic though recursive solution that yields a vector of digits:
void collect_digits(std::vector<int>& digits, unsigned long num) {
if (num > 9) {
collect_digits(digits, num / 10);
}
digits.push_back(num % 10);
}
Being that there are is a relatively small number of digits, the recursion is neatly bounded.
Here is the way to perform this action, but by this you will get in reverse order.
int num;
short temp = 0;
cin>>num;
while(num!=0){
temp = num%10;
//here you will get its element one by one but in reverse order
//you can perform your action here.
num /= 10;
}
I don't test it just write what is in my head. excuse for any syntax error
Here is online ideone demo
vector <int> v;
int i = ....
while(i != 0 ){
cout << i%10 << " - "; // reverse order
v.push_back(i%10);
i = i/10;
}
cout << endl;
for(int i=v.size()-1; i>=0; i--){
cout << v[i] << " - "; // linear
}
To get digit at "pos" position (starting at position 1 as Least Significant Digit (LSD)):
digit = (int)(number/pow(10,(pos-1))) % 10;
Example: number = 57820 --> pos = 4 --> digit = 7
To sequentially get digits:
int num_digits = floor( log10(abs(number?number:1)) + 1 );
for(; num_digits; num_digits--, number/=10) {
std::cout << number % 10 << " ";
}
Example: number = 57820 --> output: 0 2 8 7 5
You can do it with this function:
void printDigits(int number) {
if (number < 0) { // Handling negative number
printf('-');
number *= -1;
}
if (number == 0) { // Handling zero
printf('0');
}
while (number > 0) { // Printing the number
printf("%d-", number % 10);
number /= 10;
}
}
Drawn from D.Shawley's answer, can go a bit further to completely answer by outputing the result:
void stream_digits(std::ostream& output, int num, const std::string& delimiter = "")
{
if (num) {
stream_digits(output, num/10, delimiter);
output << static_cast<char>('0' + (num % 10)) << delimiter;
}
}
void splitDigits()
{
int num = 12476;
stream_digits(std::cout, num, "-");
std::cout << std::endl;
}
I don't know if this is faster or slower or worthless, but this would be an alternative:
int iNums = 12476;
string numString;
stringstream ss;
ss << iNums;
numString = ss.str();
for (int i = 0; i < numString.length(); i++) {
int myInt = static_cast<int>(numString[i] - '0'); // '0' = 48
printf("%i-", myInt);
}
I point this out as iNums alludes to possibly being user input, and if the user input was a string in the first place you wouldn't need to go through the hassle of converting the int to a string.
(to_string could be used in c++11)
I know this is an old post, but all of these answers were unacceptable to me, so I wrote my own!
My purpose was for rendering a number to a screen, hence the function names.
void RenderNumber(int to_print)
{
if (to_print < 0)
{
RenderMinusSign()
RenderNumber(-to_print);
}
else
{
int digits = 1; // Assume if 0 is entered we want to print 0 (i.e. minimum of 1 digit)
int max = 10;
while (to_print >= max) // find how many digits the number is
{
max *= 10;
digits ++;
}
for (int i = 0; i < digits; i++) // loop through each digit
{
max /= 10;
int num = to_print / max; // isolate first digit
to_print -= num * max; // subtract first digit from number
RenderDigit(num);
}
}
}
Based on #Abyx's answer, but uses div so that only 1 division is done per digit.
#include <cstdlib>
#include <iostream>
void print_each_digit(int x)
{
div_t q = div(x, 10);
if (q.quot)
print_each_digit(q.quot);
std::cout << q.rem << '-';
}
int main()
{
print_each_digit(12476);
std::cout << std::endl;
return 0;
}
Output:
1-2-4-7-6-
N.B. Only works for non-negative ints.
My solution:
void getSumDigits(int n) {
std::vector<int> int_to_vec;
while(n>0)
{
int_to_vec.push_back(n%10);
n=n/10;
}
int sum;
for(int i=0;i<int_to_vec.size();i++)
{
sum+=int_to_vec.at(i);
}
std::cout << sum << ' ';
}
The answer I've used is this simple function:
int getDigit(int n, int position) {
return (n%(int)pow(10, position) - (n % (int)pow(10, position-1))) / (int)pow(10, position-1);
}
Hope someone finds this helpful!
// Online C++ compiler to run C++ program online
#include <iostream>
#include <cmath>
int main() {
int iNums = 123458;
// int iNumsSize = 5;
int iNumsSize = trunc(log10(iNums)) + 1; // Find length of int value
for (int i=iNumsSize-1; i>=0; i--) {
int y = pow(10, i);
// The pow() function returns the result of the first argument raised to
the power of the second argument.
int z = iNums/y;
int x2 = iNums / (y * 10);
printf("%d ",z - x2*10 ); // Print Values
}
return 0;
}
You can do it using a while loop and the modulo operators.
It just gives the digits in the revese order.
int main() {
int iNums = 12476;
int iNum = 0;
while(iNums > 0) {
iNum = iNums % 10;
cout << iNum;
iNums = iNums / 10;
}
}
int a;
cout << "Enter a number: ";
cin >> a;
while (a > 0) {
cout << a % 10 << endl;
a = a / 10;
}
int iNums = 12345;
int iNumsSize = 5;
for (int i=iNumsSize-1; i>=0; i--) {
int y = pow(10, i);
int z = iNums/y;
int x2 = iNums / (y * 10);
printf("%d-",z - x2*10 );
}