I am working on decimal to binary conversion. I can convert them, using
char bin_x [10];
itoa (x,bin_x,2);
but the problem is, i want answer in 8 bits. And it gives me as, for example x =5, so output will be 101, but i want 00000101.
Is there any way to append zeros in the start of array? or is it possible to get answer in 8 bits straight away? I am doing this in C++
In C++, the easiest way is probably to use a std::bitset:
#include <iostream>
#include <bitset>
int main() {
int x = 5;
std::bitset<8> bin_x(x);
std::cout << bin_x;
return 0;
}
Result:
00000101
To print out the bits of a single digit, you need to do the following:
//get the digit (in this case, the least significant digit)
short digit = number % 10; //shorts are 8 bits
//print out each bit of the digit
for(int i = 0; i < 8; i++){
if(0x80 & digit) //if the high bit is on, print 1
cout << 1;
else
cout << 0; //otherwise print 0
digit = digit << 1; //shift the bits left by one to get the next highest bit.
}
itoa() is not a standard function so it's not good to use it if you want to write portable code.
You can also use something like that:
std::string printBinary(int num, int bits) {
std::vector<char> digits(bits);
for (int i = 0; i < bits; ++i) {
digits.push_back(num % 2 + '0');
num >>= 1;
}
return std::string(digits.rbegin(), digits.rend());
}
std:: cout << printBinary(x, 8) << std::endl;
However I must agree that using bitset would be better.
Related
void decimaltobin()
{
binaryNum = 0;
m = 1;
while (num != 0)
{
rem = num % 2;
num /= 2;
binaryNum += rem * m;
m *= 10;
}
}
Just wondering if there was an easy fix to get this function to print an 8-bit binary number instead of a 4-bit number, e.g. 0000 0101 instead of 0101.
As mentioned in the comments, your code does not print anything yet and the data type of binaryNum is not clear. Here is a working solution.
#include <iostream>
using namespace std;
void decToBinary(int n)
{
// array to store binary number
int binaryNum[32];
// counter for binary array
int i = 0;
while (n > 0) {
// storing remainder in binary array
binaryNum[i] = n % 2;
n = n / 2;
i++;
}
// printing the required number of zeros
int zeros = 8 - i;
for(int m = 0; m < zeros; m++){
cout<<0;
}
// printing binary array in reverse order
for (int j = i - 1; j >= 0; j--)
cout << binaryNum[j];
}
// Driver program to test above function
int main()
{
int n = 17;
decToBinary(n);
return 0;
}
The code implements the following:
Store the remainder when the number is divided by 2 in an array.
Divide the number by 2
Repeat the above two steps until the number is greater than zero.
Print the required number of zeros. That is 8 - length of the binary number. Note that this code will work for numbers that can be expressed in 8 bits only.
Print the array in reverse order now
Ref
Maybe I am missing your reason but why do you want to code from scratch instead of using a standard library?
You may use standard c++ without having to code a conversion from scratch using for instance std::bitset<NB_OF_BITS>.
Here is a simple example:
#include <iostream>
#include <bitset>
std::bitset<8> decimalToBin(int numberToConvert)
{
return std::bitset<8>(numberToConvert);
}
int main() {
int a = 4, b=8, c=12;
std::cout << decimalToBin(a)<< std::endl;
std::cout << decimalToBin(b)<< std::endl;
std::cout << decimalToBin(c)<< std::endl;
}
It outputs:
00000100
00001000
00001100
I have a rand number generator.
Now my question is how do I get for instance the first/second/third/fourth digit of the generated random number.
Implementing this so the user can use a hint when guessing the number.
example: result(rand): 9876
print hint 1: second number 8
print hint 2: fourth number 6
Whats the best way to tackle this? I've been thinking to convert the string to an char array in order to print out the certain locations where the values are being kept but that won't work I guess.
Correct me if my way of asking this questions is very bold.
int nummer = 4;
std :: string result = "";
for (int i = 0; i < nummer; i++)
{
result.push_back(rand()%10 + '0');
}
You have two options for solving this problem:
Create one random number between 0 and 9999 and then calculate the individual digits.
Create four random numbers between 0 and 9 which represent the individual digits and then, if necessary, calculate the whole number from the individual digits.
Normally, doing option #1 would be more straightforward. You seem to be going for option #2. Both ways are possible and which one is better probably depends on whether your program works more with the number as a whole or with individual digits.
If you decide to do option #2, then the question arises whether you want to work with ASCII character codes between '0' and '9' (i.e. codes between 48 and 57) or with actual numbers between 0 and 9. Normally, it is a bit easier to work with the actual numbers instead of ASCII character codes, but both ways are feasible.
Personally, I would solve it the following way:
#include <iostream>
#include <cstdlib>
#include <ctime>
int main()
{
//seed random number generator
std::srand( std::time(nullptr) );
//note that NUM_DIGITS should not be set so high
//that MAX_NUMBER is not representable as an `int`
constexpr int NUM_DIGITS = 4;
//set this to the highest possible number that
//corresponds to NUM_DIGITS
constexpr int MAX_NUMBER = 9999;
//this variable holds the entire number
int random;
//this array holds the individual digits
int digits[NUM_DIGITS];
///generate the random number
random = std::rand()%(MAX_NUMBER+1);
//calculate the individual digits of the random number
int remaining = random;
int divisor = (MAX_NUMBER+1) / 10;
for ( int i = 0; i < NUM_DIGITS; i++ )
{
digits[i] = remaining / divisor;
remaining = remaining % divisor;
divisor/= 10;
}
//print the whole random number
std::cout << "Whole random number:\n" << random << "\n\n";
//print the individual digits
std::cout << "Individual digits:\n";
for ( int i = 0; i < NUM_DIGITS; i++ )
{
std::cout << digits[i] << "\n";
}
}
Sample output (actual output depends on random number seed):
Whole random number:
8695
Individual digits:
8
6
9
5
I've been thinking to convert the string to an char array but that won't work I guess.
I see no reason to do this. In your posted code, you can simply write
result[0]
result[1]
result[2]
result[3]
to access the individual digits of the number.
In that respect, converting a std::string to a C-style character array has no advantages.
The code you've written is fine.
Whether you store the characters in an array or a string, you can access the elements using result[i] where i starts at 0.
I've been thinking to convert the string to an char array but that won't work I guess.
Using std::string is usually a better idea - still possible but a little harder to screw them up, and they're more powerful generally, but you could use a char array if you wanted:
char result[] = "0000"; // will have null terminator
for (int i = 0; i < 4; ++i)
result[i] = rand() % 10 + '0';
Letting result be a 5-character array - with a null terminator - means you can still print all four digits easily with std::cout << result.
Alternatively, you could pick a random 4-digit number and convert it to a string of a particular width, using '0's to pad:
#include <iostream>
#include <iomanip>
int main() {
std::ostringstream oss;
oss << std::setw(4) << std::setfill('0') << rand() % 10000;
std::cout << '[' << oss.str() << "]\n";
}
convert a positive integer number in C++ (0 to 2,147,483,647) to a 32 bit binary and display.
I want do it in traditional "mathematical" way (rather than use bitset or use vector *.pushback* or recursive function or some thing special in C++...), (one reason is so that you can implement it in different languages, well maybe)
So I go ahead and implement a simple program like this:
#include <iostream>
using namespace std;
int main()
{
int dec,rem,i=1,sum=0;
cout << "Enter the decimal to be converted: ";
cin>>dec;
do
{
rem=dec%2;
sum=sum + (i*rem);
dec=dec/2;
i=i*10;
} while(dec>0);
cout <<"The binary of the given number is: " << sum << endl;
system("pause");
return 0;
}
Problem is when you input a large number such as 9999, result will be a negative or some weird number because sum is integer and it can't handle more than its max range, so you know that a 32 bit binary will have 32 digits so is it too big for any number type in C++?. Any suggestions here and about display 32 bit number as question required?
What you get in sum as a result is hardly usable for anything but printing. It's a decimal number which just looks like a binary.
If the decimal-binary conversion is not an end in itself, note that numbers in computer memory are already represented in binary (and it's not the property of C++), and the only thing you need is a way to print it. One of the possible ways is as follows:
int size = 0;
for (int tmp = dec; tmp; tmp >>= 1)
size++;
for (int i = size - 1; i >= 0; --i)
cout << ((dec >> i) & 1);
Another variant using a character array:
char repr[33] = { 0 };
int size = 0;
for (int tmp = dec; tmp; tmp >>= 1)
size++;
for (int i = 0; i < size; ++i)
repr[i] = ((dec >> (size - i - 1)) & 1) ? '1' : '0';
cout << repr << endl;
Note that both variants don't work if dec is negative.
You have a number and want its binary representation, i.e, a string. So, use a string, not an numeric type, to store your result.
Using a for-loop, and a predefined array of zero-chars:
#include <iostream>
using namespace std;
int main()
{
int dec;
cout << "Enter the decimal to be converted: ";
cin >> dec;
char bin32[] = "00000000000000000000000000000000";
for (int pos = 31; pos >= 0; --pos)
{
if (dec % 2)
bin32[pos] = '1';
dec /= 2;
}
cout << "The binary of the given number is: " << bin32 << endl;
}
For performance reasons, you may prematurely suspend the for loop:
for (int pos = 31; pos >= 0 && dec; --pos)
Note, that in C++, you can treat an integer as a boolean - everything != 0 is considered true.
You could use an unsigned integer type. However, even with a larger type you will eventually run out of space to store binary representations. You'd probably be better off storing them in a string.
As others have pointed out, you need to generate the results in a
string. The classic way to do this (which works for any base between 2 and 36) is:
std::string
toString( unsigned n, int precision, unsigned base )
{
assert( base >= 2 && base <= 36 );
static char const digits[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
std::string retval;
while ( n != 0 ) {
retval += digits[ n % base ];
n /= base;
}
while ( retval.size() < precision ) {
retval += ' ';
}
std::reverse( retval.begin(), retval.end() );
return retval;
}
You can then display it.
Recursion. In pseudocode:
function toBinary(integer num)
if (num < 2)
then
print(num)
else
toBinary(num DIV 2)
print(num MOD 2)
endif
endfunction
This does not handle leading zeros or negative numbers. The recursion stack is used to reverse the binary bits into the standard order.
Just write:
long int dec,rem,i=1,sum=0
Instead of:
int dec,rem,i=1,sum=0;
That should solve the problem.
I am currently working on a basic program which converts a binary number to an octal. Its task is to print a table with all the numbers between 0-256, with their binary, octal and hexadecimal equivalent. The task requires me only to use my own code (i.e. using loops etc and not in-built functions). The code I have made (it is quite messy at the moment) is as following (this is only a snippit):
int counter = ceil(log10(fabs(binaryValue)+1));
int iter;
if (counter%3 == 0)
{
iter = counter/3;
}
else if (counter%3 != 0)
{
iter = ceil((counter/3));
}
c = binaryValue;
for (int h = 0; h < iter; h++)
{
tempOctal = c%1000;
c /= 1000;
int count = ceil(log10(fabs(tempOctal)+1));
for (int counter = 0; counter < count; counter++)
{
if (tempOctal%10 != 0)
{
e = pow(2.0, counter);
tempDecimal += e;
}
tempOctal /= 10;
}
octalValue += (tempDecimal * pow(10.0, h));
}
The output is completely wrong. When for example the binary code is 1111 (decimal value 15), it outputs 7. I can understand why this happens (the last three digits in the binary number, 111, is 7 in decimal format), but can't be able to identify the problem in the code. Any ideas?
Edit: After some debugging and testing I figured the answer.
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
while (true)
{
int binaryValue, c, tempOctal, tempDecimal, octalValue = 0, e;
cout << "Enter a binary number to convert to octal: ";
cin >> binaryValue;
int counter = ceil(log10(binaryValue+1));
cout << "Counter " << counter << endl;
int iter;
if (counter%3 == 0)
{
iter = counter/3;
}
else if (counter%3 != 0)
{
iter = (counter/3)+1;
}
cout << "Iterations " << iter << endl;
c = binaryValue;
cout << "C " << c << endl;
for (int h = 0; h < iter; h++)
{
tempOctal = c%1000;
cout << "3 digit binary part " << tempOctal << endl;
int count = ceil(log10(tempOctal+1));
cout << "Digits " << count << endl;
tempDecimal = 0;
for (int counterr = 0; counterr < count; counterr++)
{
if (tempOctal%10 != 0)
{
e = pow(2.0, counterr);
tempDecimal += e;
cout << "Temp Decimal value 0-7 " << tempDecimal << endl;
}
tempOctal /= 10;
}
octalValue += (tempDecimal * pow(10.0, h));
cout << "Octal Value " << octalValue << endl;
c /= 1000;
}
cout << "Final Octal Value: " << octalValue << endl;
}
system("pause");
return 0;
}
This looks overly complex. There's no need to involve floating-point math, and it can very probably introduce problems.
Of course, the obvious solution is to use a pre-existing function to do this (like { char buf[32]; snprintf(buf, sizeof buf, "%o", binaryValue); } and be done, but if you really want to do it "by hand", you should look into using bit-operations:
Use binaryValue & 3 to mask out the three lowest bits. These will be your next octal digit (three bits is 0..7, which is one octal digit).
use binaryValue >>= 3 to shift the number to get three new bits into the lowest position
Reverse the number afterwards, or (if possible) start from the end of the string buffer and emit digits backwards
It don't understand your code; it seems far too complicated. But one
thing is sure, if you are converting an internal representation into
octal, you're going to have to divide by 8 somewhere, and do a % 8
somewhere. And I don't see them. On the other hand, I see a both
operations with both 10 and 1000, neither of which should be present.
For starters, you might want to write a simple function which converts
a value (preferably an unsigned of some type—get unsigned
right before worrying about the sign) to a string using any base, e.g.:
//! \pre
//! base >= 2 && base < 36
//!
//! Digits are 0-9, then A-Z.
std::string convert(unsigned value, unsigned base);
This shouldn't take more than about 5 or 6 lines of code. But attention,
the normal algorithm generates the digits in reverse order: if you're
using std::string, the simplest solution is to push_back each digit,
then call std::reverse at the end, before returning it. Otherwise: a
C style char[] works well, provided that you make it large enough.
(sizeof(unsigned) * CHAR_BITS + 2 is more than enough, even for
signed, and even with a '\0' at the end, which you won't need if you
return a string.) Just initialize the pointer to buffer +
sizeof(buffer), and pre-decrement each time you insert a digit. To
construct the string you return:
std::string( pointer, buffer + sizeof(buffer) ) should do the trick.
As for the loop, the end condition could simply be value == 0.
(You'll be dividing value by base each time through, so you're
guaranteed to reach this condition.) If you use a do ... while,
rather than just a while, you're also guaranteed at least one digit in
the output.
(It would have been a lot easier for me to just post the code, but since
this is obviously homework, I think it better to just give indications
concerning what needs to be done.)
Edit: I've added my implementation, and some comments on your new
code:
First for the comments: there's a very misleading prompt: "Enter a
binary number" sounds like the user should enter binary; if you're
reading into an int, the value input should be decimal. And there are
still the % 1000 and / 1000 and % 10 and / 10 that I don't
understand. Whatever you're doing, it can't be right if there's no %
8 and / 8. Try it: input "128", for example, and see what you get.
If you're trying to input binary, then you really have to input a
string, and parse it yourself.
My code for the conversion itself would be:
//! \pre
//! base >= 2 && base <= 36
//!
//! Digits are 0-9, then A-Z.
std::string toString( unsigned value, unsigned base )
{
assert( base >= 2 && base <= 36 );
static char const digits[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
char buffer[sizeof(unsigned) * CHAR_BIT];
char* dst = buffer + sizeof(buffer);
do
{
*--dst = digits[value % base];
value /= base;
} while (value != 0);
return std::string(dst, buffer + sizeof(buffer));
}
If you want to parse input (e.g. for binary), then something like the
following should do the trick:
unsigned fromString( std::string const& value, unsigned base )
{
assert( base >= 2 && base <= 36 );
static char const digits[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
unsigned results = 0;
for (std::string::const_iterator iter = value.begin();
iter != value.end();
++ iter)
{
unsigned digit = std::find
( digits, digits + sizeof(digits) - 1,
toupper(static_cast<unsigned char>( *iter ) ) ) - digits;
if ( digit >= base )
throw std::runtime_error( "Illegal character" );
if ( results >= UINT_MAX / base
&& (results > UINT_MAX / base || digit > UINT_MAX % base) )
throw std::runtime_error( "Overflow" );
results = base * results + digit;
}
return results;
}
It's more complicated than toString because it has to handle all sorts
of possible error conditions. It's also still probably simpler than you
need; you probably want to trim blanks, etc., as well (or even ignore
them: entering 01000000 is more error prone than 0100 0000).
(Also, the end iterator for find has a - 1 because of the trailing
'\0' the compiler inserts into digits.)
Actually I don't understand why do you need so complex code to accomplish what you need.
First of all there is no such a thing as conversion from binary to octal (same is true for converting to/from decimal and etc.). The machine always works in binary, there's nothing you can (or should) do about this.
This is actually a question of formatting. That is, how do you print a number as octal, and how do you parse the textual representation of the octal number.
Edit:
You may use the following code for printing a number in any base:
const int PRINT_NUM_TXT_MAX = 33; // worst-case for binary
void PrintNumberInBase(unsigned int val, int base, PSTR szBuf)
{
// calculate the number of digits
int digits = 0;
for (unsigned int x = val; x; digits++)
x /= base;
if (digits < 1)
digits = 1; // will emit zero
// Print the value from right to left
szBuf[digits] = 0; // zero-term
while (digits--)
{
int dig = val % base;
val /= base;
char ch = (dig <= 9) ?
('0' + dig) :
('a' + dig - 0xa);
szBuf[digits] = ch;
}
}
Example:
char sz[PRINT_NUM_TXT_MAX];
PrintNumberInBase(19, 8, sz);
The code the OP is asking to produce is what your scientific calculator would do when you want a number in a different base.
I think your algorithm is wrong. Just looking over it, I see a function that is squared towards the end. why? There is a simple mathematical way to do what you are talking about. Once you get the math part, then you can convert it to code.
If you had pencil and paper, and no calculator (similar to not using pre built functions), the method is to take the base you are in, change it to base 10, then change to the base you require. In your case that would be base 8, to base 10, to base 2.
This should get you started. All you really need are if/else statements with modulus to get the remainders.
http://www.purplemath.com/modules/numbbase3.htm
Then you have to figure out how to get your desired output. Maybe store the remainders in an array or output to a txt file.
(For problems like this is the reason why I want to double major with applied math)
Since you want conversion from decimal 0-256, it would be easiest to make functions, say call them int binary(), char hex(), and int octal(). Do the binary and octal first as that would be the easiest since they can represented by only integers.
#include <cmath>
#include <iostream>
#include <string>
#include <cstring>
#include <cctype>
#include <cstdlib>
using namespace std;
char* toBinary(char* doubleDigit)
{
int digit = atoi(doubleDigit);
char* binary = new char();
int x = 0 ;
binary[x]='(';
//int tempDigit = digit;
int k=1;
for(int i = 9 ; digit != 0; i--)
{
k=1;//cout << digit << endl;
//cout << "i"<< i<<endl;
if(digit-k *pow(8,i)>=0)
{
k =1;
cout << "i" << i << endl;
cout << k*pow(8,i)<< endl;
while((k*pow(8,i)<=digit))
{
//cout << k <<endl;
k++;
}
k= k-1;
digit = digit -k*pow(8,i);
binary[x+1]= k+'0';
binary[x+2]= '*';
binary[x+3]= '8';
binary[x+4]='^';
binary[x+5]=i+'0';
binary[x+6]='+';
x+=6;
}
}
binary[x]=')';
return binary;
}
int main()
{
char value[6]={'4','0','9','8','7','9'};
cout<< toBinary(value);
return 0 ;
}
I am writing a program that needs to take an array of size n and convert that into it's hex value as follows:
int a[] = { 0, 1, 1, 0 };
I would like to take each value of the array to represent it as binary and convert it to a hex value. In this case:
0x6000000000000000; // 0110...0
it also has to be packed to the right with 0's to be 64 bits (i am on a 64 bit machine).
Or i could also take the array elements, convert to decimal and convert to hexadecimal it that's easier... What you be the best way of doing this in C++?
(this is not homework)
The following assumes that your a[] will only ever use 0 and 1 to represent bits. You'll also need to specify the array length, sizeof(a)/sizeof(int) can be used in this case, but not for heap allocated arrays. Also, result will need to be a 64bit integer type.
for (int c=0; c<array_len; c++)
result |= a[c] << (63-c);
If you want to see what it looks like in hex, you can use (s)printf( "%I64x", result )
std::bitset<64>::to_ulong() might be your friend. The order will probably be backwards (it is unspecified, but typically index 3 will be fetched by right-shifting the word by 3 and masking with 1), but you can remedy that by subtracting the desired index from 63.
#include <bitset>
std::bitset<64> bits;
for ( int index = 0; index < sizeof a/sizeof *a, ++ index ) {
bits[ 63 - index ] = a[ index ];
}
std::cout << std::hex << std::setw(64) << std::setfill('0')
<< bits.to_ulong() << std::endl;
unsigned long long answer= 0;
for (int i= 0; i<sizeof(a)/sizeof(a[0]); ++i)
{
answer= (answer << 1) | a[i];
}
answer<<= (64 - sizeof(a)/sizeof(a[0]));
Assumptions: a[] is at most 64 entries, is defined at compile time, and only contains 1 or 0. Being defined at compile time sidesteps issues of shifting left by 64, as you cannot declare an empty array.
Here's a rough answer:
int ConvertBitArrayToInt64(int a[])
{
int answer = 0;
for(int i=0; i<64; ++i)
{
if (isValidIndex(i))
{
answer = answer << 1 | a[i];
}
else
{
answer = answer << 1;
}
}
return answer;
}
byte hexValues[16];
for(int i = 15; i >= 0; i--)
{
hexValues = a[i*4] * 8 + a[i*4-1] * 4 + [i*4-2] * 2 + a[i*4-3];
}
This will give you an array of bytes where each byte represents one of your hex values.
Note that each byte in hexValues will be a value from 0 to 16.