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Fastest way to Convert String to Binary?
(3 answers)
Closed 5 years ago.
How do I convert a string like
string a = "hello";
to it's bit representation which is stored in a int
int b = 0110100001100101011011000110110001101111
here a and b being equivalent.
You cannot store a long character sequence (e.g. an std::string) inside an int (or inside a long int) because the size of a character is usually 8-bit and the length of an int is usually 32-bit, therefore a 32-bit long int can store only 4 characters.
If you limit the length of the number of characters, you can store them as the following example shows:
#include <iostream>
#include <string>
#include <climits>
int main() {
std::string foo = "Hello";
unsigned long bar = 0ul;
for(std::size_t i = 0; i < foo.size() && i < sizeof(bar); ++i)
bar |= static_cast<unsigned long>(foo[i]) << (CHAR_BIT * i);
std::cout << "Test: " << std::hex << bar << std::endl;
}
Seems like a daft thing to do, bit I think the following (untested) code should work.
#include <string>
#include <climits>
int foo(std::string const & s) {
int result = 0;
for (int i = 0; i < std::min(sizeof(int), s.size()); ++i) {
result = (result << CHAR_BIT) || s[i];
}
return result;
}
int output[CHAR_BIT];
char c;
int i;
for (i = 0; i < CHAR_BIT; ++i) {
output[i] = (c >> i) & 1;
}
More info in this link: how to convert a char to binary?
Related
I tried to build a function that calculates a binary number stored in a string into a decimal number stored in a long long. I'm thinking that my code should work but it doesn't.
In this example for the binary number 101110111 the decimal number is 375. But my output is completely confusing.
Here is my code:
#include <string>
#include <stdio.h>
#include <math.h>
#include <iostream>
#include <string.h>
int main() {
std::string stringNumber = "101110111";
const char *array = stringNumber.c_str();
int subtrahend = 1;
int potency = 0;
long long result = 0;
for(int i = 0; i < strlen(array); i++) {
result += pow(array[strlen(array) - subtrahend] * 2, potency);
subtrahend++;
potency++;
std::cout << result << std::endl;
}
}
Here is the output:
1
99
9703
894439
93131255
9132339223
894974720087
76039722530902
8583669948348758
What I'm doing wrong here?
'1' != 1 as mentioned in the comments by #churill. '1' == 49. If you are on linux type man ascii in terminal to get the ascii table.
Try this, it is the same code. I just used the stringNumber directly instead of using const char* to it. And I subtracted '0' from the current index. '0' == 48, so if you subtract it, you get the actual 1 or 0 integer value:
auto sz = stringNumber.size();
for(int i = 0; i < sz; i++) {
result += pow((stringNumber[sz - subtrahend] - '0') * 2, potency);
subtrahend++;
potency++;
std::cout << result << std::endl;
}
Moreover, use the methods provided by std::string like .size() instead of doing strlen() on every iteration. Much faster.
In a production environment, I would highly recommend using std::bitset instead of rolling your own solution:
std::string stringNumber = "1111";
std::bitset<64> bits(stringNumber);
bits.to_ulong();
You're forgetting to convert your digits into integers. Plus you really don't need to use C strings.
Here's a better version of the code
int main() {
std::string stringNumber = "101110111";
int subtrahend = 1;
int potency = 0;
long long result = 0;
for(int i = 0; i < stringNumber.size(); i++) {
result += pow(2*(stringNumber[stringNumber.size() - subtrahend] - '0'), potency);
subtrahend++;
potency++;
std::cout << result << std::endl;
}
}
Subtracting '0' from the string digits converts the digit into an integer.
Now for extra credit write a version that doesn't use pow (hint: potency *= 2; instead of potency++;)
c++ way
#include <string>
#include <math.h>
#include <iostream>
using namespace std;
int main() {
std::string stringNumber = "101110111";
long long result = 0;
uint string_length = stringNumber.length();
for(int i = 0; i <string_length; i++) {
if(stringNumber[i]=='1')
{
long pose_value = pow(2, string_length-1-i);
result += pose_value;
}
}
std::cout << result << std::endl;
}
I am writing a function in C++ to convert a number from some base to decimal.
It works fine when the number of digits is even, but when it is odd it gives wrong answer.
For example:
Number to convert : 100
Base to convert to: 10
Correct answer : 100
Function's output : 99
Here is the code:
unsigned long long convertToDecimal(const std::string& number, const unsigned base)
{
std::string characters = "0123456789abcdef";
unsigned long long res = 0;
for(int i = 0, len = number.size(); i<len; ++i)
{
res += characters.find(number.at(i))*std::pow(base, len-1-i);
}
return res;
}
I'm using g++ C++11.
I can't reproduce your particular issue, but std::pow returns a floating point number and your implementation may have introduced some sort of rounding error which leaded to a wrong result when converted to unsigned long long.
To avoid those errors, when dealing with integer numbers, you should consider to avoid std::pow at all. Your function, for example, could have been written like this:
#include <iostream>
#include <string>
#include <cmath>
unsigned long long convertToDecimal(const std::string& number, const unsigned base)
{
std::string characters = "0123456789abcdef";
unsigned long long res = 0;
unsigned long long power = 1;
for(auto i = number.crbegin(); i != number.crend(); ++i)
{
// As in your code, I'm not checking for erroneous input
res += characters.find(*i) * power;
power *= base;
}
return res;
}
int main ()
{
std::cout << convertToDecimal("100", 2) << '\n'; // --> 4
std::cout << convertToDecimal("1234", 8) << '\n'; // --> 668
std::cout << convertToDecimal("99999", 10) << '\n'; // --> 99999
std::cout << convertToDecimal("fedcba", 16) << '\n'; // --> 16702650
}
Is it possible to take an array filled with 2 digit numbers e.g.
[10,11,12,13,...]
and multiply each element in the list by 100^(position in the array) and sum the result so that:
mysteryFunction[10,11,12] //The function performs 10*100^0 + 11*100^1 + 12*100^3
= 121110
and also
mysteryFunction[10,11,12,13]
= 13121110
when I do not know the number of elements in the array?
(yes, the reverse of order is intended but not 100% necessary, and just in case you missed it the first time the numbers will always be 2 digits)
Just for a bit of background to the problem: this is to try to improve my attempt at an RSA encryption program, at the moment I am multiplying each member of the array by 100^(the position of the number) written out each time which means that each word which I use to encrypt must be a certain length.
For example to encrypt "ab" I have converted it to an array [10,11] but need to convert it to 1110 before I can put it through the RSA algorithm. I would need to adjust my code for if I then wanted to use a three letter word, again for a four letter word etc. which I'm sure you will agree is not ideal. My code is nothing like industry standard but I am happy to upload it should anyone want to see it (I have also already managed this in Haskell if anyone would like to see that). I thought that the background information was necessary just so that I don't get hundreds of downvotes from people thinking that I'm trying to trick them into doing homework for me. Thank you very much for any help, I really do appreciate it!
EDIT: Thank you for all of the answers! They perfectly answer the question that I asked but I am having problems incorporating them into my current program, if I post my code so far would you be able to help? When I tried to include the answer provided I got an error message (I can't vote up because I don't have enough reputation, sorry that I haven't accepted any answers yet).
#include <iostream>
#include <string>
#include <math.h>
int returnVal (char x)
{
return (int) x;
}
unsigned long long modExp(unsigned long long b, unsigned long long e, unsigned long long m)
{
unsigned long long remainder;
int x = 1;
while (e != 0)
{
remainder = e % 2;
e= e/2;
if (remainder == 1)
x = (x * b) % m;
b= (b * b) % m;
}
return x;
}
int main()
{
unsigned long long p = 80001;
unsigned long long q = 70021;
int e = 7;
unsigned long long n = p * q;
std::string foo = "ab";
for (int i = 0; i < foo.length(); i++);
{
std::cout << modExp (returnVal((foo[0]) - 87) + returnVal (foo[1] -87) * 100, e, n);
}
}
If you want to use plain C-style arrays, you will have to separately know the number of entries. With this approach, your mysterious function might be defined like this:
unsigned mysteryFunction(unsigned numbers[], size_t n)
{
unsigned result = 0;
unsigned factor = 1;
for (size_t i = 0; i < n; ++i)
{
result += factor * numbers[i];
factor *= 100;
}
return result;
}
You can test this code with the following:
#include <iostream>
int main()
{
unsigned ar[] = {10, 11, 12, 13};
std::cout << mysteryFunction(ar, 4) << "\n";
return 0;
}
On the other hand, if you want to utilize the STL's vector class, you won't separately need the size. The code itself won't need too many changes.
Also note that the built-in integer types cannot handle very large numbers, so you might want to look into an arbitrary precision number library, like GMP.
EDIT: Here's a version of the function which accepts a std::string and uses the characters' ASCII values minus 87 as the numbers:
unsigned mysteryFunction(const std::string& input)
{
unsigned result = 0;
unsigned factor = 1;
for (size_t i = 0; i < input.size(); ++i)
{
result += factor * (input[i] - 87);
factor *= 100;
}
return result;
}
The test code becomes:
#include <iostream>
#include <string>
int main()
{
std::string myString = "abcde";
std::cout << mysteryFunction(myString) << "\n";
return 0;
}
The program prints: 1413121110
As benedek mentioned, here's an implementation using dynamic arrays via std::vector.
unsigned mystery(std::vector<unsigned> vect)
{
unsigned result = 0;
unsigned factor = 1;
for (auto& item : vect)
{
result += factor * item;
factor *= 100;
}
return result;
}
void main(void)
{
std::vector<unsigned> ar;
ar.push_back(10);
ar.push_back(11);
ar.push_back(12);
ar.push_back(13);
std::cout << mystery(ar);
}
I would like to suggest the following solutions.
You could use standard algorithm std::accumulate declared in header <numeric>
For example
#include <iostream>
#include <numeric>
int main()
{
unsigned int a[] = { 10, 11, 12, 13 };
unsigned long long i = 1;
unsigned long long s =
std::accumulate( std::begin( a ), std::end( a ), 0ull,
[&]( unsigned long long acc, unsigned int x )
{
return ( acc += x * i, i *= 100, acc );
} );
std::cout << "s = " << s << std::endl;
return 0;
}
The output is
s = 13121110
The same can be done with using the range based for statement
#include <iostream>
#include <numeric>
int main()
{
unsigned int a[] = { 10, 11, 12, 13 };
unsigned long long i = 1;
unsigned long long s = 0;
for ( unsigned int x : a )
{
s += x * i; i *= 100;
}
std::cout << "s = " << s << std::endl;
return 0;
}
You could also write a separate function
unsigned long long mysteryFunction( const unsigned int a[], size_t n )
{
unsigned long long s = 0;
unsigned long long i = 1;
for ( size_t k = 0; k < n; k++ )
{
s += a[k] * i; i *= 100;
}
return s;
}
Also think about using std::string instead of integral numbers to keep an encrypted result.
I have a IEEE754 Double precision 64-bit binary string representation of a double number.
example : double value = 0.999;
Its binary representation is "0011111111101111111101111100111011011001000101101000011100101011"
I want to convert this string back to a double number in c++.
I dont want to use any external libraries or .dll's as my program would operate in any platform.
C string solution:
#include <cstring> // needed for all three solutions because of memcpy
double bitstring_to_double(const char* p)
{
unsigned long long x = 0;
for (; *p; ++p)
{
x = (x << 1) + (*p - '0');
}
double d;
memcpy(&d, &x, 8);
return d;
}
std::string solution:
#include <string>
double bitstring_to_double(const std::string& s)
{
unsigned long long x = 0;
for (std::string::const_iterator it = s.begin(); it != s.end(); ++it)
{
x = (x << 1) + (*it - '0');
}
double d;
memcpy(&d, &x, 8);
return d;
}
generic solution:
template<typename InputIterator>
double bitstring_to_double(InputIterator begin, InputIterator end)
{
unsigned long long x = 0;
for (; begin != end; ++begin)
{
x = (x << 1) + (*begin - '0');
}
double d;
memcpy(&d, &x, 8);
return d;
}
example calls:
#include <iostream>
int main()
{
const char * p = "0011111111101111111101111100111011011001000101101000011100101011";
std::cout << bitstring_to_double(p) << '\n';
std::string s(p);
std::cout << bitstring_to_double(s) << '\n';
std::cout << bitstring_to_double(s.begin(), s.end()) << '\n';
std::cout << bitstring_to_double(p + 0, p + 64) << '\n';
}
Note: I assume unsigned long long has 64 bits. A cleaner solution would be to include <cstdint> and use uint64_t instead, assuming your compiler is up to date and provides that C++11 header.
A starting point would be to iterate through the individual characters in the string and set individual bits of an existing double.
Is it really a character string of binary bits? If so, first convert to a 64-bit int. Then either use a library routine (probably there is one somewhere), or more simply, use a double aliased over the 64-bit int to convert to double.
(If it's already a 64-bit int then skip the first step.)
Ignoring byte-ordering issues, but I suppose this should be a viable option:
The below has an outcome of .999 on i386 with gcc. See it live: https://ideone.com/i4ygJ
#include <cstdint>
#include <sstream>
#include <iostream>
#include <bitset>
int main()
{
std::istringstream iss("0011111111101111111101111100111011011001000101101000011100101011");
std::bitset<32> hi, lo;
if (iss >> hi >> lo)
{
struct { uint32_t lo, hi; } words = { lo.to_ulong(), hi.to_ulong() };
double converted = *reinterpret_cast<double*>(&words);
std::cout << hi << std::endl;
std::cout << lo << std::endl;
std::cout << converted << std::endl;
}
}
my program would operate in any platform
Included I assume those whose double format isn't IEEE. Something like this should work:
#include <math.h>
...
int const dbl_exponent_bits = 11;
int const dbl_exponent_offset = 1023;
int const dbl_significand_bits = 52;
bool negative = (*num++ == '1');
int exponent = 0;
for (int i = 0; i < dbl_exponent_bits; ++i, ++num) {
exponent = 2*exponent + (*num == '1' ? 1 : 0);
}
double significand = 1;
for (int i = 0; i < dbl_significand_bits; ++i, ++num) {
significand = 2*significand + (*num == '1' ? 1 : 0);
}
assert(*num == '\0');
double result = ldexp(significand, exponent-(dbl_exponent_offset+dbl_significand_bits));
if (negative)
result = -result;
I've been working on a program that converts numbers into binary. As you can see my program here, I've written so that it can scale for larger numbers then a traditional binary code, such as 2 lines (16-bits) for numbers bigger then 255. However, going larger requires long instead of int, but that doesn't seem to be playing well, producing output such as this. Would anyone mind helping me change the program to use long? Or would it require a fundamental change in the code instead of some minor edits?
#include <iostream>
#include <math.h>
using namespace std;
int main(int argc, char **argv)
{
int j=0;
int c=8;
long a = 1;
int i=1;
cin >> a;
while (a >= (pow(2,c))) {
c = c+8;
i++;
}
long block[i*8];
for (long tw;tw<(i*8);tw++)
{
block[tw] = 0;
}
j=((i*8)-1);
long b = 0;
while (j != -1)
{
if (b+(pow(2,j))<=a)
{
block[j]=1;
b=b+(pow(2,j));
}
j--;
}
long q=0;
cout << endl;
int y=1;
long z = 0;
for (y;y<=i;y++) {
for (z;z<8;z++) {
cout << block[z+q];
}
cout << endl;
z = 0;
q = q + (8*y);
}
}
You are making your code far more complicated than it needs to be. This will print out a single 32-bit integer in binary:
const unsigned int bit_count = sizeof(int) * 8 - 1;
int a;
std::cin >> a;
for (unsigned int i = bit_count; i > 0; --i)
{
unsigned int t = (1 << i);
std::cout << (a & t ? "1" : "0");
}
std::cout << (a & 1 ? "1" : "0");
std::cout << std::endl;
If you want to block it off by ranges to make it easier to read, you simply need to place range on the loop (or move it to a function that takes a range).
Why not something simple like this? You could store the intermediate bits in an array or a string instead of using cout.
int convert(long n)
{
long k=1;
while(k<n)//find the most significant bit
{
k*=2;
}
if(k>n)//fix the overshoot
{
k/=2;
}
while(k>0)
{
if(int(n/k)%2==0)
{
cout<<0;//find the (next) most
}
else
{
cout<<1;//significant binary digit
}
k/=2;//go to the next column to the right and repeat
}
}
For a bit more flexibly, here's another way to do it with templates. Template instantiations with signed types are omitted intentionally due to extension issues.
template <typename T>
void print_binary(const T input, const short grouping = 4)
{
unsigned int bit_count = sizeof(T) * 8;
T nth_bit = 1 << (bit_count - 1);
for(int i = 0; i < bit_count; i++, nth_bit >>= 1 )
{
cout << (input & nth_bit ? "1" : "0");
if( i % grouping == grouping-1 ) // print binary in groups
cout << ' ';
}
cout << endl;
}
template <>
void print_binary<signed>(const signed input, const short grouping);
template <>
void print_binary<signed short>(const signed short input, const short grouping);
template <>
void print_binary<signed long>(const signed long input, const short grouping);
template <>
void print_binary<signed char>(const signed char input, const short grouping);