Can I use itoa() for converting long long int to a binary string?
I have seen various examples for conversion of int to binary using itoa. Is there a risk of overflow or perhaps loss of precision, if I use long long int?
Edit
Thanks all of you for replying. I achieved what I was trying to do. itoa() was not useful enough, as it does not support long long int. Moreover I can't use itoa() in gcc as it is not a standard library function.
To convert an integer to a string containing only binary digits, you can do it by checking each bit in the integer with a one-bit mask, and append it to the string.
Something like this:
std::string convert_to_binary_string(const unsigned long long int value,
bool skip_leading_zeroes = false)
{
std::string str;
bool found_first_one = false;
const int bits = sizeof(unsigned long long) * 8; // Number of bits in the type
for (int current_bit = bits - 1; current_bit >= 0; current_bit--)
{
if ((value & (1ULL << current_bit)) != 0)
{
if (!found_first_one)
found_first_one = true;
str += '1';
}
else
{
if (!skip_leading_zeroes || found_first_one)
str += '0';
}
}
return str;
}
Edit:
A more general way of doing it might be done with templates:
#include <type_traits>
#include <cassert>
template<typename T>
std::string convert_to_binary_string(const T value, bool skip_leading_zeroes = false)
{
// Make sure the type is an integer
static_assert(std::is_integral<T>::value, "Not integral type");
std::string str;
bool found_first_one = false;
const int bits = sizeof(T) * 8; // Number of bits in the type
for (int current_bit = bits - 1; current_bit >= 0; current_bit--)
{
if ((value & (1ULL << current_bit)) != 0)
{
if (!found_first_one)
found_first_one = true;
str += '1';
}
else
{
if (!skip_leading_zeroes || found_first_one)
str += '0';
}
}
return str;
}
Note: Both static_assert and std::is_integral is part of C++11, but is supported in both Visual C++ 2010 and GCC from at least 4.4.5.
Yes, you can. As you showed yourself, itoa can be called with base 2, which means binary.
#include <stdio.h>
#include <stdlib.h>
int main()
{
int i;
char str[33];
i = 37; /* Just some number. */
itoa (i, str, 2);
printf("binary: %s\n", str);
return 0;
}
Also, yes, there will be truncation if you use an integer type larger than int, since itoa() takes only plain "int" as a value. long long is on your compiler probably 64 bit while int is probably 32 bit, so the compiler would truncate the 64 bit value to a 32 bit value before conversion.
Your wording is a bit confusing,
normally if you state 'decimal' I would take that to mean: 'a number represented as a string of decimal digits', while you seem to mean 'integer'.
and with 'binary' I would take that to mean: 'a number represented as bytes - as directly usable by the CPU'.
a better way of phrasing your subject would be: converting 64-bit integer to string of binary digits.
some systems have a _i64toa function.
You can use std::bitset for this purpose
template<typename T>
inline std::string to_binary_string(const T value)
{
return std::bitset<sizeof(T)>(value).to_string();
}
std::cout << to_binary_string(10240);
std::cout << to_binary_string(123LL);
Related
I want to convert a hex string to a 32 bit signed integer in C++.
So, for example, I have the hex string "fffefffe". The binary representation of this is 11111111111111101111111111111110. The signed integer representation of this is: -65538.
How do I do this conversion in C++? This also needs to work for non-negative numbers. For example, the hex string "0000000A", which is 00000000000000000000000000001010 in binary, and 10 in decimal.
use std::stringstream
unsigned int x;
std::stringstream ss;
ss << std::hex << "fffefffe";
ss >> x;
the following example produces -65538 as its result:
#include <sstream>
#include <iostream>
int main() {
unsigned int x;
std::stringstream ss;
ss << std::hex << "fffefffe";
ss >> x;
// output it as a signed type
std::cout << static_cast<int>(x) << std::endl;
}
In the new C++11 standard, there are a few new utility functions which you can make use of! specifically, there is a family of "string to number" functions (http://en.cppreference.com/w/cpp/string/basic_string/stol and http://en.cppreference.com/w/cpp/string/basic_string/stoul). These are essentially thin wrappers around C's string to number conversion functions, but know how to deal with a std::string
So, the simplest answer for newer code would probably look like this:
std::string s = "0xfffefffe";
unsigned int x = std::stoul(s, nullptr, 16);
NOTE: Below is my original answer, which as the edit says is not a complete answer. For a functional solution, stick the code above the line :-).
It appears that since lexical_cast<> is defined to have stream conversion semantics. Sadly, streams don't understand the "0x" notation. So both the boost::lexical_cast and my hand rolled one don't deal well with hex strings. The above solution which manually sets the input stream to hex will handle it just fine.
Boost has some stuff to do this as well, which has some nice error checking capabilities as well. You can use it like this:
try {
unsigned int x = lexical_cast<int>("0x0badc0de");
} catch(bad_lexical_cast &) {
// whatever you want to do...
}
If you don't feel like using boost, here's a light version of lexical cast which does no error checking:
template<typename T2, typename T1>
inline T2 lexical_cast(const T1 &in) {
T2 out;
std::stringstream ss;
ss << in;
ss >> out;
return out;
}
which you can use like this:
// though this needs the 0x prefix so it knows it is hex
unsigned int x = lexical_cast<unsigned int>("0xdeadbeef");
For a method that works with both C and C++, you might want to consider using the standard library function strtol().
#include <cstdlib>
#include <iostream>
using namespace std;
int main() {
string s = "abcd";
char * p;
long n = strtol( s.c_str(), & p, 16 );
if ( * p != 0 ) { //my bad edit was here
cout << "not a number" << endl;
}
else {
cout << n << endl;
}
}
Andy Buchanan, as far as sticking to C++ goes, I liked yours, but I have a few mods:
template <typename ElemT>
struct HexTo {
ElemT value;
operator ElemT() const {return value;}
friend std::istream& operator>>(std::istream& in, HexTo& out) {
in >> std::hex >> out.value;
return in;
}
};
Used like
uint32_t value = boost::lexical_cast<HexTo<uint32_t> >("0x2a");
That way you don't need one impl per int type.
Working example with strtoul will be:
#include <cstdlib>
#include <iostream>
using namespace std;
int main() {
string s = "fffefffe";
char * p;
long n = strtoul( s.c_str(), & p, 16 );
if ( * p != 0 ) {
cout << "not a number" << endl;
} else {
cout << n << endl;
}
}
strtol converts string to long. On my computer numeric_limits<long>::max() gives 0x7fffffff. Obviously that 0xfffefffe is greater than 0x7fffffff. So strtol returns MAX_LONG instead of wanted value. strtoul converts string to unsigned long that's why no overflow in this case.
Ok, strtol is considering input string not as 32-bit signed integer before convertation. Funny sample with strtol:
#include <cstdlib>
#include <iostream>
using namespace std;
int main() {
string s = "-0x10002";
char * p;
long n = strtol( s.c_str(), & p, 16 );
if ( * p != 0 ) {
cout << "not a number" << endl;
} else {
cout << n << endl;
}
}
The code above prints -65538 in console.
Here's a simple and working method I found elsewhere:
string hexString = "7FF";
int hexNumber;
sscanf(hexString.c_str(), "%x", &hexNumber);
Please note that you might prefer using unsigned long integer/long integer, to receive the value.
Another note, the c_str() function just converts the std::string to const char* .
So if you have a const char* ready, just go ahead with using that variable name directly, as shown below [I am also showing the usage of the unsigned long variable for a larger hex number. Do not confuse it with the case of having const char* instead of string]:
const char *hexString = "7FFEA5"; //Just to show the conversion of a bigger hex number
unsigned long hexNumber; //In case your hex number is going to be sufficiently big.
sscanf(hexString, "%x", &hexNumber);
This works just perfectly fine (provided you use appropriate data types per your need).
I had the same problem today, here's how I solved it so I could keep lexical_cast<>
typedef unsigned int uint32;
typedef signed int int32;
class uint32_from_hex // For use with boost::lexical_cast
{
uint32 value;
public:
operator uint32() const { return value; }
friend std::istream& operator>>( std::istream& in, uint32_from_hex& outValue )
{
in >> std::hex >> outValue.value;
}
};
class int32_from_hex // For use with boost::lexical_cast
{
uint32 value;
public:
operator int32() const { return static_cast<int32>( value ); }
friend std::istream& operator>>( std::istream& in, int32_from_hex& outValue )
{
in >> std::hex >> outvalue.value;
}
};
uint32 material0 = lexical_cast<uint32_from_hex>( "0x4ad" );
uint32 material1 = lexical_cast<uint32_from_hex>( "4ad" );
uint32 material2 = lexical_cast<uint32>( "1197" );
int32 materialX = lexical_cast<int32_from_hex>( "0xfffefffe" );
int32 materialY = lexical_cast<int32_from_hex>( "fffefffe" );
// etc...
(Found this page when I was looking for a less sucky way :-)
Cheers,
A.
just use stoi/stol/stoll
for example:
std::cout << std::stol("fffefffe", nullptr, 16) << std::endl;
output: 4294901758
This worked for me:
string string_test = "80123456";
unsigned long x;
signed long val;
std::stringstream ss;
ss << std::hex << string_test;
ss >> x;
// ss >> val; // if I try this val = 0
val = (signed long)x; // However, if I cast the unsigned result I get val = 0x80123456
Try this. This solution is a bit risky. There are no checks. The string must only have hex values and the string length must match the return type size. But no need for extra headers.
char hextob(char ch)
{
if (ch >= '0' && ch <= '9') return ch - '0';
if (ch >= 'A' && ch <= 'F') return ch - 'A' + 10;
if (ch >= 'a' && ch <= 'f') return ch - 'a' + 10;
return 0;
}
template<typename T>
T hextot(char* hex)
{
T value = 0;
for (size_t i = 0; i < sizeof(T)*2; ++i)
value |= hextob(hex[i]) << (8*sizeof(T)-4*(i+1));
return value;
};
Usage:
int main()
{
char str[4] = {'f','f','f','f'};
std::cout << hextot<int16_t>(str) << "\n";
}
Note: the length of the string must be divisible by 2
For those looking to convert number base for unsigned numbers, it is pretty trivial to do yourself in both C/C++ with minimal dependency (only operator not provided by the language itself is pow() function).
In mathematical terms, a positive ordinal number d in base b with n number of digits can be converted to base 10 using:
Example: Converting base 16 number 00f looks like:
= 0*16^2 + 0*16^1 + 16*16^0 = 15
C/C++ Example:
#include <math.h>
unsigned int to_base10(char *d_str, int len, int base)
{
if (len < 1) {
return 0;
}
char d = d_str[0];
// chars 0-9 = 48-57, chars a-f = 97-102
int val = (d > 57) ? d - ('a' - 10) : d - '0';
int result = val * pow(base, (len - 1));
d_str++; // increment pointer
return result + to_base10(d_str, len - 1, base);
}
int main(int argc, char const *argv[])
{
char n[] = "00f"; // base 16 number of len = 3
printf("%d\n", to_base10(n, 3, 16));
}
I want to convert a hex string to a 32 bit signed integer in C++.
So, for example, I have the hex string "fffefffe". The binary representation of this is 11111111111111101111111111111110. The signed integer representation of this is: -65538.
How do I do this conversion in C++? This also needs to work for non-negative numbers. For example, the hex string "0000000A", which is 00000000000000000000000000001010 in binary, and 10 in decimal.
use std::stringstream
unsigned int x;
std::stringstream ss;
ss << std::hex << "fffefffe";
ss >> x;
the following example produces -65538 as its result:
#include <sstream>
#include <iostream>
int main() {
unsigned int x;
std::stringstream ss;
ss << std::hex << "fffefffe";
ss >> x;
// output it as a signed type
std::cout << static_cast<int>(x) << std::endl;
}
In the new C++11 standard, there are a few new utility functions which you can make use of! specifically, there is a family of "string to number" functions (http://en.cppreference.com/w/cpp/string/basic_string/stol and http://en.cppreference.com/w/cpp/string/basic_string/stoul). These are essentially thin wrappers around C's string to number conversion functions, but know how to deal with a std::string
So, the simplest answer for newer code would probably look like this:
std::string s = "0xfffefffe";
unsigned int x = std::stoul(s, nullptr, 16);
NOTE: Below is my original answer, which as the edit says is not a complete answer. For a functional solution, stick the code above the line :-).
It appears that since lexical_cast<> is defined to have stream conversion semantics. Sadly, streams don't understand the "0x" notation. So both the boost::lexical_cast and my hand rolled one don't deal well with hex strings. The above solution which manually sets the input stream to hex will handle it just fine.
Boost has some stuff to do this as well, which has some nice error checking capabilities as well. You can use it like this:
try {
unsigned int x = lexical_cast<int>("0x0badc0de");
} catch(bad_lexical_cast &) {
// whatever you want to do...
}
If you don't feel like using boost, here's a light version of lexical cast which does no error checking:
template<typename T2, typename T1>
inline T2 lexical_cast(const T1 &in) {
T2 out;
std::stringstream ss;
ss << in;
ss >> out;
return out;
}
which you can use like this:
// though this needs the 0x prefix so it knows it is hex
unsigned int x = lexical_cast<unsigned int>("0xdeadbeef");
For a method that works with both C and C++, you might want to consider using the standard library function strtol().
#include <cstdlib>
#include <iostream>
using namespace std;
int main() {
string s = "abcd";
char * p;
long n = strtol( s.c_str(), & p, 16 );
if ( * p != 0 ) { //my bad edit was here
cout << "not a number" << endl;
}
else {
cout << n << endl;
}
}
Andy Buchanan, as far as sticking to C++ goes, I liked yours, but I have a few mods:
template <typename ElemT>
struct HexTo {
ElemT value;
operator ElemT() const {return value;}
friend std::istream& operator>>(std::istream& in, HexTo& out) {
in >> std::hex >> out.value;
return in;
}
};
Used like
uint32_t value = boost::lexical_cast<HexTo<uint32_t> >("0x2a");
That way you don't need one impl per int type.
Working example with strtoul will be:
#include <cstdlib>
#include <iostream>
using namespace std;
int main() {
string s = "fffefffe";
char * p;
long n = strtoul( s.c_str(), & p, 16 );
if ( * p != 0 ) {
cout << "not a number" << endl;
} else {
cout << n << endl;
}
}
strtol converts string to long. On my computer numeric_limits<long>::max() gives 0x7fffffff. Obviously that 0xfffefffe is greater than 0x7fffffff. So strtol returns MAX_LONG instead of wanted value. strtoul converts string to unsigned long that's why no overflow in this case.
Ok, strtol is considering input string not as 32-bit signed integer before convertation. Funny sample with strtol:
#include <cstdlib>
#include <iostream>
using namespace std;
int main() {
string s = "-0x10002";
char * p;
long n = strtol( s.c_str(), & p, 16 );
if ( * p != 0 ) {
cout << "not a number" << endl;
} else {
cout << n << endl;
}
}
The code above prints -65538 in console.
Here's a simple and working method I found elsewhere:
string hexString = "7FF";
int hexNumber;
sscanf(hexString.c_str(), "%x", &hexNumber);
Please note that you might prefer using unsigned long integer/long integer, to receive the value.
Another note, the c_str() function just converts the std::string to const char* .
So if you have a const char* ready, just go ahead with using that variable name directly, as shown below [I am also showing the usage of the unsigned long variable for a larger hex number. Do not confuse it with the case of having const char* instead of string]:
const char *hexString = "7FFEA5"; //Just to show the conversion of a bigger hex number
unsigned long hexNumber; //In case your hex number is going to be sufficiently big.
sscanf(hexString, "%x", &hexNumber);
This works just perfectly fine (provided you use appropriate data types per your need).
I had the same problem today, here's how I solved it so I could keep lexical_cast<>
typedef unsigned int uint32;
typedef signed int int32;
class uint32_from_hex // For use with boost::lexical_cast
{
uint32 value;
public:
operator uint32() const { return value; }
friend std::istream& operator>>( std::istream& in, uint32_from_hex& outValue )
{
in >> std::hex >> outValue.value;
}
};
class int32_from_hex // For use with boost::lexical_cast
{
uint32 value;
public:
operator int32() const { return static_cast<int32>( value ); }
friend std::istream& operator>>( std::istream& in, int32_from_hex& outValue )
{
in >> std::hex >> outvalue.value;
}
};
uint32 material0 = lexical_cast<uint32_from_hex>( "0x4ad" );
uint32 material1 = lexical_cast<uint32_from_hex>( "4ad" );
uint32 material2 = lexical_cast<uint32>( "1197" );
int32 materialX = lexical_cast<int32_from_hex>( "0xfffefffe" );
int32 materialY = lexical_cast<int32_from_hex>( "fffefffe" );
// etc...
(Found this page when I was looking for a less sucky way :-)
Cheers,
A.
just use stoi/stol/stoll
for example:
std::cout << std::stol("fffefffe", nullptr, 16) << std::endl;
output: 4294901758
This worked for me:
string string_test = "80123456";
unsigned long x;
signed long val;
std::stringstream ss;
ss << std::hex << string_test;
ss >> x;
// ss >> val; // if I try this val = 0
val = (signed long)x; // However, if I cast the unsigned result I get val = 0x80123456
Try this. This solution is a bit risky. There are no checks. The string must only have hex values and the string length must match the return type size. But no need for extra headers.
char hextob(char ch)
{
if (ch >= '0' && ch <= '9') return ch - '0';
if (ch >= 'A' && ch <= 'F') return ch - 'A' + 10;
if (ch >= 'a' && ch <= 'f') return ch - 'a' + 10;
return 0;
}
template<typename T>
T hextot(char* hex)
{
T value = 0;
for (size_t i = 0; i < sizeof(T)*2; ++i)
value |= hextob(hex[i]) << (8*sizeof(T)-4*(i+1));
return value;
};
Usage:
int main()
{
char str[4] = {'f','f','f','f'};
std::cout << hextot<int16_t>(str) << "\n";
}
Note: the length of the string must be divisible by 2
For those looking to convert number base for unsigned numbers, it is pretty trivial to do yourself in both C/C++ with minimal dependency (only operator not provided by the language itself is pow() function).
In mathematical terms, a positive ordinal number d in base b with n number of digits can be converted to base 10 using:
Example: Converting base 16 number 00f looks like:
= 0*16^2 + 0*16^1 + 16*16^0 = 15
C/C++ Example:
#include <math.h>
unsigned int to_base10(char *d_str, int len, int base)
{
if (len < 1) {
return 0;
}
char d = d_str[0];
// chars 0-9 = 48-57, chars a-f = 97-102
int val = (d > 57) ? d - ('a' - 10) : d - '0';
int result = val * pow(base, (len - 1));
d_str++; // increment pointer
return result + to_base10(d_str, len - 1, base);
}
int main(int argc, char const *argv[])
{
char n[] = "00f"; // base 16 number of len = 3
printf("%d\n", to_base10(n, 3, 16));
}
A shell of the desired code:
#include <iostream>
#include <string>
std::string str_to_bin(const std::string& str)
{
//...
}
int main()
{
std::string str = "123";
std::cout << str_to_bin(str); //would print 1111011
}
Question title says it all. I've been stuck on this for a while. Is there a solution for this in the STL? Or something simple that I'm missing? If not, how would I go about doing this? Maybe a direction you could point me to? Also, speed is of great importance.
EDIT: The number can be of any size (larger than long long as well), so std::stoi and std::bitset<> are off the table.
You can do it using GMP (GNU Multi-Precision). Something like this:
#include <gmpxx.h>
std::string str_to_bin(const std::string& str)
{
mpz_class bignum;
int rc = bignum.set_str(str, 10);
if (rc != 0)
throw std::invalid_argument("bad number: " + str);
return bignum.get_str(2);
}
Or using the traditional C API:
#include <gmp.h>
std::string str_to_bin(const std::string& str)
{
mpz_t bignum;
int rc = mpz_set_str(bignum, str.c_str(), 10);
if (rc != 0)
throw std::invalid_argument("bad number: " + str);
char* rawstr = mpz_get_str(nullptr, 2, bignum);
std::string result(rawstr);
free(rawstr);
return result;
}
Okay let's break down the process you require here. (only one of an infinite number of ways to do this)
Conversion of a number represented as a string type into an integer type.
Conversion of the intermediary integer type into a binary number which is held in another string type. (judging by the return type of your function, which could just as easily return an integer by the way and save the headache of representing the binary equivalent as a string)
For step 1:
Use the standard library function stoi(). It does what you might imagine, extracts the numerical data from the string and stores it in an integer.
std::string numberstr = "123";
int numberint = std::stoi(numberstr);
std::cout << numberint << "\n";
Now you have the number as an integer.
For step 2:
This process involves the conversion of a number from base 10 (decimal) to base 2 (binary).
Divide the number by 2.
Store the remainder and the quotient of this division operation for further use.
The remainder becomes part of the binary representation, while the quotient is used as the next dividend.
This process repeats until the dividend becomes 1, at which point it too is included in the binary representation.
Reverse the string, and voila! You now have the binary representation of a number.
If you want to handle negative numbers (which I imagine you might), simply perform a check before the conversion to see if the converted integer is negative, and set a flag to true if it is.
Check this flag before reversing, and add a negative sign to end of the string before reversing.
The final function looks like this:
std::string str_to_bin(const std::string& str)
{
std::string binarystr = ""; // Output string
int remainder;
int numberint = std::stoi(str);
bool flagnegative = false;
// If negative number, beginning of binary equivalent is 1
if (numberint < 0)
{
numberint = abs(numberint);
flagnegative = true;
}
// If number is 0, don't perform conversion simply return 0
if (numberint == 0)
{
binarystr = "0";
return binarystr;
}
std::cout << numberint << "\n";
while (numberint != 1)
{
remainder = numberint % 2;
numberint /= 2;
std::ostringstream convert; // stream used for the conversion
convert << remainder; // insert the textual representation of 'remainder' in the characters in the stream
binarystr += convert.str();
}
std::ostringstream final;
final << numberint; // To insert the last (or rather first once reversed) binary number
binarystr += final.str();
if (flagnegative == true)
binarystr += "-";
std::reverse(binarystr.begin(), binarystr.end());
return binarystr;
}
Other people have posted the STL method using bitset, which might be of value to you, but I believe there's no fun in simply copy pasting a function found online.
This way, you understand exactly whats going on under the hood!
However I cannot provide a guarantee for speed, especially since this is using streams. Bit operations would definitely be more efficient.
Anywho, hope this helps! I had quite a bit of fun writing this.
I want to use the numbers that strlen() returns in a function that requires an int value.
functionName((strlen(word)+strlen(otherWord)+1));
This piece of code doesn't work because they return size_t.
Is there any way to convert that result to int so I can make addition operations and use them as int?
Most of the other answers are using C-style casts which you shouldn't do in C++:
You should static cast.
functionName( static_cast<int>((strlen(word)+strlen(otherWord)+1));
First you need to check if you can add the values, then you need to check if you can cast to int, and the you cast..
sizt_t size = strlen(word);
if (std::numeric_limits<size_t>::max()-size >= strlen(otherword))
{
size += strlen(otherword);
if (std::numeric_limits<size_t>::max()-size >= 1))
{
size += 1;
if (size <= std::numeric_limits<int>::max())
{
functionName(int(size));
}
else
{
//failed
}
}
else
{
//failed
}
}
else
{
//failed
}
This piece of code does work, unless your compiler does something very strange or you haven't shown us the actual code.
The only thing that could possibly go wrong is an overflow error (unsigned ints can hold larger values than signed ints). You could check that with std::numeric_limits<int> if you really think your strings can become that incredibly large.
size_t is internally unsigned int on most sensible platforms. Typecast it to int and be done with it:
functionName(static_cast<int>((strlen(word)+strlen(otherWord)+1));
or C-style (which is discouraged but legal):
functionName((int)((strlen(word)+strlen(otherWord)+1));
How likely is it that the sum total of those lengths would be over 0x7fffffff (that's the limit of int on 32-bit platforms)? Probably not very likely.
Unless you have veeery long strings here it is safe to use cast:
functionName((int)(strlen(word)+strlen(otherWord)+1));
if you want you can use static_cast<int> but there is no differences now.
functionName(static_cast<int>(strlen(word)+strlen(otherWord)+1));
In C++ use otherWord.length(): functionName( word.length() + otherWord.length() ).
This is one way to do a strlen() of functionName((int)(strlen(word)+strlen(otherWord)+1));.
#include <iostream>
#include <string>
using namespace std;
int functionName(int rstr )
{
return rstr;
}
int main ()
{
char word[] = "aaa";
char otherWord[] = "bbb";
cout<< "word = "<<word<<"\n" ;
cout<< "otherWord = "<<otherWord<<"\n" ;
cout<< "The sentence entered is "<< functionName((int)(strlen(word)+strlen(otherWord) ))<<" characters long \n\n";
return 0;
}
I am writing a compression program, and need to write bit data to a binary file using c++. If anyone could advise on the write statement, or a website with advice, I would be very grateful.
Apologies if this is a simple or confusing question, I am struggling to find answers on web.
Collect the bits into whole bytes, such as an unsigned char or std::bitset (where the bitset size is a multiple of CHAR_BIT), then write whole bytes at a time. Computers "deal with bits", but the available abstraction – especially for IO – is that you, as a programmer, deal with individual bytes. Bitwise manipulation can be used to toggle specific bits, but you're always handling byte-sized objects.
At the end of the output, if you don't have a whole byte, you'll need to decide how that should be stored. Both iostreams and stdio can write unformatted data using ostream::write and fwrite, respectively.
Instead of a single char or bitset<8> (8 being the most common value for CHAR_BIT), you might consider using a larger block size, such as an array of 4-32, or more, chars or the equivalent sized bitset.
For writing binary, the trick I have found most helpful is to store all the binary as a single array in memory and then move it all over to the hard drive. Doing a bit at a time, or a byte at a time, or an unsigned long long at a time is not as fast as having all the data stored in an array and using one instance of "fwrite()" to store it to the hard drive.
size_t fwrite ( const void * ptr, size_t size, size_t count, FILE * stream );
Ref: http://www.cplusplus.com/reference/clibrary/cstdio/fwrite/
In English:
fwrite( [array* of stored data], [size in bytes of array OBJECT. For unsigned chars -> 1, for unsigned long longs -> 8], [number of instances in array], [FILE*])
Always check your returns for validation of success!
Additionally, an argument can be made that having the object type be as large as possible is the fastest way to go ([unsigned long long] > [char]). While I am not versed in the coding behind "fwrite()", I feel the time to convert from the natural object used in your code to [unsigned long long] will take more time when combined with the writing than the "fwrite()" making due with what you have.
Back when I was learning Huffman Coding, it took me a few hours to realize that there was a difference between [char] and [unsigned char]. Notice for this method that you should always use unsigned variables to store the pure binary.
by below class you can write and read bit by bit
class bitChar{
public:
unsigned char* c;
int shift_count;
string BITS;
bitChar()
{
shift_count = 0;
c = (unsigned char*)calloc(1, sizeof(char));
}
string readByBits(ifstream& inf)
{
string s ="";
char buffer[1];
while (inf.read (buffer, 1))
{
s += getBits(*buffer);
}
return s;
}
void setBITS(string X)
{
BITS = X;
}
int insertBits(ofstream& outf)
{
int total = 0;
while(BITS.length())
{
if(BITS[0] == '1')
*c |= 1;
*c <<= 1;
++shift_count;
++total;
BITS.erase(0, 1);
if(shift_count == 7 )
{
if(BITS.size()>0)
{
if(BITS[0] == '1')
*c |= 1;
++total;
BITS.erase(0, 1);
}
writeBits(outf);
shift_count = 0;
free(c);
c = (unsigned char*)calloc(1, sizeof(char));
}
}
if(shift_count > 0)
{
*c <<= (7 - shift_count);
writeBits(outf);
free(c);
c = (unsigned char*)calloc(1, sizeof(char));
}
outf.close();
return total;
}
string getBits(unsigned char X)
{
stringstream itoa;
for(unsigned s = 7; s > 0 ; s--)
{
itoa << ((X >> s) & 1);
}
itoa << (X&1) ;
return itoa.str();
}
void writeBits(ofstream& outf)
{
outf << *c;
}
~bitChar()
{
if(c)
free(c);
}
};
for example
#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <stdlib.h>
using namespace std;
int main()
{
ofstream outf("Sample.dat");
ifstream inf("Sample.dat");
string enCoded = "101000001010101010";
//write to file
cout << enCoded << endl ; //print 101000001010101010
bitChar bchar;
bchar.setBITS(enCoded);
bchar.insertBits(outf);
//read from file
string decoded =bchar.readByBits(inf);
cout << decoded << endl ; //print 101000001010101010000000
return 0;
}