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?
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.
current_bit = 63 to current_bit = ((sizeof value) * 8) - 1bits variable come from?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);
itoaconverts an integer to a string..... Could you provide some sample code from those examples that you mentioned?itoais not a standard function. Also, do you mean you want to convert an integer to a string with only binary digits?