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.