Printing in binary format (user defined function) [duplicate] - c++

Question

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How do I use an uint8_t with I/O streams while avoiding the char behavior?
(2 answers)
Closed 2 years ago.
I have implemented in C++ print_in_binary_format() function, which (as the name states) print 8-bit number in binary:
void print_in_binary_format(std::uint8_t number)
{
for (int bit_index = 7; bit_index >= 0; --bit_index) {
std::cout << ((number & (1U << bit_index)) ? '1' : '0');
}
std::cout << '\n';
}
But it doesn't work as supposed, i.e.:
Input: 2
Output: 00110010
Input: 0
Output: 00110000
I know that in C++ I could use bitset library, but it is not the case.
I would like to know what is wrong with this function, my brain is stuck!
There is whole program to test that:
#include <cstdint>
#include <iostream>
std::uint8_t read_user_number();
void print_in_binary_format(std::uint8_t number);
int main()
{
std::uint8_t number {};
number = read_user_number();
print_in_binary_format(number);
return EXIT_SUCCESS;
}
std::uint8_t read_user_number()
{
std::cout << "Enter a number (0-255): ";
std::uint8_t user_number {};
std::cin >> user_number;
return user_number;
}
void print_in_binary_format(std::uint8_t number)
{
for (int bit_index = 7; bit_index >= 0; --bit_index) {
std::cout << ((number & (1U << bit_index)) ? '1' : '0');
}
std::cout << '\n';
}

Answer 1

The problem is your input data. You are providing 2 or 0 as a character, not as a number.
Let me explain:
character '2' -> ASCII 0x32, 32 is the real input of your function, 00110010 is the bit representation of 0x32.
character '0' -> ASCII 0x30 has 00110000 as binary representation.
Hence the function above is working.
In order to solve the issue look at the way you are collecting the input data.
My 2 cents, Ste.