int x = 5;
cout<<(char)x;
the code above outputs an int x in raw binary, but only 1 byte. what I need it to do is output the x as 4-bytes in binary, because in my code, x can be anywhere between 0 and 2^32-1, since
cout<<(int)x;
doesn't do the trick, how would I do it?
A bit late, but, as Katy shows in her blog, this might be an elegant solution:
#include <bitset>
#include <iostream>
int main(){
int x=5;
std::cout<<std::bitset<32>(x)<<std::endl;
}
taken from: https://katyscode.wordpress.com/2012/05/12/printing-numbers-in-binary-format-in-c/
You can use the std::ostream::write() member function:
std::cout.write(reinterpret_cast<const char*>(&x), sizeof x);
Note that you would usually want to do this with a stream that has been opened in binary mode.
Try:
int x = 5;
std::cout.write(reinterpret_cast<const char*>(&x),sizeof(x));
Note: That writting data in binary format is non portable.
If you want to read it on an alternative machine you need to either have exactly the same architecture or you need to standardise the format and make sure all machines use the standard format.
If you want to write binary the easiest way to standardise the format is to convert data to network format (there is a set of functions for that htonl() <--> ntohl() etc)
int x = 5;
u_long transport = htonl(x);
std::cout.write(reinterpret_cast<const char*>(&transport), sizeof(u_long));
But the most transportable format is to just convert to text.
std::cout << x;
and what about this?
int x = 5;
cout<<(char) ((0xff000000 & x) >> 24);
cout<<(char) ((0x00ff0000 & x) >> 16);
cout<<(char) ((0x0000ff00 & x) >> 8);
cout<<(char) (0x000000ff & x);
A couple of hints.
First, to be between 0 and 2^32 - 1 you'll need an unsigned four-byte int.
Second, the four bytes starting at the address of x (&x) already have the bytes you want.
Does that help?
Starting with C++20, one can also use std::format and the binary format specifier b:
#include <format>
#include <iostream>
int x = 5;
std::cout << std::format("In binary: {:b}\n", x); // 101
std::cout << std::format("In binary: {:08b}\n", x); // 00000101
std::cout << std::format("In binary: {:#08b}\n", x); // 0b000101
cout, because you don't have control over how it's opened. If you want to do binary output, open your own stream and include theios_base::binaryflag in youropenmmodeargument.