Printing numbers in binary format in C++
The problem
You want to display or output a number in binary format. Iostream only has output manipulators for decimal, hexadecimal and octal.
The bad solution
People usually write a loop to do this that iterates over each bit in the integer:
int v = 0x12345678;
for (int i = 31; i >= 0; i--)
std::cout << ((v >> i) & 1);
There are many ways to do this; the loop above shifts the bit to print into the least significant bit (bit 0) of a temporary and ANDs it with 1 to remove all the other (higher order) bits and leave a 0 or 1. It starts with the most significant bit (31) and iterates down to the least significant bit (0), ensuring that the bits are printed in the correct order.
The good solution
While a looping solution still has applicability in some other languages, there is a much more elegant way to do it in C++ that is often overlooked:
int v = 0x12345678; std::cout << std::bitset<32>(v);
The C++ standard library includes a container type bitset whose first non-type template parameter specifies the number of bits to store. It includes an overload for operator<< and a conversion constructor for ints, making printing binary numbers a piece of cake! This is a much preferred solution to the messy for loop construct above.
To use bitset, remember to add:
#include <bitset>
at the top of your code.
Good luck!
Share your thoughts! Note: to post source code, enclose it in [code lang=...] [/code] tags. Valid values for 'lang' are cpp, csharp, xml, javascript, php etc. To post compiler errors or other text that is best read monospaced, use 'text' as the value for lang.
Katy would love it if you could help keep her stocked with cookies while she codes!
Katy's Code 
it’s a good explanation. just to let you know the bitset examplet is okay, but the syntax at the
int v = 0x12345678;
std::cout << std::bitset(v); is the other way around, cout << bitset(32); 😀 took me a while to figure out why wasn’t working for me.
BOOOH!
not using data words, unsign or format string.
here:
int getValuePNTR(const char* memory, int &start, int size)
{
DWORD retVal = 0;
//now just add up array fields
for (int i = start + size-1,j = size-1; j >= 0; –j, i–)
{
//fprintf(stdout, “\ncycle: %d, memory: [%x]”, j, memory[i]);
if ((unsigned char)memory[i] == 00 && j > 0)
retVal <<= 8;
else
retVal |= ((unsigned char)(memory[i]) << (8 * j));
}
//get the next field after this one
start += size;
return retVal;
}
call:
fprintf(stdout, "\n[DWORD] LoaderFlag (zeroVal) :0x%08x", getValuePNTR(fileContent, readPointer, DWORD_L));
This code automatically assumes the size of a unit of memory. The two assignments to retVal should replace 8 with sizeof(char). The function signature should also replace const char* with const unsigned char* for consistency and to allow you to remove the two unnecessary casts in the inner loop.
😂
So simple to understand the concept of left and right shifting now. Thanks! 😀
it helps me =) Thanks =)
Thank you! It was really helpful! 🙂
Thanks!!
Thank you for your hint! I’m currently struggling with the SPI interface of my Raspi and found out that the bit-endianness (MSB first vs. LSB first) was the problem, so I employed this algo to reverse all the data to get it LSB first: http://stackoverflow.com/questions/2602823/in-c-c-whats-the-simplest-way-to-reverse-the-order-of-bits-in-a-byte
To test this algo, I used your code snippet, and yaaaay, everything works just as it should! *_*
Thanks that’s really useful.
bitset obitset.h is not found when I compile
what do I do if I want to write these numbers in a text file??
I have to say that I have been impressed. A very nice tip this page is.
I’m always dumping binary numbers, and bitset is much cleaner than the looping method. Very nice!