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Consider the following code for integral types:
template <class T>
std::string as_binary_string( T value ) {
return std::bitset<sizeof( T ) * 8>( value ).to_string();
}
int main() {
unsigned char a(2);
char b(4);
unsigned short c(2);
short d(4);
unsigned int e(2);
int f(4);
unsigned long long g(2);
long long h(4);
std::cout << "a = " << +a << " " << as_binary_string( a ) << std::endl;
std::cout << "b = " << +b << " " << as_binary_string( b ) << std::endl;
std::cout << "c = " << c << " " << as_binary_string( c ) << std::endl;
std::cout << "d = " << c << " " << as_binary_string( d ) << std::endl;
std::cout << "e = " << e << " " << as_binary_string( e ) << std::endl;
std::cout << "f = " << f << " " << as_binary_string( f ) << std::endl;
std::cout << "g = " << g << " " << as_binary_string( g ) << std::endl;
std::cout << "h = " << h << " " << as_binary_string( h ) << std::endl;
std::cout << "\nPress any key and enter to quit.\n";
char q;
std::cin >> q;
return 0;
}
Pretty straight forward, works well and is quite simple.
EDIT
How would one go about writing a function to extract the binary or bit pattern of arbitrary floating point types at compile time?
When it comes to floats I have not found anything similar in any existing libraries of my own knowledge. I've searched google for days looking for one, so then I resorted into trying to write my own function without any success. I no longer have the attempted code available since I've originally asked this question so I can not exactly show you all of the different attempts of implementations along with their compiler - build errors. I was interested in trying to generate the bit pattern for floats in a generic way during compile time and wanted to integrate that into my existing class that seamlessly does the same for any integral type. As for the floating types themselves, I have taken into consideration the different formats as well as architecture endian. For my general purposes the standard IEEE versions of the floating point types is all that I should need to be concerned with.
iBug had suggested for me to write my own function when I originally asked this question, while I was in the attempt of trying to do so. I understand binary numbers, memory sizes, and the mathematics, but when trying to put it all together with how floating point types are stored in memory with their different parts {sign bit, base & exp } is where I was having the most trouble.
Since then with the suggestions those who have given a great answer - example I was able to write a function that would fit nicely into my already existing class template and now it works for my intended purposes.
What about writing one by yourself?
static_assert(sizeof(float) == sizeof(uint32_t));
static_assert(sizeof(double) == sizeof(uint64_t));
std::string as_binary_string( float value ) {
std::uint32_t t;
std::memcpy(&t, &value, sizeof(value));
return std::bitset<sizeof(float) * 8>(t).to_string();
}
std::string as_binary_string( double value ) {
std::uint64_t t;
std::memcpy(&t, &value, sizeof(value));
return std::bitset<sizeof(double) * 8>(t).to_string();
}
You may need to change the helper variable t in case the sizes for the floating point numbers are different.
You can alternatively copy them bit-by-bit. This is slower but serves for arbitrarily any type.
template <typename T>
std::string as_binary_string( T value )
{
const std::size_t nbytes = sizeof(T), nbits = nbytes * CHAR_BIT;
std::bitset<nbits> b;
std::uint8_t buf[nbytes];
std::memcpy(buf, &value, nbytes);
for(int i = 0; i < nbytes; ++i)
{
std::uint8_t cur = buf[i];
int offset = i * CHAR_BIT;
for(int bit = 0; bit < CHAR_BIT; ++bit)
{
b[offset] = cur & 1;
++offset; // Move to next bit in b
cur >>= 1; // Move to next bit in array
}
}
return b.to_string();
}
You said it doesn't need to be standard. So, here is what works in clang on my computer:
#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
char *result;
result=new char[33];
fill(result,result+32,'0');
float input;
cin >>input;
asm(
"mov %0,%%eax\n"
"mov %1,%%rbx\n"
".intel_syntax\n"
"mov rcx,20h\n"
"loop_begin:\n"
"shr eax\n"
"jnc loop_end\n"
"inc byte ptr [rbx+rcx-1]\n"
"loop_end:\n"
"loop loop_begin\n"
".att_syntax\n"
:
: "m" (input), "m" (result)
);
cout <<result <<endl;
delete[] result;
return 0;
}
This code makes a bunch of assumptions about the computer architecture and I am not sure on how many computers it would work.
EDIT:
My computer is a 64-bit Mac-Air. This program basically works by allocating a 33-byte string and filling the first 32 bytes with '0' (the 33rd byte will automatically be '\0').
Then it uses inline assembly to store the float into a 32-bit register and then it repeatedly shifts it to the right by one bit.
If the last bit in the register was 1 before the shift, it gets stored into the carry flag.
The assembly code then checks the carry flag and, if it contains 1, it increases the corresponding byte in the string by 1.
Since it was previously initialized to '0', it will turn to '1'.
So, effectively, when the loop in the assembly is finished, the binary representation of a float is stored into a string.
This code only works for x64 (it uses 64-bit registers "rbx" and "rcx" to store the pointer and the counter for the loop), but I think it's easy to tweak it to work on other processors.
An IEEE floating point number looks like the following
sign exponent mantissa
1 bit 11 bits 52 bits
Note that there's a hidden 1 before the mantissa, and the exponent
is biased so 1023 = 0, not two's complement.
By memcpy()ing to a 64 bit unsigned integer you can then apply AND and
OR masks to get the bit pattern. The arrangement could be big endian
or little endian.
You can easily work out which arrangement you have by passing easy numbers
such as 1 or 2.
Generally people either use std::hexfloat or cast a pointer to the floating-point value to a pointer to an unsigned integer of the same size and print the indirected value in hex format. Both methods facilitate bit-level analysis of floating-point in a productive fashion.
You could roll your by casting the address of the float/double to a char and iterating it that way:
#include <memory>
#include <iostream>
#include <limits>
#include <iomanip>
template <typename T>
std::string getBits(T t) {
std::string returnString{""};
char *base{reinterpret_cast<char *>(std::addressof(t))};
char *tail{base + sizeof(t) - 1};
do {
for (int bits = std::numeric_limits<unsigned char>::digits - 1; bits >= 0; bits--) {
returnString += ( ((*tail) & (1 << bits)) ? '1' : '0');
}
} while (--tail >= base);
return returnString;
}
int main() {
float f{10.0};
double d{100.0};
double nd{-100.0};
std::cout << std::setprecision(1);
std::cout << getBits(f) << std::endl;
std::cout << getBits(d) << std::endl;
std::cout << getBits(nd) << std::endl;
}
Output on my machine (note the sign flip in the third output):
01000001001000000000000000000000
0100000001011001000000000000000000000000000000000000000000000000
1100000001011001000000000000000000000000000000000000000000000000
I have a number stored as a ulong. I want the bits stored in memory to be interpreted in a 2's complement fashion. So I want the first bit to be the sign bit etc. If I want to convert to a long, so that the number is interpreted correctly as a 2's complement , how do I do this?
I tried creating pointers of different data types that all pointed to the same buffer. I then stored the ulong into the buffer. I then dereferenced a long pointer. This however is giving me a bad result?
I did :
#include <iostream>
using namespace std;
int main() {
unsigned char converter_buffer[4];//
unsigned long *pulong;
long *plong;
pulong = (unsigned long*)&converter_buffer;
plong = (long*)&converter_buffer;
unsigned long ulong_num = 65535; // this has a 1 as the first bit
*pulong = ulong_num;
std:: cout << "the number as a long is" << *plong << std::endl;
return 0;
}
For some reason this is giving me the same positive number.
Would casting help?
Actually using pointers was a good start but you have to cast your unsigned long* to void* first, then you can cast the result to long* and dereference it:
#include <iostream>
#include <climits>
int main() {
unsigned long ulongValue = ULONG_MAX;
long longValue = *((long*)((void*)&ulongValue));
std::cout << "ulongValue: " << ulongValue << std::endl;
std::cout << "longValue: " << longValue << std::endl;
return 0;
}
The code above will results the following:
ulongValue: 18446744073709551615
longValue: -1
With templates you can make it more readable in your code:
#include <iostream>
#include <climits>
template<typename T, typename U>
T unsafe_cast(const U& from) {
return *((T*)((void*)&from));
}
int main() {
unsigned long ulongValue = ULONG_MAX;
long longValue = unsafe_cast<long>(ulongValue);
std::cout << "ulongValue: " << ulongValue << std::endl;
std::cout << "longValue: " << longValue << std::endl;
return 0;
}
Keep in mind that this solution is absolutely unsafe due to the fact that you can cast anyithing to void*. This practicle was common in C but I do not recommend to use it in C++. Consider the following cases:
#include <iostream>
template<typename T, typename U>
T unsafe_cast(const U& from) {
return *((T*)((void*)&from));
}
int main() {
std::cout << std::hex << std::showbase;
float fValue = 3.14;
int iValue = unsafe_cast<int>(fValue); // OK, they have same size.
std::cout << "Hexadecimal representation of " << fValue
<< " is: " << iValue << std::endl;
std::cout << "Converting back to float results: "
<< unsafe_cast<float>(iValue) << std::endl;
double dValue = 3.1415926535;
int lossyValue = unsafe_cast<int>(dValue); // Bad, they have different size.
std::cout << "Lossy hexadecimal representation of " << dValue
<< " is: " << lossyValue << std::endl;
std::cout << "Converting back to double results: "
<< unsafe_cast<double>(lossyValue) << std::endl;
return 0;
}
The code above results for me the following:
Hexadecimal representation of 3.14 is: 0x4048f5c3
Converting back to float results: 3.14
Lossy hexadecimal representation of 3.14159 is: 0x54411744
Converting back to double results: 6.98387e-315
And for last line you can get anything because the conversion will read garbage from the memory.
Edit
As lorro commented bellow, using memcpy() is safer and can prevent the overflow. So, here is another version of type casting which is safer:
template<typename T, typename U>
T safer_cast(const U& from) {
T to;
memcpy(&to, &from, (sizeof(T) > sizeof(U) ? sizeof(U) : sizeof(T)));
return to;
}
You can do this:
uint32_t u;
int32_t& s = (int32_t&) u;
Then you can use s and u interchangeably with 2's complement, e.g.:
s = -1;
std::cout << u << '\n'; // 4294967295
In your question you ask about 65535 but that is a positive number. You could do:
uint16_t u;
int16_t& s = (int16_t&) u;
u = 65535;
std::cout << s << '\n'; // -1
Note that assigning 65535 (a positive number) to int16_t would implementation-defined behaviour, it does not necessarily give -1.
The problem with your original code is that it is not permitted to alias a char buffer as long. (And that you might overflow your buffer). However, it is OK to alias an integer type as its corresponding signed/unsigned type.
In general, when you have two arithmetic types that are the same size and you want to reinterpret the bit representation of one using the type of the other, you do it with a union:
#include <stdint.h>
union reinterpret_u64_d_union {
uint64_t u64;
double d;
};
double
reinterpret_u64_as_double(uint64_t v)
{
union reinterpret_u64_d_union u;
u.u64 = v;
return u.d;
}
For the special case of turning an unsigned number into a signed type with the same size (or vice versa), however, you can just use a traditional cast:
int64_t
reinterpret_u64_as_i64(uint64_t v)
{
return (int64_t)v;
}
(The cast is not strictly required for [u]int64_t, but if you don't explicitly write a cast, and the types you're converting between are small, the "integer promotions" may get involved, which is usually undesirable.)
The way you were trying to do it violates the pointer-aliasing rules and provokes undefined behavior.
In C++, note that reinterpret_cast<> does not do what the union does; it is the same as static_cast<> when applied to arithmetic types.
In C++, also note that the use of a union above relies on a rule in the 1999 C standard (with corrigienda) that has not been officially incorporated into the C++ standard last I checked; however, all compilers I am familiar with will do what you expect.
And finally, in both C and C++, long and unsigned long are guaranteed to be able to represent at least −2,147,483,647 ... 214,7483,647 and 0 ... 4,294,967,295, respectively. Your test program used 65535, which is guaranteed to be representable by both long and unsigned long, so the value would have been unchanged however you did it. Well, unless you used invalid pointer aliasing and the compiler decided to make demons fly out of your nose instead.
I'm following a college course about operating systems and we're learning how to convert from binary to hexadecimal, decimal to hexadecimal, etc. and today we just learned how signed/unsigned numbers are stored in memory using the two's complement (~number + 1).
We have a couple of exercises to do on paper and I would like to be able to verify my answers before submitting my work to the teacher. I wrote a C++ program for the first few exercises but now I'm stuck as to how I could verify my answer with the following problem:
char a, b;
short c;
a = -58;
c = -315;
b = a >> 3;
and we need to show the binary representation in memory of a, b and c.
I've done it on paper and it gives me the following results (all the binary representations in memory of the numbers after the two's complement):
a = 00111010 (it's a char, so 1 byte)
b = 00001000 (it's a char, so 1 byte)
c = 11111110 11000101 (it's a short, so 2 bytes)
Is there a way to verify my answer? Is there a standard way in C++ to show the binary representation in memory of a number, or do I have to code each step myself (calculate the two's complement and then convert to binary)? I know the latter wouldn't take so long but I'm curious as to if there is a standard way to do so.
The easiest way is probably to create an std::bitset representing the value, then stream that to cout.
#include <bitset>
...
char a = -58;
std::bitset<8> x(a);
std::cout << x << '\n';
short c = -315;
std::bitset<16> y(c);
std::cout << y << '\n';
Use on-the-fly conversion to std::bitset. No temporary variables, no loops, no functions, no macros.
Live On Coliru
#include <iostream>
#include <bitset>
int main() {
int a = -58, b = a>>3, c = -315;
std::cout << "a = " << std::bitset<8>(a) << std::endl;
std::cout << "b = " << std::bitset<8>(b) << std::endl;
std::cout << "c = " << std::bitset<16>(c) << std::endl;
}
Prints:
a = 11000110
b = 11111000
c = 1111111011000101
In C++20 you can use std::format to do this:
unsigned char a = -58;
std::cout << std::format("{:b}", a);
Output:
11000110
On older systems you can use the {fmt} library, std::format is based on. {fmt} also provides the print function that makes this even easier and more efficient (godbolt):
unsigned char a = -58;
fmt::print("{:b}", a);
Disclaimer: I'm the author of {fmt} and C++20 std::format.
If you want to display the bit representation of any object, not just an integer, remember to reinterpret as a char array first, then you can print the contents of that array, as hex, or even as binary (via bitset):
#include <iostream>
#include <bitset>
#include <climits>
template<typename T>
void show_binrep(const T& a)
{
const char* beg = reinterpret_cast<const char*>(&a);
const char* end = beg + sizeof(a);
while(beg != end)
std::cout << std::bitset<CHAR_BIT>(*beg++) << ' ';
std::cout << '\n';
}
int main()
{
char a, b;
short c;
a = -58;
c = -315;
b = a >> 3;
show_binrep(a);
show_binrep(b);
show_binrep(c);
float f = 3.14;
show_binrep(f);
}
Note that most common systems are little-endian, so the output of show_binrep(c) is not the 1111111 011000101 you expect, because that's not how it's stored in memory. If you're looking for value representation in binary, then a simple cout << bitset<16>(c) works.
Is there a standard way in C++ to show the binary representation in memory of a number [...]?
No. There's no std::bin, like std::hex or std::dec, but it's not hard to output a number binary yourself:
You output the left-most bit by masking all the others, left-shift, and repeat that for all the bits you have.
(The number of bits in a type is sizeof(T) * CHAR_BIT.)
Similar to what is already posted, just using bit-shift and mask to get the bit; usable for any type, being a template (only not sure if there is a standard way to get number of bits in 1 byte, I used 8 here).
#include<iostream>
#include <climits>
template<typename T>
void printBin(const T& t){
size_t nBytes=sizeof(T);
char* rawPtr((char*)(&t));
for(size_t byte=0; byte<nBytes; byte++){
for(size_t bit=0; bit<CHAR_BIT; bit++){
std::cout<<(((rawPtr[byte])>>bit)&1);
}
}
std::cout<<std::endl;
};
int main(void){
for(int i=0; i<50; i++){
std::cout<<i<<": ";
printBin(i);
}
}
Reusable function:
template<typename T>
static std::string toBinaryString(const T& x)
{
std::stringstream ss;
ss << std::bitset<sizeof(T) * 8>(x);
return ss.str();
}
Usage:
int main(){
uint16_t x=8;
std::cout << toBinaryString(x);
}
This works with all kind of integers.
#include <iostream>
#include <cmath> // in order to use pow() function
using namespace std;
string show_binary(unsigned int u, int num_of_bits);
int main()
{
cout << show_binary(128, 8) << endl; // should print 10000000
cout << show_binary(128, 5) << endl; // should print 00000
cout << show_binary(128, 10) << endl; // should print 0010000000
return 0;
}
string show_binary(unsigned int u, int num_of_bits)
{
string a = "";
int t = pow(2, num_of_bits); // t is the max number that can be represented
for(t; t>0; t = t/2) // t iterates through powers of 2
if(u >= t){ // check if u can be represented by current value of t
u -= t;
a += "1"; // if so, add a 1
}
else {
a += "0"; // if not, add a 0
}
return a ; // returns string
}
Using the std::bitset answers and convenience templates:
#include <iostream>
#include <bitset>
#include <climits>
template<typename T>
struct BinaryForm {
BinaryForm(const T& v) : _bs(v) {}
const std::bitset<sizeof(T)*CHAR_BIT> _bs;
};
template<typename T>
inline std::ostream& operator<<(std::ostream& os, const BinaryForm<T>& bf) {
return os << bf._bs;
}
Using it like this:
auto c = 'A';
std::cout << "c: " << c << " binary: " << BinaryForm{c} << std::endl;
unsigned x = 1234;
std::cout << "x: " << x << " binary: " << BinaryForm{x} << std::endl;
int64_t z { -1024 };
std::cout << "z: " << z << " binary: " << BinaryForm{z} << std::endl;
Generates output:
c: A binary: 01000001
x: 1234 binary: 00000000000000000000010011010010
z: -1024 binary: 1111111111111111111111111111111111111111111111111111110000000000
Using old C++ version, you can use this snippet :
template<typename T>
string toBinary(const T& t)
{
string s = "";
int n = sizeof(T)*8;
for(int i=n-1; i>=0; i--)
{
s += (t & (1 << i))?"1":"0";
}
return s;
}
int main()
{
char a, b;
short c;
a = -58;
c = -315;
b = a >> 3;
cout << "a = " << a << " => " << toBinary(a) << endl;
cout << "b = " << b << " => " << toBinary(b) << endl;
cout << "c = " << c << " => " << toBinary(c) << endl;
}
a = => 11000110
b = => 11111000
c = -315 => 1111111011000101
I have had this problem when playing competitive coding games online. Here is a solution that is quick to implement and is fairly intuitive. It also avoids outputting leading zeros or relying on <bitset>
std::string s;
do {
s = std::to_string(r & 1) + s;
} while ( r>>=1 );
std::cout << s;
You should note however that this solution will increase your runtime, so if you are competing for optimization or not competing at all you should use one of the other solutions on this page.
Here is the true way to get binary representation of a number:
unsigned int i = *(unsigned int*) &x;
Is this what you're looking for?
std::cout << std::hex << val << std::endl;
I'm following a college course about operating systems and we're learning how to convert from binary to hexadecimal, decimal to hexadecimal, etc. and today we just learned how signed/unsigned numbers are stored in memory using the two's complement (~number + 1).
We have a couple of exercises to do on paper and I would like to be able to verify my answers before submitting my work to the teacher. I wrote a C++ program for the first few exercises but now I'm stuck as to how I could verify my answer with the following problem:
char a, b;
short c;
a = -58;
c = -315;
b = a >> 3;
and we need to show the binary representation in memory of a, b and c.
I've done it on paper and it gives me the following results (all the binary representations in memory of the numbers after the two's complement):
a = 00111010 (it's a char, so 1 byte)
b = 00001000 (it's a char, so 1 byte)
c = 11111110 11000101 (it's a short, so 2 bytes)
Is there a way to verify my answer? Is there a standard way in C++ to show the binary representation in memory of a number, or do I have to code each step myself (calculate the two's complement and then convert to binary)? I know the latter wouldn't take so long but I'm curious as to if there is a standard way to do so.
The easiest way is probably to create an std::bitset representing the value, then stream that to cout.
#include <bitset>
...
char a = -58;
std::bitset<8> x(a);
std::cout << x << '\n';
short c = -315;
std::bitset<16> y(c);
std::cout << y << '\n';
Use on-the-fly conversion to std::bitset. No temporary variables, no loops, no functions, no macros.
Live On Coliru
#include <iostream>
#include <bitset>
int main() {
int a = -58, b = a>>3, c = -315;
std::cout << "a = " << std::bitset<8>(a) << std::endl;
std::cout << "b = " << std::bitset<8>(b) << std::endl;
std::cout << "c = " << std::bitset<16>(c) << std::endl;
}
Prints:
a = 11000110
b = 11111000
c = 1111111011000101
In C++20 you can use std::format to do this:
unsigned char a = -58;
std::cout << std::format("{:b}", a);
Output:
11000110
On older systems you can use the {fmt} library, std::format is based on. {fmt} also provides the print function that makes this even easier and more efficient (godbolt):
unsigned char a = -58;
fmt::print("{:b}", a);
Disclaimer: I'm the author of {fmt} and C++20 std::format.
If you want to display the bit representation of any object, not just an integer, remember to reinterpret as a char array first, then you can print the contents of that array, as hex, or even as binary (via bitset):
#include <iostream>
#include <bitset>
#include <climits>
template<typename T>
void show_binrep(const T& a)
{
const char* beg = reinterpret_cast<const char*>(&a);
const char* end = beg + sizeof(a);
while(beg != end)
std::cout << std::bitset<CHAR_BIT>(*beg++) << ' ';
std::cout << '\n';
}
int main()
{
char a, b;
short c;
a = -58;
c = -315;
b = a >> 3;
show_binrep(a);
show_binrep(b);
show_binrep(c);
float f = 3.14;
show_binrep(f);
}
Note that most common systems are little-endian, so the output of show_binrep(c) is not the 1111111 011000101 you expect, because that's not how it's stored in memory. If you're looking for value representation in binary, then a simple cout << bitset<16>(c) works.
Is there a standard way in C++ to show the binary representation in memory of a number [...]?
No. There's no std::bin, like std::hex or std::dec, but it's not hard to output a number binary yourself:
You output the left-most bit by masking all the others, left-shift, and repeat that for all the bits you have.
(The number of bits in a type is sizeof(T) * CHAR_BIT.)
Similar to what is already posted, just using bit-shift and mask to get the bit; usable for any type, being a template (only not sure if there is a standard way to get number of bits in 1 byte, I used 8 here).
#include<iostream>
#include <climits>
template<typename T>
void printBin(const T& t){
size_t nBytes=sizeof(T);
char* rawPtr((char*)(&t));
for(size_t byte=0; byte<nBytes; byte++){
for(size_t bit=0; bit<CHAR_BIT; bit++){
std::cout<<(((rawPtr[byte])>>bit)&1);
}
}
std::cout<<std::endl;
};
int main(void){
for(int i=0; i<50; i++){
std::cout<<i<<": ";
printBin(i);
}
}
Reusable function:
template<typename T>
static std::string toBinaryString(const T& x)
{
std::stringstream ss;
ss << std::bitset<sizeof(T) * 8>(x);
return ss.str();
}
Usage:
int main(){
uint16_t x=8;
std::cout << toBinaryString(x);
}
This works with all kind of integers.
#include <iostream>
#include <cmath> // in order to use pow() function
using namespace std;
string show_binary(unsigned int u, int num_of_bits);
int main()
{
cout << show_binary(128, 8) << endl; // should print 10000000
cout << show_binary(128, 5) << endl; // should print 00000
cout << show_binary(128, 10) << endl; // should print 0010000000
return 0;
}
string show_binary(unsigned int u, int num_of_bits)
{
string a = "";
int t = pow(2, num_of_bits); // t is the max number that can be represented
for(t; t>0; t = t/2) // t iterates through powers of 2
if(u >= t){ // check if u can be represented by current value of t
u -= t;
a += "1"; // if so, add a 1
}
else {
a += "0"; // if not, add a 0
}
return a ; // returns string
}
Using the std::bitset answers and convenience templates:
#include <iostream>
#include <bitset>
#include <climits>
template<typename T>
struct BinaryForm {
BinaryForm(const T& v) : _bs(v) {}
const std::bitset<sizeof(T)*CHAR_BIT> _bs;
};
template<typename T>
inline std::ostream& operator<<(std::ostream& os, const BinaryForm<T>& bf) {
return os << bf._bs;
}
Using it like this:
auto c = 'A';
std::cout << "c: " << c << " binary: " << BinaryForm{c} << std::endl;
unsigned x = 1234;
std::cout << "x: " << x << " binary: " << BinaryForm{x} << std::endl;
int64_t z { -1024 };
std::cout << "z: " << z << " binary: " << BinaryForm{z} << std::endl;
Generates output:
c: A binary: 01000001
x: 1234 binary: 00000000000000000000010011010010
z: -1024 binary: 1111111111111111111111111111111111111111111111111111110000000000
Using old C++ version, you can use this snippet :
template<typename T>
string toBinary(const T& t)
{
string s = "";
int n = sizeof(T)*8;
for(int i=n-1; i>=0; i--)
{
s += (t & (1 << i))?"1":"0";
}
return s;
}
int main()
{
char a, b;
short c;
a = -58;
c = -315;
b = a >> 3;
cout << "a = " << a << " => " << toBinary(a) << endl;
cout << "b = " << b << " => " << toBinary(b) << endl;
cout << "c = " << c << " => " << toBinary(c) << endl;
}
a = => 11000110
b = => 11111000
c = -315 => 1111111011000101
I have had this problem when playing competitive coding games online. Here is a solution that is quick to implement and is fairly intuitive. It also avoids outputting leading zeros or relying on <bitset>
std::string s;
do {
s = std::to_string(r & 1) + s;
} while ( r>>=1 );
std::cout << s;
You should note however that this solution will increase your runtime, so if you are competing for optimization or not competing at all you should use one of the other solutions on this page.
Here is the true way to get binary representation of a number:
unsigned int i = *(unsigned int*) &x;
Is this what you're looking for?
std::cout << std::hex << val << std::endl;
I'm following a college course about operating systems and we're learning how to convert from binary to hexadecimal, decimal to hexadecimal, etc. and today we just learned how signed/unsigned numbers are stored in memory using the two's complement (~number + 1).
We have a couple of exercises to do on paper and I would like to be able to verify my answers before submitting my work to the teacher. I wrote a C++ program for the first few exercises but now I'm stuck as to how I could verify my answer with the following problem:
char a, b;
short c;
a = -58;
c = -315;
b = a >> 3;
and we need to show the binary representation in memory of a, b and c.
I've done it on paper and it gives me the following results (all the binary representations in memory of the numbers after the two's complement):
a = 00111010 (it's a char, so 1 byte)
b = 00001000 (it's a char, so 1 byte)
c = 11111110 11000101 (it's a short, so 2 bytes)
Is there a way to verify my answer? Is there a standard way in C++ to show the binary representation in memory of a number, or do I have to code each step myself (calculate the two's complement and then convert to binary)? I know the latter wouldn't take so long but I'm curious as to if there is a standard way to do so.
The easiest way is probably to create an std::bitset representing the value, then stream that to cout.
#include <bitset>
...
char a = -58;
std::bitset<8> x(a);
std::cout << x << '\n';
short c = -315;
std::bitset<16> y(c);
std::cout << y << '\n';
Use on-the-fly conversion to std::bitset. No temporary variables, no loops, no functions, no macros.
Live On Coliru
#include <iostream>
#include <bitset>
int main() {
int a = -58, b = a>>3, c = -315;
std::cout << "a = " << std::bitset<8>(a) << std::endl;
std::cout << "b = " << std::bitset<8>(b) << std::endl;
std::cout << "c = " << std::bitset<16>(c) << std::endl;
}
Prints:
a = 11000110
b = 11111000
c = 1111111011000101
In C++20 you can use std::format to do this:
unsigned char a = -58;
std::cout << std::format("{:b}", a);
Output:
11000110
On older systems you can use the {fmt} library, std::format is based on. {fmt} also provides the print function that makes this even easier and more efficient (godbolt):
unsigned char a = -58;
fmt::print("{:b}", a);
Disclaimer: I'm the author of {fmt} and C++20 std::format.
If you want to display the bit representation of any object, not just an integer, remember to reinterpret as a char array first, then you can print the contents of that array, as hex, or even as binary (via bitset):
#include <iostream>
#include <bitset>
#include <climits>
template<typename T>
void show_binrep(const T& a)
{
const char* beg = reinterpret_cast<const char*>(&a);
const char* end = beg + sizeof(a);
while(beg != end)
std::cout << std::bitset<CHAR_BIT>(*beg++) << ' ';
std::cout << '\n';
}
int main()
{
char a, b;
short c;
a = -58;
c = -315;
b = a >> 3;
show_binrep(a);
show_binrep(b);
show_binrep(c);
float f = 3.14;
show_binrep(f);
}
Note that most common systems are little-endian, so the output of show_binrep(c) is not the 1111111 011000101 you expect, because that's not how it's stored in memory. If you're looking for value representation in binary, then a simple cout << bitset<16>(c) works.
Is there a standard way in C++ to show the binary representation in memory of a number [...]?
No. There's no std::bin, like std::hex or std::dec, but it's not hard to output a number binary yourself:
You output the left-most bit by masking all the others, left-shift, and repeat that for all the bits you have.
(The number of bits in a type is sizeof(T) * CHAR_BIT.)
Similar to what is already posted, just using bit-shift and mask to get the bit; usable for any type, being a template (only not sure if there is a standard way to get number of bits in 1 byte, I used 8 here).
#include<iostream>
#include <climits>
template<typename T>
void printBin(const T& t){
size_t nBytes=sizeof(T);
char* rawPtr((char*)(&t));
for(size_t byte=0; byte<nBytes; byte++){
for(size_t bit=0; bit<CHAR_BIT; bit++){
std::cout<<(((rawPtr[byte])>>bit)&1);
}
}
std::cout<<std::endl;
};
int main(void){
for(int i=0; i<50; i++){
std::cout<<i<<": ";
printBin(i);
}
}
Reusable function:
template<typename T>
static std::string toBinaryString(const T& x)
{
std::stringstream ss;
ss << std::bitset<sizeof(T) * 8>(x);
return ss.str();
}
Usage:
int main(){
uint16_t x=8;
std::cout << toBinaryString(x);
}
This works with all kind of integers.
#include <iostream>
#include <cmath> // in order to use pow() function
using namespace std;
string show_binary(unsigned int u, int num_of_bits);
int main()
{
cout << show_binary(128, 8) << endl; // should print 10000000
cout << show_binary(128, 5) << endl; // should print 00000
cout << show_binary(128, 10) << endl; // should print 0010000000
return 0;
}
string show_binary(unsigned int u, int num_of_bits)
{
string a = "";
int t = pow(2, num_of_bits); // t is the max number that can be represented
for(t; t>0; t = t/2) // t iterates through powers of 2
if(u >= t){ // check if u can be represented by current value of t
u -= t;
a += "1"; // if so, add a 1
}
else {
a += "0"; // if not, add a 0
}
return a ; // returns string
}
Using the std::bitset answers and convenience templates:
#include <iostream>
#include <bitset>
#include <climits>
template<typename T>
struct BinaryForm {
BinaryForm(const T& v) : _bs(v) {}
const std::bitset<sizeof(T)*CHAR_BIT> _bs;
};
template<typename T>
inline std::ostream& operator<<(std::ostream& os, const BinaryForm<T>& bf) {
return os << bf._bs;
}
Using it like this:
auto c = 'A';
std::cout << "c: " << c << " binary: " << BinaryForm{c} << std::endl;
unsigned x = 1234;
std::cout << "x: " << x << " binary: " << BinaryForm{x} << std::endl;
int64_t z { -1024 };
std::cout << "z: " << z << " binary: " << BinaryForm{z} << std::endl;
Generates output:
c: A binary: 01000001
x: 1234 binary: 00000000000000000000010011010010
z: -1024 binary: 1111111111111111111111111111111111111111111111111111110000000000
Using old C++ version, you can use this snippet :
template<typename T>
string toBinary(const T& t)
{
string s = "";
int n = sizeof(T)*8;
for(int i=n-1; i>=0; i--)
{
s += (t & (1 << i))?"1":"0";
}
return s;
}
int main()
{
char a, b;
short c;
a = -58;
c = -315;
b = a >> 3;
cout << "a = " << a << " => " << toBinary(a) << endl;
cout << "b = " << b << " => " << toBinary(b) << endl;
cout << "c = " << c << " => " << toBinary(c) << endl;
}
a = => 11000110
b = => 11111000
c = -315 => 1111111011000101
I have had this problem when playing competitive coding games online. Here is a solution that is quick to implement and is fairly intuitive. It also avoids outputting leading zeros or relying on <bitset>
std::string s;
do {
s = std::to_string(r & 1) + s;
} while ( r>>=1 );
std::cout << s;
You should note however that this solution will increase your runtime, so if you are competing for optimization or not competing at all you should use one of the other solutions on this page.
Here is the true way to get binary representation of a number:
unsigned int i = *(unsigned int*) &x;
Is this what you're looking for?
std::cout << std::hex << val << std::endl;