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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 have a bitset in which I need to store a # of randomly generated integers (store its bit representation of course). So, the thing is that I am confuse on how to do that.
i.e suppose that I generate the integers (all unsigned int) 8, 15 , 20, one at a time. How can I store the recently generated integer in my existing bit set.
Say that I start by generating "8" and store in the bitset, then I generate "15" and store it in the bitset.
I don't know or don't understand how to store those values within the bitset.
Note: I know in advance the size of the bitset, the size is based on the number of integers that I am going to generate, and that I know too. So, at the end what I need is a bitset with all the bits set matching the bits of all the generated integers.
I'll Appreciate your help.
How can I store the recently generated integer in my existing bit set.
You can generate a temporary bitset form the integer and then assign values between the two bitsets.
Example program:
#include <iostream>
#include <bitset>
#include <cstdlib>
int main()
{
const int size = sizeof(int)*8;
std::bitset<2*size> res;
std::bitset<size> res1(rand());
std::bitset<size> res2(rand());
for ( size_t i = 0; i < size; ++i )
{
res[i] = res1[i];
res[size+i] = res2[i];
}
std::cout << "res1: " << res1 << std::endl;
std::cout << "res2: " << res2 << std::endl;
std::cout << "res: " << res << std::endl;
return 0;
}
Output:
res1: 01101011100010110100010101100111
res2: 00110010011110110010001111000110
res: 0011001001111011001000111100011001101011100010110100010101100111
Update
A function to set the bitset values given an integer can be used to avoid the cost of creating temporary bitsets.
#include <iostream>
#include <bitset>
#include <cstdlib>
#include <climits>
const int size = sizeof(int)*8;
void setBitsetValue(std::bitset<2*size>& res,
int num,
size_t bitsetIndex,
size_t numIndex)
{
if ( numIndex < size )
{
res[bitsetIndex] = (num >> numIndex) & 0x1;
setBitsetValue(res, num, bitsetIndex+1, numIndex+1);
}
}
int main()
{
std::bitset<2*size> res;
int num1 = rand()%INT_MAX;
int num2 = rand()%INT_MAX;
std::bitset<size> res1(num1);
std::bitset<size> res2(num2);
std::cout << "res1: " << res1 << std::endl;
std::cout << "res2: " << res2 << std::endl;
setBitsetValue(res, num1, 0, 0);
setBitsetValue(res, num2, size, 0);
std::cout << "res: " << res << std::endl;
return 0;
}
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 have this array : BYTE set[6] = { 0xA8,0x12,0x84,0x03,0x00,0x00, } and i need to insert this value : "" int Value = 1200; "" ....on last 4 bytes. Practically to convert from int to hex and then to write inside the array...
Is this possible ?
I already have BitConverter::GetBytes function, but that's not enough.
Thank you,
To answer original quesion: sure you can.
As soon as your sizeof(int) == 4 and sizeof(BYTE) == 1.
But I'm not sure what you mean by "converting int to hex". If you want a hex string representation, you'll be much better off just using one of standard methods of doing it.
For example, on last line I use std::hex to print numbers as hex.
Here is solution to what you've been asking for and a little more (live example: http://codepad.org/rsmzngUL):
#include <iostream>
using namespace std;
int main() {
const int value = 1200;
unsigned char set[] = { 0xA8,0x12,0x84,0x03,0x00,0x00 };
for (const unsigned char* c = set; c != set + sizeof(set); ++c) {
cout << static_cast<int>(*c) << endl;
}
cout << endl << "Putting value into array:" << endl;
*reinterpret_cast<int*>(&set[2]) = value;
for (const unsigned char* c = set; c != set + sizeof(set); ++c) {
cout << static_cast<int>(*c) << endl;
}
cout << endl << "Printing int's bytes one by one: " << endl;
for (int byteNumber = 0; byteNumber != sizeof(int); ++byteNumber) {
const unsigned char oneByte = reinterpret_cast<const unsigned char*>(&value)[byteNumber];
cout << static_cast<int>(oneByte) << endl;
}
cout << endl << "Printing value as hex: " << hex << value << std::endl;
}
UPD: From comments to your question:
1. If you need just getting separate digits out of the number in separate bytes, it's a different story.
2. Little vs Big endianness matters as well, I did not account for that in my answer.
did you mean this ?
#include <stdio.h>
#include <stdlib.h>
#define BYTE unsigned char
int main ( void )
{
BYTE set[6] = { 0xA8,0x12,0x84,0x03,0x00,0x00, } ;
sprintf ( &set[2] , "%d" , 1200 ) ;
printf ( "\n%c%c%c%c", set[2],set[3],set[4],set[5] ) ;
return 0 ;
}
output :
1200
First off, I’m not concerned with portability, and can safely assume that the endianness will not change. Assuming I read a hardware register value, I would like to overlay that register value over bitfields so that I can refer to the individual fields in the register without using bit masks.
EDIT: Fixed problems pointed out by GMan, and adjusted the code so it's clearer for future readers.
SEE: Anders K. & Michael J's answers below for a more eloquent solution.
#include <iostream>
/// \class HardwareRegister
/// Abstracts out bitfields in a hardware register.
/// \warning This is non-portable code.
class HardwareRegister
{
public:
/// Constructor.
/// \param[in] registerValue - the value of the entire register. The
/// value will be overlayed onto the bitfields
/// defined in this class.
HardwareRegister(unsigned long registerValue = 0)
{
/// Lots of casting to get registerValue to overlay on top of the
/// bitfields
*this = *(reinterpret_cast<HardwareRegister*>(®isterValue));
}
/// Bitfields of this register.
/// The data type of this field should be the same size as the register
/// unsigned short for 16 bit register
/// unsigned long for 32 bit register.
///
/// \warning Remember endianess! Order of the following bitfields are
/// important.
/// Big Endian - Start with the most signifcant bits first.
/// Little Endian - Start with the least signifcant bits first.
unsigned long field1: 8;
unsigned long field2:16;
unsigned long field3: 8;
}; //end class Hardware
int main()
{
unsigned long registerValue = 0xFFFFFF00;
HardwareRegister testRegister(registerValue);
// Prints out for little endianess machine
// Field 1 = 0
// Field 2 = 65535
// Field 3 = 255
std::cout << "Field 1 = " << testRegister.field1 << std::endl;
std::cout << "Field 2 = " << testRegister.field2 << std::endl;
std::cout << "Field 3 = " << testRegister.field3 << std::endl;
}
don't do this
*this = *(reinterpret_cast<HW_Register*>(®isterValue));
the 'this' pointer shouldn't be fiddled with in that way:
HW_Register reg(val)
HW_Register *reg = new HW_Register(val)
here 'this' is in two different places in memory
instead have an internal union/struct to hold the value, that way its easy to convert
back and forth (since you are not interested in portability)
e.g.
union
{
struct {
unsigned short field1:2;
unsigned short field2:4;
unsigned short field3:2;
...
} bits;
unsigned short value;
} reg
edit: true enough with the name 'register'
Bitfields don't work that way. You can't assign a scalar value to a struct full of bitfields. It looks like you already know this since you used reinterpret_cast, but since reinterpret_cast isn't guaranteed to do very much, it's just rolling the dice.
You need to encode and decode the values if you want to translate between bitfield structs and scalars.
HW_Register(unsigned char value)
: field1( value & 3 ),
field2( value >> 2 & 3 ),
field3( value >> 4 & 7 )
{}
Edit: The reason you don't get any output is that the ASCII characters corresponding to the numbers in the fields are non-printing. Try this:
std::cout << "Field 1 = " << (int) testRegister.field1 << std::endl;
std::cout << "Field 2 = " << (int) testRegister.field2 << std::endl;
std::cout << "Field 3 = " << (int) testRegister.field3 << std::endl;
Try this:
class HW_Register
{
public:
HW_Register(unsigned char nRegisterValue=0)
{
Init(nRegisterValue);
}
~HW_Register(void){};
void Init(unsigned char nRegisterValue)
{
nVal = nRegisterValue;
}
unsigned Field1() { return nField1; }
unsigned Field2() { return nField2; }
unsigned Field3() { return nField3; }
private:
union
{
struct
{
unsigned char nField1:2;
unsigned char nField2:4;
unsigned char nField3:2;
};
unsigned char nVal;
};
};
int main()
{
unsigned char registerValue = 0xFF;
HW_Register testRegister(registerValue);
std::cout << "Field 1 = " << testRegister.Field1() << std::endl;
std::cout << "Field 2 = " << testRegister.Field2() << std::endl;
std::cout << "Field 3 = " << testRegister.Field3() << std::endl;
return 0;
}
HW_Register(unsigned char registerValue) : field1(0), field2(0), field3(0)