How do I convert an integer to a hex string in C++?
I can find some ways to do it, but they mostly seem targeted towards C. It doesn't seem there's a native way to do it in C++. It is a pretty simple problem though; I've got an int which I'd like to convert to a hex string for later printing.
Just print it as an hexadecimal number:
int i = /* ... */;
std::cout << std::hex << i;
int num = 30;
std::cout << std::hex << num << endl; // This should give you hexa- decimal of 30
My solution. Only integral types are allowed.
Update. You can set optional prefix 0x in second parameter.
definition.h
#include <iomanip>
#include <sstream>
template <class T, class T2 = typename std::enable_if<std::is_integral<T>::value>::type>
static std::string ToHex(const T & data, bool addPrefix = true);
template<class T, class>
inline std::string Convert::ToHex(const T & data, bool addPrefix)
{
std::stringstream sstream;
sstream << std::hex;
std::string ret;
if (typeid(T) == typeid(char) || typeid(T) == typeid(unsigned char) || sizeof(T)==1)
{
sstream << static_cast<int>(data);
ret = sstream.str();
if (ret.length() > 2)
{
ret = ret.substr(ret.length() - 2, 2);
}
}
else
{
sstream << data;
ret = sstream.str();
}
return (addPrefix ? u8"0x" : u8"") + ret;
}
main.cpp
#include <definition.h>
int main()
{
std::cout << ToHex<unsigned char>(254) << std::endl;
std::cout << ToHex<char>(-2) << std::endl;
std::cout << ToHex<int>(-2) << std::endl;
std::cout << ToHex<long long>(-2) << std::endl;
std::cout<< std::endl;
std::cout << ToHex<unsigned char>(254, false) << std::endl;
std::cout << ToHex<char>(-2, false) << std::endl;
std::cout << ToHex<int>(-2, false) << std::endl;
std::cout << ToHex<long long>(-2, false) << std::endl;
return 0;
}
Results:
0xfe
0xfe
0xfffffffe
0xfffffffffffffffe
fe
fe
fffffffe
fffffffffffffffe
To make it lighter and faster I suggest to use direct filling of a string.
template <typename I> std::string n2hexstr(I w, size_t hex_len = sizeof(I)<<1) {
static const char* digits = "0123456789ABCDEF";
std::string rc(hex_len,'0');
for (size_t i=0, j=(hex_len-1)*4 ; i<hex_len; ++i,j-=4)
rc[i] = digits[(w>>j) & 0x0f];
return rc;
}
#include <iostream>
#include <sstream>
int main()
{
unsigned int i = 4967295; // random number
std::string str1, str2;
unsigned int u1, u2;
std::stringstream ss;
Using void pointer:
// INT to HEX
ss << (void*)i; // <- FULL hex address using void pointer
ss >> str1; // giving address value of one given in decimals.
ss.clear(); // <- Clear bits
// HEX to INT
ss << std::hex << str1; // <- Capitals doesn't matter so no need to do extra here
ss >> u1;
ss.clear();
Adding 0x:
// INT to HEX with 0x
ss << "0x" << (void*)i; // <- Same as above but adding 0x to beginning
ss >> str2;
ss.clear();
// HEX to INT with 0x
ss << std::hex << str2; // <- 0x is also understood so need to do extra here
ss >> u2;
ss.clear();
Outputs:
std::cout << str1 << std::endl; // 004BCB7F
std::cout << u1 << std::endl; // 4967295
std::cout << std::endl;
std::cout << str2 << std::endl; // 0x004BCB7F
std::cout << u2 << std::endl; // 4967295
return 0;
}
for fixed number of digits, for instance 2:
static const char* digits = "0123456789ABCDEF";//dec 2 hex digits positional map
char value_hex[3];//2 digits + terminator
value_hex[0] = digits[(int_value >> 4) & 0x0F]; //move of 4 bit, that is an HEX digit, and take 4 lower. for higher digits use multiple of 4
value_hex[1] = digits[int_value & 0x0F]; //no need to move the lower digit
value_hex[2] = '\0'; //terminator
you can also write a for cycle variant to handle variable digits amount
benefits:
For those of you who figured out that many/most of the ios::fmtflags don't work with std::stringstream yet like the template idea that Kornel posted way back when, the following works and is relatively clean:
#include <iomanip>
#include <sstream>
template< typename T >
std::string hexify(T i)
{
std::stringbuf buf;
std::ostream os(&buf);
os << "0x" << std::setfill('0') << std::setw(sizeof(T) * 2)
<< std::hex << i;
return buf.str().c_str();
}
int someNumber = 314159265;
std::string hexified = hexify< int >(someNumber);
Use std::stringstream to convert integers into strings and its special manipulators to set the base. For example like that:
std::stringstream sstream;
sstream << std::hex << my_integer;
std::string result = sstream.str();
I would like to add an answer to enjoy the beauty of C ++ language. Its adaptability to work at high and low levels. Happy programming.
public:template <class T,class U> U* Int2Hex(T lnumber, U* buffer)
{
const char* ref = "0123456789ABCDEF";
T hNibbles = (lnumber >> 4);
unsigned char* b_lNibbles = (unsigned char*)&lnumber;
unsigned char* b_hNibbles = (unsigned char*)&hNibbles;
U* pointer = buffer + (sizeof(lnumber) << 1);
*pointer = 0;
do {
*--pointer = ref[(*b_lNibbles++) & 0xF];
*--pointer = ref[(*b_hNibbles++) & 0xF];
} while (pointer > buffer);
return buffer;
}
Examples:
char buffer[100] = { 0 };
Int2Hex(305419896ULL, buffer);//returns "0000000012345678"
Int2Hex(305419896UL, buffer);//returns "12345678"
Int2Hex((short)65533, buffer);//returns "FFFD"
Int2Hex((char)18, buffer);//returns "12"
wchar_t buffer[100] = { 0 };
Int2Hex(305419896ULL, buffer);//returns L"0000000012345678"
Int2Hex(305419896UL, buffer);//returns L"12345678"
Int2Hex((short)65533, buffer);//returns L"FFFD"
Int2Hex((char)18, buffer);//returns L"12"
Code for your reference:
#include <iomanip>
#include <sstream>
...
string intToHexString(int intValue) {
string hexStr;
/// integer value to hex-string
std::stringstream sstream;
sstream << "0x"
<< std::setfill ('0') << std::setw(2)
<< std::hex << (int)intValue;
hexStr= sstream.str();
sstream.clear(); //clears out the stream-string
return hexStr;
}
Thanks to Lincoln's comment below, I've changed this answer.
The following answer properly handles 8-bit ints at compile time. It doees, however, require C++17. If you don't have C++17, you'll have to do something else (e.g. provide overloads of this function, one for uint8_t and one for int8_t, or use something besides "if constexpr", maybe enable_if).
template< typename T >
std::string int_to_hex( T i )
{
// Ensure this function is called with a template parameter that makes sense. Note: static_assert is only available in C++11 and higher.
static_assert(std::is_integral<T>::value, "Template argument 'T' must be a fundamental integer type (e.g. int, short, etc..).");
std::stringstream stream;
stream << "0x" << std::setfill ('0') << std::setw(sizeof(T)*2) << std::hex;
// If T is an 8-bit integer type (e.g. uint8_t or int8_t) it will be
// treated as an ASCII code, giving the wrong result. So we use C++17's
// "if constexpr" to have the compiler decides at compile-time if it's
// converting an 8-bit int or not.
if constexpr (std::is_same_v<std::uint8_t, T>)
{
// Unsigned 8-bit unsigned int type. Cast to int (thanks Lincoln) to
// avoid ASCII code interpretation of the int. The number of hex digits
// in the returned string will still be two, which is correct for 8 bits,
// because of the 'sizeof(T)' above.
stream << static_cast<int>(i);
}
else if (std::is_same_v<std::int8_t, T>)
{
// For 8-bit signed int, same as above, except we must first cast to unsigned
// int, because values above 127d (0x7f) in the int will cause further issues.
// if we cast directly to int.
stream << static_cast<int>(static_cast<uint8_t>(i));
}
else
{
// No cast needed for ints wider than 8 bits.
stream << i;
}
return stream.str();
}
Original answer that doesn't handle 8-bit ints correctly as I thought it did:
Kornel Kisielewicz's answer is great. But a slight addition helps catch cases where you're calling this function with template arguments that don't make sense (e.g. float) or that would result in messy compiler errors (e.g. user-defined type).
template< typename T >
std::string int_to_hex( T i )
{
// Ensure this function is called with a template parameter that makes sense. Note: static_assert is only available in C++11 and higher.
static_assert(std::is_integral<T>::value, "Template argument 'T' must be a fundamental integer type (e.g. int, short, etc..).");
std::stringstream stream;
stream << "0x"
<< std::setfill ('0') << std::setw(sizeof(T)*2)
<< std::hex << i;
// Optional: replace above line with this to handle 8-bit integers.
// << std::hex << std::to_string(i);
return stream.str();
}
I've edited this to add a call to std::to_string because 8-bit integer types (e.g. std::uint8_t values passed) to std::stringstream are treated as char, which doesn't give you the result you want. Passing such integers to std::to_string handles them correctly and doesn't hurt things when using other, larger integer types. Of course you may possibly suffer a slight performance hit in these cases since the std::to_string call is unnecessary.
Note: I would have just added this in a comment to the original answer, but I don't have the rep to comment.
You can try the following. It's working...
#include <iostream>
#include <fstream>
#include <string>
#include <sstream>
using namespace std;
template <class T>
string to_string(T t, ios_base & (*f)(ios_base&))
{
ostringstream oss;
oss << f << t;
return oss.str();
}
int main ()
{
cout<<to_string<long>(123456, hex)<<endl;
system("PAUSE");
return 0;
}
Just have a look on my solution,[1] that I verbatim copied from my project, so there a German is API doc included. My goal was to combine flexibility and safety within my actual needs:[2]
0x prefix added: caller may decidelong long#include <string>
#include <sstream>
#include <iomanip>
/// Vertextet einen Ganzzahlwert val im Hexadezimalformat.
/// Auf die Minimal-Breite width wird mit führenden Nullen aufgefüllt;
/// falls nicht angegeben, wird diese Breite aus dem Typ des Arguments
/// abgeleitet. Funktion geeignet von char bis long long.
/// Zeiger, Fließkommazahlen u.ä. werden nicht unterstützt, ihre
/// Übergabe führt zu einem (beabsichtigten!) Compilerfehler.
/// Grundlagen aus: http://stackoverflow.com/a/5100745/2932052
template <typename T>
inline std::string int_to_hex(T val, size_t width=sizeof(T)*2)
{
std::stringstream ss;
ss << std::setfill('0') << std::setw(width) << std::hex << (val|0);
return ss.str();
}
[1] based on the answer by Kornel Kisielewicz
[2] Translated into the language of CppTest, this is how it reads:
TEST_ASSERT(int_to_hex(char(0x12)) == "12");
TEST_ASSERT(int_to_hex(short(0x1234)) == "1234");
TEST_ASSERT(int_to_hex(long(0x12345678)) == "12345678");
TEST_ASSERT(int_to_hex((long long)(0x123456789abcdef0)) == "123456789abcdef0");
TEST_ASSERT(int_to_hex(0x123, 1) == "123");
TEST_ASSERT(int_to_hex(0x123, 8) == "00000123");
// with deduction test as suggested by Lightness Races in Orbit:
TEST_ASSERT(int_to_hex(short(0x12)) == "0012");
int var = 20;
cout << &var << endl;
cout << (int)&var << endl;
cout << std::hex << "0x" << (int)&var << endl << std::dec; // output in hex, reset back to dec
0x69fec4 (address)
6946500 (address to dec)
0x69fec4 (address to dec, output in hex)
instinctively went with this...
int address = (int)&var;
saw this elsewhere...
unsigned long address = reinterpret_cast(&var);
comment told me this is correct...
int address = (int)&var;
speaking of well covered lightness, where are you at? they're getting too many likes!
This question is old, but I'm surprised why no one mentioned boost::format:
cout << (boost::format("%x") % 1234).str(); // output is: 4d2
Source: Stackoverflow.com