hex << setw() << setfill() doesn't work with pointer value output - c++

Win7 - 64
cygwin
Netbeans 7.4
gcc (cygwin) 4.8.3
Compiler call
g++ -Wall -Wno-reorder -Wno-unused-value -c -g -MMD -MP -MF
Any reason that the output and the expected output are different? I did not expect the "0x" prefix.
Output << SlipHashEntry::create 0x22a000 list3
Expected << SlipHashEntry::create 0x000000000022a000 list3
Output << SlipDescription::delete 0x600062d50 list5
Expected << SlipDescription::delete 0x0000000600062d50 list5
Here's the relevant code snippet
SlipHashEntry::SlipHashEntry
( const string& name, void* ptr, Type type, int debugFlag )
: completeFlag(false)
, debugFlag(debugFlag)
, hashDebugFlag((bool)(debugFlag & SlipRead::HASH))
, inputDebugFlag((bool)(debugFlag & SlipRead::INPUT))
, leakDebugFlag((bool)(debugFlag & SlipRead::LEAK))
, descriptorChain(NULL)
, link(NULL)
, name(new string(name))
, nestedPtr(NULL)
, ptr(ptr)
, type(type) {
DEBUG(leakDebugFlag,
cout << left << setw(27) << setfill(' ') << "SlipHashEntry::create "
<< hex << setw(sizeof(void*))
<< setfill('0') << (void*)this << ' ' << name <<endl;)
};

"Any reason that the output and the expected output are different? I did not expect the "0x" prefix."
Regarding the 0x prefix in the output, this is done by the intrinsic operator overload for std::ostream& operator<<(std::ostream&, const void*) see (7)
Also you have a little misconception about the setw(): You should remember, that setw() sets the fieldsize in characters (i.e. digits). You want to display a 64-bit pointer which is equivalent to 8 bytes with 2 digits for representation of each byte
cout << left << setw(27) << setfill(' ') << "SlipHashEntry::create "
<< hex << setw(2 * sizeof(void*))
// ^^^
<< setfill('0') << (void*)this << ' ' << name <<endl;)
just multiply the pointer size (in bytes) by 2, to get the correct field width for the full number of digits.
Though the intrinsic output operator definition for void* doesn't exactly do what you want.
When I've been trying to make an online compilable sample for what I was stating above, I noticed that you can't actually manipulate the intrinsic pointer output format. I have made up a simple sample, how to achieve the fixed 8 byte format you want:
#include <iostream>
#include <iomanip>
// A simple manipulator to format pointers in a fixed length format (according
// 64 bit) with leading zeroes and a "0x" prefix
class fmt_longptr {
public:
fmt_longptr(void* ptr) : ptr_(ptr) {}
void put(std::ostream& os) const {
os << "0x" << std::hex << std::setw(2 * sizeof(void*))
<< std::setfill('0') << (unsigned long)ptr_;
}
private:
friend std::ostream& operator<<(std::ostream& os, const fmt_longptr& fmt);
void* ptr_;
};
std::ostream& operator<<(std::ostream& os, const fmt_longptr& fmt) {
fmt.put(os);
return os;
}
Here's how to use it, and showing what the difference vs the intrinsic pointer formatting is
int main() {
int a;
std::cout << std::setfill('0') << std::setw(2 * sizeof(void*)) << &a
<< std::endl;
std::cout << fmt_longptr(&a) << std::endl;
return 0;
}
Output
000x7fff270d987c
0x00007fff270d987c
Here's the working online sample.

Related

How to change the cout format for pointers

I have to write a C++ program on in VS which does the same as a previously written programm for Solaris which is compiled with gcc.
The following "problem" occured:
int var1 = 42;
int* var1Ptr = &var1;
cout << "Address of pointer " << var1Ptr << endl;
This code returns in the solaris program a 0x indexed address (0x08FFAFC).
In my VS code it returns as 008FFAFC.
Since we only do comparison within the code it would be fine, but the supportteam have their own tools which extract data from the logs which is looking for those 0x indexed values. Is there a way to format it like this without adding the 0x prefix everytime we write into the logs?
cout << "Maybe this way: " << hex << int(&var1Ptr) << endl;
doesn't have the effect I wanted.
A little helper class:
namespace detail {
template<class T>
struct debug_pointer
{
constexpr static std::size_t pointer_digits()
{
return sizeof(void*) * 2;
};
static constexpr std::size_t width = pointer_digits();
std::ostream& operator()(std::ostream& os) const
{
std::uintptr_t i = reinterpret_cast<std::uintptr_t>(p);
return os << "0x" << std::hex << std::setw(width) << std::setfill('0') << i;
}
T* p;
friend
std::ostream& operator<<(std::ostream& os, debug_pointer const& dp) {
return dp(os);
}
};
}
offered via a custom manipulator...
template<class T>
auto debug_pointer(T* p)
{
return detail::debug_pointer<T>{ p };
}
allows us this expression:
int i;
std::cout << debug_pointer(&i) << std::endl;
Which will yield either an 8-digit or 16-digit hex pointer value, depending on your architecture (mine is 64-bit):
0x00007fff5fbff3bc

c++ insert to a stream with formatting for hex

I often do something like:
uint8_t c=some_value;
std::cout << std::setfill('0') << std::setw(2);
std::cout << std::hex << int(c);
std::cout << std::setfill(' ');
(in particular while dumping debugging information). Wouldn't it be nice to have something manipulatorish that I could put in a stream like this:
std::cout << "c value: 0x" << hexb(c) << '\n';
that would do all of that? Does anyone know how to do that?
I've gotten this to work but would love to have a simpler way:
#include <iostream>
#include <iomanip>
class hexcdumper{
public:
hexcdumper(uint8_t c):c(c){};
std::ostream&
operator( )(std::ostream& os) const
{
// set fill and width and save the previous versions to be restored later
char fill=os.fill('0');
std::streamsize ss=os.width(2);
// save the format flags so we can restore them after setting std::hex
std::ios::fmtflags ff=os.flags();
// output the character with hex formatting
os << std::hex << int(c);
// now restore the fill, width and flags
os.fill(fill);
os.width(ss);
os.flags(ff);
return os;
}
private:
uint8_t c;
};
hexcdumper
hexb(uint8_t c)
{
// dump a hex byte with width 2 and a fill character of '0'
return(hexcdumper(c));
}
std::ostream& operator<<(std::ostream& os, const hexcdumper& hcd)
{
return(hcd(os));
}
When I do this:
std::cout << "0x" << hexb(14) << '\n';
hexb(c) is invoked and returns a hexcdumper whose constructor saves c
the overloaded operator<< for hexcdumper invokes
hexcdumper::operator() passing it the stream
hexcdumper's operator() does all the magic for us
after hexcdumper::operator() returns, the overloaded operator<<
returns the stream as returned from hexcdumper::operator() so chaining works.
On the output, I see:
0x0e
Is there a simpler way to do this?
Patrick
You can do this directly on the stream pipe:
std::cout << "Hex = 0x" << hex << 14 << ", decimal = #" << dec << 14 << endl;
Output:
Hex = 0xe, decimal = #14

How to print a bunch of integers with the same formatting?

I would like to print a bunch of integers on 2 fields with '0' as fill character. I can do it but it leads to code duplication. How should I change the code so that the code duplication can be factored out?
#include <ctime>
#include <sstream>
#include <iomanip>
#include <iostream>
using namespace std;
string timestamp() {
time_t now = time(0);
tm t = *localtime(&now);
ostringstream ss;
t.tm_mday = 9; // cheat a little to test it
t.tm_hour = 8;
ss << (t.tm_year+1900)
<< setw(2) << setfill('0') << (t.tm_mon+1) // Code duplication
<< setw(2) << setfill('0') << t.tm_mday
<< setw(2) << setfill('0') << t.tm_hour
<< setw(2) << setfill('0') << t.tm_min
<< setw(2) << setfill('0') << t.tm_sec;
return ss.str();
}
int main() {
cout << timestamp() << endl;
return 0;
}
I have tried
std::ostream& operator<<(std::ostream& s, int i) {
return s << std::setw(2) << std::setfill('0') << i;
}
but it did not work, the operator<< calls are ambigous.
EDIT I got 4 awesome answers and I picked the one that is perhaps the simplest and the most generic one (that is, doesn't assume that we are dealing with timestamps). For the actual problem, I will probably use std::put_time or strftime though.
In C++20 you'll be able to do this with std::format in a less verbose way:
ss << std::format("{}{:02}{:02}{:02}{:02}{:02}",
t.tm_year + 1900, t.tm_mon + 1, t.tm_mday,
t.tm_hour, t.tm_min, t.tm_sec);
and it's even easier with the {fmt} library that supports tm formatting directly:
auto s = fmt::format("{:%Y%m%d%H%M%S}", t);
You need a proxy for your string stream like this:
struct stream{
std::ostringstream ss;
stream& operator<<(int i){
ss << std::setw(2) << std::setfill('0') << i;
return *this; // See Note below
}
};
Then your formatting code will just be this:
stream ss;
ss << (t.tm_year+1900)
<< (t.tm_mon+1)
<< t.tm_mday
<< t.tm_hour
<< t.tm_min
<< t.tm_sec;
return ss.ss.str();
ps. Note the general format of my stream::operator<<() which does its work first, then returns something.
The "obvious" solution is to use a manipulator to install a custom std::num_put<char> facet which just formats ints as desired.
The above statement may be a bit cryptic although it entirely describes the solution. Below is the code to actually implement the logic. The first ingredient is a special std::num_put<char> facet which is just a class derived from std::num_put<char> and overriding one of its virtual functions. The used facet is a filtering facet which looks at a flag stored with the stream (using iword()) to determine whether it should change the behavior or not. Here is the code:
class num_put
: public std::num_put<char>
{
std::locale loc_;
static int index() {
static int rc(std::ios_base::xalloc());
return rc;
}
friend std::ostream& twodigits(std::ostream&);
friend std::ostream& notwodigits(std::ostream&);
public:
num_put(std::locale loc): loc_(loc) {}
iter_type do_put(iter_type to, std::ios_base& fmt,
char fill, long value) const {
if (fmt.iword(index())) {
fmt.width(2);
return std::use_facet<std::num_put<char> >(this->loc_)
.put(to, fmt, '0', value);
}
else {
return std::use_facet<std::num_put<char> >(this->loc_)
.put(to, fmt, fill, value);
}
}
};
The main part is the do_put() member function which decides how the value needs to be formatted: If the flag in fmt.iword(index()) is non-zero, it sets the width to 2 and calls the formatting function with a fill character of 0. The width is going to be reset anyway and the fill character doesn't get stored with the stream, i.e., there is no need for any clean-up.
Normally, the code would probably live in a separate translation unit and it wouldn't be declared in a header. The only functions really declared in a header would be twodigits() and notwodigits() which are made friends in this case to provide access to the index() member function. The index() member function just allocates an index usable with std::ios_base::iword() when called the time and it then just returns this index. The manipulators twodigits() and notwodigits() primarily set this index. If the num_put facet isn't installed for the stream twodigits() also installs the facet:
std::ostream& twodigits(std::ostream& out)
{
if (!dynamic_cast<num_put const*>(
&std::use_facet<std::num_put<char> >(out.getloc()))) {
out.imbue(std::locale(out.getloc(), new num_put(out.getloc())));
}
out.iword(num_put::index()) = true;
return out;
}
std::ostream& notwodigits(std::ostream& out)
{
out.iword(num_put::index()) = false;
return out;
}
The twodigits() manipulator allocates the num_put facet using new num_put(out.getloc()). It doesn't require any clean-up because installing a facet in a std::locale object does the necessary clean-up. The original std::locale of the stream is accessed using out.getloc(). It is changed by the facet. In theory the notwodigits could restore the original std::locale instead of using a flag. However, imbue() can be a relatively expensive operation and using a flag should be a lot cheaper. Of course, if there are lots of similar formatting flags, things may become different...
To demonstrate the use of the manipulators there is a simple test program below. It sets up the formatting flag twodigits twice to verify that facet is only created once (it would be a bit silly to create a chain of std::locales to pass through the formatting:
int main()
{
std::cout << "some-int='" << 1 << "' "
<< twodigits << '\n'
<< "two-digits1='" << 1 << "' "
<< "two-digits2='" << 2 << "' "
<< "two-digits3='" << 3 << "' "
<< notwodigits << '\n'
<< "some-int='" << 1 << "' "
<< twodigits << '\n'
<< "two-digits4='" << 4 << "' "
<< '\n';
}
Besides formatting integers with std::setw / std::setfill or ios_base::width / basic_ios::fill, if you want to format a date/time object you may want to consider using std::put_time / std::gettime
For convenient output formatting you may use boost::format() with sprintf-like formatting options:
#include <boost/format.hpp>
#include <iostream>
int main() {
int i1 = 1, i2 = 10, i3 = 100;
std::cout << boost::format("%03i %03i %03i\n") % i1 % i2 % i3;
// output is: 001 010 100
}
Little code duplication, additional implementation effort is marginal.
If all you want to do is output formatting of your timestamp, you should obviously use strftime(). That's what it's made for:
#include <ctime>
#include <iostream>
std::string timestamp() {
char buf[20];
const char fmt[] = "%Y%m%d%H%M%S";
time_t now = time(0);
strftime(buf, sizeof(buf), fmt, localtime(&now));
return buf;
}
int main() {
std::cout << timestamp() << std::endl;
}
operator<<(std::ostream& s, int i) is "ambiguous" because such a function already exists.
All you need to do is give that function a signature that doesn't conflict.

format, iomanip, c++

I'm trying to learn to use namespaces declarations more definitive than not just say "using namespace std". I'm trying to format my data to 2 decimal places, and set the format to be fixed and not scientific. This is my main file:
#include <iostream>
#include <iomanip>
#include "SavingsAccount.h"
using std::cout;
using std::setprecision;
using std::ios_base;
int main()
{
SavingsAccount *saver1 = new SavingsAccount(2000.00);
SavingsAccount *saver2 = new SavingsAccount(3000.00);
SavingsAccount::modifyInterestRate(.03);
saver1->calculateMonthlyInterest();
saver2->calculateMonthlyInterest();
cout << ios_base::fixed << "saver1\n" << "monthlyInterestRate: " << saver1->getMonthlyInterest()
<< '\n' << "savingsBalance: " << saver1->getSavingsBalance() << '\n';
cout << "saver2\n" << "monthlyInterestRate: " << saver2->getMonthlyInterest()
<< '\n' << "savingsBalance: " << saver2->getSavingsBalance() << '\n';
}
On Visual Studio 2008, when I run my program, I get an output of "8192" before the data I want. Is there a reason for that?
Also, I don't think I am setting the fixed part or 2 decimal places correctly since I seem to get scientific notation once I added the setprecision(2). Thanks.
You want std::fixed (the other one just inserts its value into the stream, which is why you see 8192), and I don't see a call to std::setprecision in your code anywhere.
This'll fix it:
#include <iostream>
#include <iomanip>
using std::cout;
using std::setprecision;
using std::fixed;
int main()
{
cout << fixed << setprecision(2)
<< "saver1\n"
<< "monthlyInterestRate: " << 5.5 << '\n'
<< "savingsBalance: " << 10928.8383 << '\n';
cout << "saver2\n"
<< "monthlyInterestRate: " << 4.7 << '\n'
<< "savingsBalance: " << 22.44232 << '\n';
}
It might not be the answer you're looking for, but floating-point numbers are not suited to financial calculations because fractions like 1/100 cannot be represented exactly. You might be better off doing the formatting yourself. This can be encapsulated:
class money {
int cents;
public:
money( int in_cents ) : cents( in_cents ) {}
friend ostream &operator<< ( ostream &os, money const &rhs )
{ return os << '$' << m.cents / 100 << '.' << m.cents % 100; }
};
cout << money( 123 ) << endl; // prints $1.23
Better(?) yet, C++ has a facility called the monetary locale category which includes a money formatter which takes cents as an argument.
locale::global( locale("") );
use_facet< money_put<char> >( locale() ).put( cout, false, cout, ' ', 123 );
This should Do the Right thing internationally, printing the user's local currency and hiding the number of decimal places from your implementation. It even accepts fractions of a cent. Unfortunately, this does not seem to work on my system (Mac OS X), which has generally poor locale support. (Linux and Windows should fare better.)
cout << setiosflags(ios::fixed) << setprecision(2) << 1/3.;
ios_base::fixed is not manipulator it is a value (1 << 13) for the ios flag.

how do I print an unsigned char as hex in c++ using ostream?

I want to work with unsigned 8-bit variables in C++. Either unsigned char or uint8_t do the trick as far as the arithmetic is concerned (which is expected, since AFAIK uint8_t is just an alias for unsigned char, or so the debugger presents it.
The problem is that if I print out the variables using ostream in C++ it treats it as char. If I have:
unsigned char a = 0;
unsigned char b = 0xff;
cout << "a is " << hex << a <<"; b is " << hex << b << endl;
then the output is:
a is ^#; b is 377
instead of
a is 0; b is ff
I tried using uint8_t, but as I mentioned before, that's typedef'ed to unsigned char, so it does the same. How can I print my variables correctly?
Edit: I do this in many places throughout my code. Is there any way I can do this without casting to int each time I want to print?
Use:
cout << "a is " << hex << (int) a <<"; b is " << hex << (int) b << endl;
And if you want padding with leading zeros then:
#include <iomanip>
...
cout << "a is " << setw(2) << setfill('0') << hex << (int) a ;
As we are using C-style casts, why not go the whole hog with terminal C++ badness and use a macro!
#define HEX( x )
setw(2) << setfill('0') << hex << (int)( x )
you can then say
cout << "a is " << HEX( a );
Edit: Having said that, MartinStettner's solution is much nicer!
I would suggest using the following technique:
struct HexCharStruct
{
unsigned char c;
HexCharStruct(unsigned char _c) : c(_c) { }
};
inline std::ostream& operator<<(std::ostream& o, const HexCharStruct& hs)
{
return (o << std::hex << (int)hs.c);
}
inline HexCharStruct hex(unsigned char _c)
{
return HexCharStruct(_c);
}
int main()
{
char a = 131;
std::cout << hex(a) << std::endl;
}
It's short to write, has the same efficiency as the original solution and it lets you choose to use the "original" character output. And it's type-safe (not using "evil" macros :-))
You can read more about this at http://cpp.indi.frih.net/blog/2014/09/tippet-printing-numeric-values-for-chars-and-uint8_t/ and http://cpp.indi.frih.net/blog/2014/08/code-critique-stack-overflow-posters-cant-print-the-numeric-value-of-a-char/. I am only posting this because it has become clear that the author of the above articles does not intend to.
The simplest and most correct technique to do print a char as hex is
unsigned char a = 0;
unsigned char b = 0xff;
auto flags = cout.flags(); //I only include resetting the ioflags because so
//many answers on this page call functions where
//flags are changed and leave no way to
//return them to the state they were in before
//the function call
cout << "a is " << hex << +a <<"; b is " << +b << endl;
cout.flags(flags);
The readers digest version of how this works is that the unary + operator forces a no op type conversion to an int with the correct signedness. So, an unsigned char converts to unsigned int, a signed char converts to int, and a char converts to either unsigned int or int depending on whether char is signed or unsigned on your platform (it comes as a shock to many that char is special and not specified as either signed or unsigned).
The only negative of this technique is that it may not be obvious what is happening to a someone that is unfamiliar with it. However, I think that it is better to use the technique that is correct and teach others about it rather than doing something that is incorrect but more immediately clear.
Well, this works for me:
std::cout << std::hex << (0xFF & a) << std::endl;
If you just cast (int) as suggested it might add 1s to the left of a if its most significant bit is 1. So making this binary AND operation guarantees the output will have the left bits filled by 0s and also converts it to unsigned int forcing cout to print it as hex.
I hope this helps.
In C++20 you'll be able to use std::format to do this:
std::cout << std::format("a is {:x}; b is {:x}\n", a, b);
Output:
a is 0; b is ff
In the meantime 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):
fmt::print("a is {:x}; b is {:x}\n", a, b);
Disclaimer: I'm the author of {fmt} and C++20 std::format.
Hm, it seems I re-invented the wheel yesterday... But hey, at least it's a generic wheel this time :) chars are printed with two hex digits, shorts with 4 hex digits and so on.
template<typename T>
struct hex_t
{
T x;
};
template<typename T>
hex_t<T> hex(T x)
{
hex_t<T> h = {x};
return h;
}
template<typename T>
std::ostream& operator<<(std::ostream& os, hex_t<T> h)
{
char buffer[2 * sizeof(T)];
for (auto i = sizeof buffer; i--; )
{
buffer[i] = "0123456789ABCDEF"[h.x & 15];
h.x >>= 4;
}
os.write(buffer, sizeof buffer);
return os;
}
I think TrungTN and anon's answer is okay, but MartinStettner's way of implementing the hex() function is not really simple, and too dark, considering hex << (int)mychar is already a workaround.
here is my solution to make "<<" operator easier:
#include <sstream>
#include <iomanip>
string uchar2hex(unsigned char inchar)
{
ostringstream oss (ostringstream::out);
oss << setw(2) << setfill('0') << hex << (int)(inchar);
return oss.str();
}
int main()
{
unsigned char a = 131;
std::cout << uchar2hex(a) << std::endl;
}
It's just not worthy implementing a stream operator :-)
I think we are missing an explanation of how these type conversions work.
char is platform dependent signed or unsigned. In x86 char is equivalent to signed char.
When an integral type (char, short, int, long) is converted to a larger capacity type, the conversion is made by adding zeros to the left in case of unsigned types and by sign extension for signed ones. Sign extension consists in replicating the most significant (leftmost) bit of the original number to the left till we reach the bit size of the target type.
Hence if I am in a signed char by default system and I do this:
char a = 0xF0; // Equivalent to the binary: 11110000
std::cout << std::hex << static_cast<int>(a);
We would obtain F...F0 since the leading 1 bit has been extended.
If we want to make sure that we only print F0 in any system we would have to make an additional intermediate type cast to an unsigned char so that zeros are added instead and, since they are not significant for a integer with only 8-bits, not printed:
char a = 0xF0; // Equivalent to the binary: 11110000
std::cout << std::hex << static_cast<int>(static_cast<unsigned char>(a));
This produces F0
I'd do it like MartinStettner but add an extra parameter for number of digits:
inline HexStruct hex(long n, int w=2)
{
return HexStruct(n, w);
}
// Rest of implementation is left as an exercise for the reader
So you have two digits by default but can set four, eight, or whatever if you want to.
eg.
int main()
{
short a = 3142;
std:cout << hex(a,4) << std::endl;
}
It may seem like overkill but as Bjarne said: "libraries should be easy to use, not easy to write".
I would suggest:
std::cout << setbase(16) << 32;
Taken from:
http://www.cprogramming.com/tutorial/iomanip.html
You can try the following code:
unsigned char a = 0;
unsigned char b = 0xff;
cout << hex << "a is " << int(a) << "; b is " << int(b) << endl;
cout << hex
<< "a is " << setfill('0') << setw(2) << int(a)
<< "; b is " << setfill('0') << setw(2) << int(b)
<< endl;
cout << hex << uppercase
<< "a is " << setfill('0') << setw(2) << int(a)
<< "; b is " << setfill('0') << setw(2) << int(b)
<< endl;
Output:
a is 0; b is ff
a is 00; b is ff
a is 00; b is FF
I use the following on win32/linux(32/64 bit):
#include <iostream>
#include <iomanip>
template <typename T>
std::string HexToString(T uval)
{
std::stringstream ss;
ss << "0x" << std::setw(sizeof(uval) * 2) << std::setfill('0') << std::hex << +uval;
return ss.str();
}
I realize this is an old question, but its also a top Google result in searching for a solution to a very similar problem I have, which is the desire to implement arbitrary integer to hex string conversions within a template class. My end goal was actually a Gtk::Entry subclass template that would allow editing various integer widths in hex, but that's beside the point.
This combines the unary operator+ trick with std::make_unsigned from <type_traits> to prevent the problem of sign-extending negative int8_t or signed char values that occurs in this answer
Anyway, I believe this is more succinct than any other generic solution. It should work for any signed or unsigned integer types, and throws a compile-time error if you attempt to instantiate the function with any non-integer types.
template <
typename T,
typename = typename std::enable_if<std::is_integral<T>::value, T>::type
>
std::string toHexString(const T v)
{
std::ostringstream oss;
oss << std::hex << +((typename std::make_unsigned<T>::type)v);
return oss.str();
}
Some example usage:
int main(int argc, char**argv)
{
int16_t val;
// Prints 'ff' instead of "ffffffff". Unlike the other answer using the '+'
// operator to extend sizeof(char) int types to int/unsigned int
std::cout << toHexString(int8_t(-1)) << std::endl;
// Works with any integer type
std::cout << toHexString(int16_t(0xCAFE)) << std::endl;
// You can use setw and setfill with strings too -OR-
// the toHexString could easily have parameters added to do that.
std::cout << std::setw(8) << std::setfill('0') <<
toHexString(int(100)) << std::endl;
return 0;
}
Update: Alternatively, if you don't like the idea of the ostringstream being used, you can combine the templating and unary operator trick with the accepted answer's struct-based solution for the following. Note that here, I modified the template by removing the check for integer types. The make_unsigned usage might be enough for compile time type safety guarantees.
template <typename T>
struct HexValue
{
T value;
HexValue(T _v) : value(_v) { }
};
template <typename T>
inline std::ostream& operator<<(std::ostream& o, const HexValue<T>& hs)
{
return o << std::hex << +((typename std::make_unsigned<T>::type) hs.value);
}
template <typename T>
const HexValue<T> toHex(const T val)
{
return HexValue<T>(val);
}
// Usage:
std::cout << toHex(int8_t(-1)) << std::endl;
If you're using prefill and signed chars, be careful not to append unwanted 'F's
char out_character = 0xBE;
cout << setfill('0') << setw(2) << hex << unsigned short(out_character);
prints: ffbe
using int instead of short results in ffffffbe
To prevent the unwanted f's you can easily mask them out.
char out_character = 0xBE;
cout << setfill('0') << setw(2) << hex << unsigned short(out_character) & 0xFF;
I'd like to post my re-re-inventing version based on #FredOverflow's. I made the following modifications.
fix:
Rhs of operator<< should be of const reference type. In #FredOverflow's code, h.x >>= 4 changes output h, which is surprisingly not compatible with standard library, and type T is requared to be copy-constructable.
Assume only CHAR_BITS is a multiple of 4. #FredOverflow's code assumes char is 8-bits, which is not always true, in some implementations on DSPs, particularly, it is not uncommon that char is 16-bits, 24-bits, 32-bits, etc.
improve:
Support all other standard library manipulators available for integral types, e.g. std::uppercase. Because format output is used in _print_byte, standard library manipulators are still available.
Add hex_sep to print separate bytes (note that in C/C++ a 'byte' is by definition a storage unit with the size of char). Add a template parameter Sep and instantiate _Hex<T, false> and _Hex<T, true> in hex and hex_sep respectively.
Avoid binary code bloat. Function _print_byte is extracted out of operator<<, with a function parameter size, to avoid instantiation for different Size.
More on binary code bloat:
As mentioned in improvement 3, no matter how extensively hex and hex_sep is used, only two copies of (nearly) duplicated function will exits in binary code: _print_byte<true> and _print_byte<false>. And you might realized that this duplication can also be eliminated using exactly the same approach: add a function parameter sep. Yes, but if doing so, a runtime if(sep) is needed. I want a common library utility which may be used extensively in the program, thus I compromised on the duplication rather than runtime overhead. I achieved this by using compile-time if: C++11 std::conditional, the overhead of function call can hopefully be optimized away by inline.
hex_print.h:
namespace Hex
{
typedef unsigned char Byte;
template <typename T, bool Sep> struct _Hex
{
_Hex(const T& t) : val(t)
{}
const T& val;
};
template <typename T, bool Sep>
std::ostream& operator<<(std::ostream& os, const _Hex<T, Sep>& h);
}
template <typename T> Hex::_Hex<T, false> hex(const T& x)
{ return Hex::_Hex<T, false>(x); }
template <typename T> Hex::_Hex<T, true> hex_sep(const T& x)
{ return Hex::_Hex<T, true>(x); }
#include "misc.tcc"
hex_print.tcc:
namespace Hex
{
struct Put_space {
static inline void run(std::ostream& os) { os << ' '; }
};
struct No_op {
static inline void run(std::ostream& os) {}
};
#if (CHAR_BIT & 3) // can use C++11 static_assert, but no real advantage here
#error "hex print utility need CHAR_BIT to be a multiple of 4"
#endif
static const size_t width = CHAR_BIT >> 2;
template <bool Sep>
std::ostream& _print_byte(std::ostream& os, const void* ptr, const size_t size)
{
using namespace std;
auto pbyte = reinterpret_cast<const Byte*>(ptr);
os << hex << setfill('0');
for (int i = size; --i >= 0; )
{
os << setw(width) << static_cast<short>(pbyte[i]);
conditional<Sep, Put_space, No_op>::type::run(os);
}
return os << setfill(' ') << dec;
}
template <typename T, bool Sep>
inline std::ostream& operator<<(std::ostream& os, const _Hex<T, Sep>& h)
{
return _print_byte<Sep>(os, &h.val, sizeof(T));
}
}
test:
struct { int x; } output = {0xdeadbeef};
cout << hex_sep(output) << std::uppercase << hex(output) << endl;
output:
de ad be ef DEADBEEF
This will also work:
std::ostream& operator<< (std::ostream& o, unsigned char c)
{
return o<<(int)c;
}
int main()
{
unsigned char a = 06;
unsigned char b = 0xff;
std::cout << "a is " << std::hex << a <<"; b is " << std::hex << b << std::endl;
return 0;
}
I have used in this way.
char strInput[] = "yourchardata";
char chHex[2] = "";
int nLength = strlen(strInput);
char* chResut = new char[(nLength*2) + 1];
memset(chResut, 0, (nLength*2) + 1);
for (int i = 0; i < nLength; i++)
{
sprintf(chHex, "%02X", strInput[i]& 0x00FF);
memcpy(&(chResut[i*2]), chHex, 2);
}
printf("\n%s",chResut);
delete chResut;
chResut = NULL;