如何在c+;中将无符号字符打印为十六进制+;使用ostream? 我想用C++中的无符号8位变量工作。无论是unsigned char还是uint8\u t都可以在算法方面使用技巧(这是意料之中的,因为AFAIKuint8\u t只是unsigned char的别名,或者调试器会显示它 问题是,如果我用C++中的OFFROW打印变量,它就把它当作char。如果我有: unsigned char a = 0; unsigned char b = 0xff; cout << "a is " << hex << a <<"; b is " << hex << b << endl;
而不是如何在c+;中将无符号字符打印为十六进制+;使用ostream? 我想用C++中的无符号8位变量工作。无论是unsigned char还是uint8\u t都可以在算法方面使用技巧(这是意料之中的,因为AFAIKuint8\u t只是unsigned char的别名,或者调试器会显示它 问题是,如果我用C++中的OFFROW打印变量,它就把它当作char。如果我有: unsigned char a = 0; unsigned char b = 0xff; cout << "a is " << hex << a <<"; b is " << hex << b << endl;,c++,formatting,ostream,C++,Formatting,Ostream,而不是 a is 0; b is ff 我尝试使用uint8\u t,但正如我前面提到的,它的类型定义为unsigned char,因此它的作用是相同的。如何正确打印变量 编辑:我在代码中的许多地方都这样做。是否有任何方法可以做到这一点,而不必每次打印时都转换为int?使用: cout << "a is " << hex << (int) a <<"; b is " << hex << (int) b << e
a is 0; b is ff
我尝试使用uint8\u t
,但正如我前面提到的,它的类型定义为unsigned char
,因此它的作用是相同的。如何正确打印变量
编辑:我在代码中的许多地方都这样做。是否有任何方法可以做到这一点,而不必每次打印时都转换为int
?使用:
cout << "a is " << hex << (int) a <<"; b is " << hex << (int) b << endl;
cout这也会起作用:
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;
}
std::ostream&operator我建议使用以下技术:
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;
}
struct HexCharStruct
{
无符号字符c;
HexCharStruct(无符号字符c):c(_c){
};
内联std::ostream&operator我会像MartinStettner那样做,但会为位数添加一个额外的参数:
inline HexStruct hex(long n, int w=2)
{
return HexStruct(n, w);
}
// Rest of implementation is left as an exercise for the reader
因此,默认情况下,您有两个数字,但如果您愿意,可以设置为4、8或其他任何数字
例如
intmain()
{
短a=3142;
std:cout我建议:
std::cout << setbase(16) << 32;
std::cout您可以尝试以下代码:
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;
无符号字符a=0;
无符号字符b=0xff;
cout我认为TrungTN和anon的答案是可以的,但是MartinStettner实现hex()函数的方法并不简单,考虑到我用这种方法使用的hex,它太暗了
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;
char strInput[]=“yourchardata”;
char chHex[2]=“”;
int NLENGHT=斯特伦(strInput);
char*chResut=新字符[(nLength*2)+1];
memset(chResut,0,(nLength*2)+1);
对于(int i=0;iHm,似乎我昨天重新发明了轮子……但是,嘿,这次至少是一个通用轮子:)char
s用两个十六进制数字打印,short
s用四个十六进制数字打印,依此类推
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;
}
模板
结构十六进制
{
tx;
};
模板
十六进制(t x)
{
hex_t h={x};
返回h;
}
模板
标准::ostream&operator=4;
}
写操作(缓冲区,缓冲区大小);
返回操作系统;
}
我想发布我基于@FredOverflow重新发明的版本。我做了以下修改
修正:
- 操作员的Rhs
operator您可以在和处阅读更多关于此的信息。我之所以发布这篇文章,是因为很明显,上述文章的作者并不打算这么做
以十六进制打印字符最简单、最正确的方法是
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);
无符号字符a=0;
无符号字符b=0xff;
自动标志=cout.flags()//我只包括重置ioflags,因为是这样
//本页上的许多答案调用函数,其中
//旗帜已更改,无法更改
//让他们回到以前的状态
//函数调用
cout我意识到这是一个老问题,但这也是谷歌搜索一个类似问题解决方案的最佳结果,这是在模板类中实现任意整数到十六进制字符串转换的愿望。我的最终目标实际上是一个Gtk::Entry
子类模板,它允许以十六进制编辑各种整数宽度,但这并不重要
这将一元运算符+
技巧与std::make_unsigned
from
相结合,以防止在
无论如何,我相信这比任何其他通用解决方案都要简洁。它应该适用于任何有符号或无符号整数类型,如果试图用任何非整数类型实例化函数,则会引发编译时错误
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();
}
我在win32/linux(32/64位)上使用以下命令:
#包括
#包括
模板
标准:字符串六角字符串(T uval)
{
std::stringstream-ss;
ss嗯,这对我来说很有用:
std::cout << std::hex << (0xFF & a) << std::endl;
std::cout如果使用预填充和带符号字符,请小心不要附加不需要的'F'
char out\u character=0xBE;
我想我们没有解释这些类型转换是如何工作的
char
依赖于平台signed
或unsigned
。在x86中char
相当于signed char
当整数类型(char
、short
、int
、long
)转换为更大容量类型时,如果是无符号
类型,则在左侧添加零,对于有符号
类型,则在左侧添加符号扩展。符号扩展包括复制最重要的类型(最左边的)位,直到达到目标类型的位大小为止
因此,如果默认情况下我在带符号字符系统中,我会这样做:
char a = 0xF0; // Equivalent to the binary: 11110000
std::cout << std::hex << static_cast<int>(a);
这将在C++20中生成F0
,您可以使用它来执行以下操作:
std::cout << std::format("a is {:x}; b is {:x}\n", a, b);
同时,您可以使用,std::format
是基于。{fmt}还提供了print
功能,使这一过程更加简单和高效():
免责声明:我是{fmt}的作者还有C++20std::format
我能说服你一劳永逸地避免邪恶的C风格的强制转换吗?不是在这种情况下-这是我认为它们是合理的唯一地方。我也会这样做,除了避免强制转换之外,我会使用cout,但如果他真的想将它们作为字符输出,他就必须使用强制转换,这似乎有点不自然L!只要我们避开宏:为了更完整地C++,你不应该写<代码>(int)h吗?
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;
}
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"
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));
}
}
struct { int x; } output = {0xdeadbeef};
cout << hex_sep(output) << std::uppercase << hex(output) << endl;
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);
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();
}
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;
}
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;
#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();
}
std::cout << std::hex << (0xFF & a) << std::endl;
char a = 0xF0; // Equivalent to the binary: 11110000
std::cout << std::hex << static_cast<int>(a);
char a = 0xF0; // Equivalent to the binary: 11110000
std::cout << std::hex << static_cast<int>(static_cast<unsigned char>(a));
std::cout << std::format("a is {:x}; b is {:x}\n", a, b);
fmt::print("a is {:x}; b is {:x}\n", a, b);