C++ 如何将字节打印为十六进制？
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How do I print bytes as hexadecimal?
提问by Danny
I know in C# you can use String.Format method. But how do you do this in C++? Is there a function that allows me to convert a byte to a Hex?? Just need to convert a 8 byte long data to Hex, how do I do that?
我知道在 C# 中你可以使用 String.Format 方法。但是你如何在 C++ 中做到这一点？是否有允许我将字节转换为十六进制的函数？只需要将 8 字节长的数据转换为十六进制，我该怎么做？
采纳答案by bentech
Well you can convert one byte (unsigned char) at a time into a array like so
好吧，您可以一次将一个字节（无符号字符）转换为这样的数组
char buffer [17];
buffer[16] = 0;
for(j = 0; j < 8; j++)
sprintf(&buffer[2*j], "%02X", data[j]);
回答by Component 10
If you want to use C++ streams rather than C functions, you can do the following:
如果要使用 C++ 流而不是 C 函数，可以执行以下操作：
int ar[] = { 20, 30, 40, 50, 60, 70, 80, 90 };
const int siz_ar = sizeof(ar) / sizeof(int);
for (int i = 0; i < siz_ar; ++i)
cout << ar[i] << " ";
cout << endl;
for (int i = 0; i < siz_ar; ++i)
cout << hex << setfill('0') << setw(2) << ar[i] << " ";
cout << endl;
Very simple.
很简单。
Output:
输出：
20 30 40 50 60 70 80 90
14 1e 28 32 3c 46 50 5a
回答by user1763487
C:
C：
static void print_buf(const char *title, const unsigned char *buf, size_t buf_len)
{
size_t i = 0;
fprintf(stdout, "%s\n", title);
for(i = 0; i < buf_len; ++i)
fprintf(stdout, "%02X%s", buf[i],
( i + 1 ) % 16 == 0 ? "\r\n" : " " );
}
C++:
C++：
void print_bytes(std::ostream& out, const char *title, const unsigned char *data, size_t dataLen, bool format = true) {
out << title << std::endl;
out << std::setfill('0');
for(size_t i = 0; i < dataLen; ++i) {
out << std::hex << std::setw(2) << (int)data[i];
if (format) {
out << (((i + 1) % 16 == 0) ? "\n" : " ");
}
}
out << std::endl;
}
回答by Lucio Paiva
Printing arbitrary structures in modern C++
在现代 C++ 中打印任意结构
All answers so far only tell you how to print an array of integers, but we can also print any arbitrary structure, given that we know its size. The example below creates such structure and iterates a pointer through its bytes, printing them to the output:
到目前为止，所有答案都只告诉您如何打印整数数组，但我们也可以打印任何任意结构，前提是我们知道它的大小。下面的例子创建了这样的结构，并通过它的字节迭代一个指针，将它们打印到输出：
#include <iostream>
#include <iomanip>
#include <cstring>
using std::cout;
using std::endl;
using std::hex;
using std::setfill;
using std::setw;
using u64 = unsigned long long;
using u16 = unsigned short;
using f64 = double;
struct Header {
u16 version;
u16 msgSize;
};
struct Example {
Header header;
u64 someId;
u64 anotherId;
bool isFoo;
bool isBar;
f64 floatingPointValue;
};
int main () {
Example example;
// fill with zeros so padding regions don't contain garbage
memset(&example, 0, sizeof(Example));
example.header.version = 5;
example.header.msgSize = sizeof(Example) - sizeof(Header);
example.someId = 0x1234;
example.anotherId = 0x5678;
example.isFoo = true;
example.isBar = true;
example.floatingPointValue = 1.1;
cout << hex << setfill('0'); // needs to be set only once
auto *ptr = reinterpret_cast<unsigned char *>(&example);
for (int i = 0; i < sizeof(Example); i++, ptr++) {
if (i % sizeof(u64) == 0) {
cout << endl;
}
cout << setw(2) << static_cast<unsigned>(*ptr) << " ";
}
return 0;
}
And here's the output:
这是输出：
05 00 24 00 00 00 00 00
34 12 00 00 00 00 00 00
78 56 00 00 00 00 00 00
01 01 00 00 00 00 00 00
9a 99 99 99 99 99 f1 3f
Notice this example also illustrates memory alignmentworking. We see versionoccupying 2 bytes (05 00), followed by msgSizewith 2 more bytes (24 00) and then 4 bytes of padding, after which comes someId(34 12 00 00 00 00 00 00) and anotherId(78 56 00 00 00 00 00 00). Then isFoo, which occupies 1 byte (01) and isBar, another byte (01), followed by 6 bytes of padding, finally ending with the IEEE 754 standard representation of the double field floatingPointValue.
请注意，此示例还说明了内存对齐工作。我们看到version占用了 2 个字节 ( 05 00)，然后是另外msgSize2 个字节 ( 24 00)，然后是 4 个字节的填充，之后是someId( 34 12 00 00 00 00 00 00) 和anotherId( 78 56 00 00 00 00 00 00)。然后isFoo，它占用 1 个字节 ( 01) 和isBar另一个字节 ( 01)，接着是 6 个字节的填充，最后以双字段的 IEEE 754 标准表示结束floatingPointValue。
Also notice that all values are represented as little endian(least significant bytes come first), since this was compiled and run on an Intel platform.
另请注意，所有值都表示为小端（最低有效字节在前），因为这是在 Intel 平台上编译和运行的。
回答by Mohamed Saad
This is a modified version of the Nibble to Hex method
这是 Nibble to Hex 方法的修改版本
void hexArrayToStr(unsigned char* info, unsigned int infoLength, char **buffer) {
const char* pszNibbleToHex = {"0123456789ABCDEF"};
int nNibble, i;
if (infoLength > 0) {
if (info != NULL) {
*buffer = (char *) malloc((infoLength * 2) + 1);
buffer[0][(infoLength * 2)] = 0;
for (i = 0; i < infoLength; i++) {
nNibble = info[i] >> 4;
buffer[0][2 * i] = pszNibbleToHex[nNibble];
nNibble = info[i] & 0x0F;
buffer[0][2 * i + 1] = pszNibbleToHex[nNibble];
}
} else {
*buffer = NULL;
}
} else {
*buffer = NULL;
}
}
回答by Rick Murtagh
I don't know of a better way than:
我不知道比以下更好的方法：
unsigned char byData[xxx];
int nLength = sizeof(byData) * 2;
char *pBuffer = new char[nLength + 1];
pBuffer[nLength] = 0;
for (int i = 0; i < sizeof(byData); i++)
{
sprintf(pBuffer[2 * i], "%02X", byData[i]);
}
You can speed it up by using a Nibble to Hex method
您可以使用 Nibble to Hex 方法加快速度
unsigned char byData[xxx];
const char szNibbleToHex = { "0123456789ABCDEF" };
int nLength = sizeof(byData) * 2;
char *pBuffer = new char[nLength + 1];
pBuffer[nLength] = 0;
for (int i = 0; i < sizeof(byData); i++)
{
// divide by 16
int nNibble = byData[i] >> 4;
pBuffer[2 * i] = pszNibbleToHex[nNibble];
nNibble = byData[i] & 0x0F;
pBuffer[2 * i + 1] = pszNibbleToHex[nNibble];
}
