I'm working with an old program and need help swapping the order of a Hex String.
Yes, a string...as in:
string hexString = "F07D0079"
string hexString2= "F07F"
I need each string to look like:
79007DF0 &
7FF0 respectively.
For the love of god i don't know why they're stored in strings, but they are.
This is a little endian/big endian issue but since it's in a string i can't use standard functions to reverse the order can I?
Is there any easy way to do this?
std::string swapValues(string originalHex)
{
string swappedHex;
//what to do here.
return swappedHex;
}
First check that the length is even (if it hasn't already been sanitised):
assert(hex.length() % 2 == 0);
Then reverse the string:
std::reverse(hex.begin(), hex.end());
Now the bytes are in the correct order, but the digits within each are wrong, so we need to swap them back:
for (auto it = hex.begin(); it != hex.end(); it += 2) {
std::swap(it[0], it[1]);
}
I might use the append member function.
std::string reverse_pairs(std::string const & src)
{
assert(src.size() % 2 == 0);
std::string result;
result.reserve(src.size());
for (std::size_t i = src.size(); i != 0; i -= 2)
{
result.append(src, i - 2, 2);
}
return result;
}
(As an exercise in extensibility, you can make the "2" a parameter, too.)
If you want to do it in-place, you can use std::rotate in a loop.
I wouldn't bother with something overly clever for this:
std::string swapValues(const std::string& o)
{
std::string s(o.length());
if (s.length() == 4) {
s[0] = o[2];
s[1] = o[3];
s[2] = o[0];
s[3] = o[1];
return s;
}
if (s.length() == 8) {
// left as an exercise
}
throw std::logic_error("You got to be kidding me...");
}
There should be library functions available (a naive string manipulation might be no good):
#include <iostream>
#include <arpa/inet.h>
int main() {
std::string hex32 = "F07D0079";
std::string hex16 = "F07F";
std::uint32_t u32 = std::strtoul(hex32.c_str(), 0, 16);
std::uint16_t u16 = std::strtoul(hex16.c_str(), 0, 16);
// Here we would need to know the endian of the sources.
u32 = ntohl(u32);
u16 = ntohs(u16);
std::cout << std::hex << u32 << ", " << u16 << '\n';
}
Linux/Little Endian
Any function operating on the strings must know the target platform (hence there is no general solution)
Related
I used Huffman encoding that we wrote to compress a file.
The function takes String and its output is String.
The problem is I want to save it as binary to get lower size than the original size, but when I take it back (0's and 1's ) as a string its size is larger than the main file. How can I convert that string of (0's and 1's) to a binary so that every character is saved in 1 bit? I am using Qt to achieve this:
string Huffman_encoding(string text)
{
buildHuffmanTree(text);
string encoded = "";
unordered_map<char, string> StringEncoded;
encoding(main_root, "", StringEncoded);
for (char ch : text) {
encoded += StringEncoded[ch];
}
return encoded;
}
The canonical solution uses a "bit packer" that accepts bitstrings and emits packed bytes. As a first start, replace encoded by an instance of the following:
class BitPacker {
QByteArray res;
quint8 bitsLeft = 8;
quint8 buf = 0;
public:
void operator+=(const std::string& s) {
for (auto c : s) {
buf = buf << 1 | c - '0';
if (--bitsLeft == 0) {
res.append(buf);
buf = 0;
bitsLeft = 8;
}
}
}
QByteArray finish() {
if (bitsLeft < 8) {
res.append(buf << bitsLeft);
buf = 0;
bitsLeft = 8;
}
return res;
}
}
operator+= will add additional bits to buf and flush complete bytes to res. At the end of the process you may be left with, say, 3 bits. finish uses a simple algorithm: it pads the buffer with zeroes to produce a final byte and hands you back the fully encoded buffer.
A more sophisticated solution might be to introduce an explicit "end of stream" token that is not present in the source character set.
Seems what you're searching for is a way to convert a string containing a sequence of 0s and 1s like "0000010010000000" to an actual binary representation (numbers 4 and 128 in this example).
This could be achieved with a function like this:
#include <iostream>
#include <string>
#include <cstdint>
#include <vector>
std::vector<uint8_t> toBinary(std::string const& binStr)
{
std::vector<uint8_t> result;
result.reserve(binStr.size() / 8);
size_t pos = 0;
size_t len = binStr.length();
while (pos < len)
{
size_t curLen = std::min(static_cast<size_t>(8), len-pos);
auto curStr = binStr.substr(pos, curLen) + std::string(8-curLen, '0');
std::cout << "curLen: " << curLen << ", curStr: " << curStr << "\n";
result.push_back(std::stoi(curStr, 0, 2));
pos += 8;
}
return result;
}
// test:
int main()
{
std::string binStr("000001001000000001");
auto bin = toBinary(binStr);
for (auto i: bin)
{
std::cout << static_cast<int>(i) << " ";
}
return 0;
}
Output:
4 128 64
You can then do whatever you want with these numbers, e.g. write them into a binary file.
Note that toBinary as above, pads the last byte, if incomplete, with zeros.
You can create a bitstream using bitwise logic like this :
#include <cassert>
#include <string>
#include <stdexcept>
#include <vector>
auto to_bit_stream(const std::string& bytes)
{
std::vector<std::uint8_t> stream;
std::uint8_t shift{ 0 };
std::uint8_t out{ 0 };
// allocate enough bytes to hold the bits
// speeds up the code a bit
stream.reserve((bytes.size() + 7) / 8);
// loop over all bytes
for (const auto c : bytes)
{
// check input
if (!((c == '0') || (c == '1'))) throw std::invalid_argument("invalid character in input");
// shift output by one to accept next bit
out <<= 1;
// keep track of number of shifts
// after 8 shifts a byte has been filled
shift++;
// or the output with a 1 if needed
out |= (c == '1');
// complete an output byte
if (shift == 8)
{
stream.push_back(out);
out = 0;
shift = 0;
}
}
return stream;
}
int main()
{
// stream is 8 bits per value, values 0,1,2,3
auto stream = to_bit_stream("00000000000000010000001000000011");
assert(stream.size() == 4ul);
assert(stream[0] == 0);
assert(stream[1] == 1);
assert(stream[2] == 2);
assert(stream[3] == 3);
return 0;
}
Use std::stoi()
int n = std::stoi("01000100", nullptr, 2);
I tried to write a script that removes extra white spaces but I didn't manage to finish it.
Basically I want to transform abc sssd g g sdg gg gf into abc sssd g g sdg gg gf.
In languages like PHP or C#, it would be very easy, but not in C++, I see. This is my code:
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <cstring>
#include <unistd.h>
#include <string.h>
char* trim3(char* s) {
int l = strlen(s);
while(isspace(s[l - 1])) --l;
while(* s && isspace(* s)) ++s, --l;
return strndup(s, l);
}
char *str_replace(char * t1, char * t2, char * t6)
{
char*t4;
char*t5=(char *)malloc(10);
memset(t5, 0, 10);
while(strstr(t6,t1))
{
t4=strstr(t6,t1);
strncpy(t5+strlen(t5),t6,t4-t6);
strcat(t5,t2);
t4+=strlen(t1);
t6=t4;
}
return strcat(t5,t4);
}
void remove_extra_whitespaces(char* input,char* output)
{
char* inputPtr = input; // init inputPtr always at the last moment.
int spacecount = 0;
while(*inputPtr != '\0')
{
char* substr;
strncpy(substr, inputPtr+0, 1);
if(substr == " ")
{
spacecount++;
}
else
{
spacecount = 0;
}
printf("[%p] -> %d\n",*substr,spacecount);
// Assume the string last with \0
// some code
inputPtr++; // After "some code" (instead of what you wrote).
}
}
int main(int argc, char **argv)
{
printf("testing 2 ..\n");
char input[0x255] = "asfa sas f f dgdgd dg ggg";
char output[0x255] = "NO_OUTPUT_YET";
remove_extra_whitespaces(input,output);
return 1;
}
It doesn't work. I tried several methods. What I am trying to do is to iterate the string letter by letter and dump it in another string as long as there is only one space in a row; if there are two spaces, don't write the second character to the new string.
How can I solve this?
There are already plenty of nice solutions. I propose you an alternative based on a dedicated <algorithm> meant to avoid consecutive duplicates: unique_copy():
void remove_extra_whitespaces(const string &input, string &output)
{
output.clear(); // unless you want to add at the end of existing sring...
unique_copy (input.begin(), input.end(), back_insert_iterator<string>(output),
[](char a,char b){ return isspace(a) && isspace(b);});
cout << output<<endl;
}
Here is a live demo. Note that I changed from c style strings to the safer and more powerful C++ strings.
Edit: if keeping c-style strings is required in your code, you could use almost the same code but with pointers instead of iterators. That's the magic of C++. Here is another live demo.
Here's a simple, non-C++11 solution, using the same remove_extra_whitespace() signature as in the question:
#include <cstdio>
void remove_extra_whitespaces(char* input, char* output)
{
int inputIndex = 0;
int outputIndex = 0;
while(input[inputIndex] != '\0')
{
output[outputIndex] = input[inputIndex];
if(input[inputIndex] == ' ')
{
while(input[inputIndex + 1] == ' ')
{
// skip over any extra spaces
inputIndex++;
}
}
outputIndex++;
inputIndex++;
}
// null-terminate output
output[outputIndex] = '\0';
}
int main(int argc, char **argv)
{
char input[0x255] = "asfa sas f f dgdgd dg ggg";
char output[0x255] = "NO_OUTPUT_YET";
remove_extra_whitespaces(input,output);
printf("input: %s\noutput: %s\n", input, output);
return 1;
}
Output:
input: asfa sas f f dgdgd dg ggg
output: asfa sas f f dgdgd dg ggg
Since you use C++, you can take advantage of standard-library features designed for that sort of work. You could use std::string (instead of char[0x255]) and std::istringstream, which will replace most of the pointer arithmetic.
First, make a string stream:
std::istringstream stream(input);
Then, read strings from it. It will remove the whitespace delimiters automatically:
std::string word;
while (stream >> word)
{
...
}
Inside the loop, build your output string:
if (!output.empty()) // special case: no space before first word
output += ' ';
output += word;
A disadvantage of this method is that it allocates memory dynamically (including several reallocations, performed when the output string grows).
There are plenty of ways of doing this (e.g., using regular expressions), but one way you could do this is using std::copy_if with a stateful functor remembering whether the last character was a space:
#include <algorithm>
#include <string>
#include <iostream>
struct if_not_prev_space
{
// Is last encountered character space.
bool m_is = false;
bool operator()(const char c)
{
// Copy if last was not space, or current is not space.
const bool ret = !m_is || c != ' ';
m_is = c == ' ';
return ret;
}
};
int main()
{
const std::string s("abc sssd g g sdg gg gf into abc sssd g g sdg gg gf");
std::string o;
std::copy_if(std::begin(s), std::end(s), std::back_inserter(o), if_not_prev_space());
std::cout << o << std::endl;
}
You can use std::unique which reduces adjacent duplicates to a single instance according to how you define what makes two elements equal is.
Here I have defined elements as equal if they are both whitespace characters:
inline std::string& remove_extra_ws_mute(std::string& s)
{
s.erase(std::unique(std::begin(s), std::end(s), [](unsigned char a, unsigned char b){
return std::isspace(a) && std::isspace(b);
}), std::end(s));
return s;
}
inline std::string remove_extra_ws_copy(std::string s)
{
return remove_extra_ws_mute(s);
}
std::unique moves the duplicates to the end of the string and returns an iterator to the beginning of them so they can be erased.
Additionally, if you must work with low level strings then you can still use std::unique on the pointers:
char* remove_extra_ws(char const* s)
{
std::size_t len = std::strlen(s);
char* buf = new char[len + 1];
std::strcpy(buf, s);
// Note that std::unique will also retain the null terminator
// in its correct position at the end of the valid portion
// of the string
std::unique(buf, buf + len + 1, [](unsigned char a, unsigned char b){
return (a && std::isspace(a)) && (b && std::isspace(b));
});
return buf;
}
for in-place modification you can apply erase-remove technic:
#include <string>
#include <iostream>
#include <algorithm>
#include <cctype>
int main()
{
std::string input {"asfa sas f f dgdgd dg ggg"};
bool prev_is_space = true;
input.erase(std::remove_if(input.begin(), input.end(), [&prev_is_space](unsigned char curr) {
bool r = std::isspace(curr) && prev_is_space;
prev_is_space = std::isspace(curr);
return r;
}), input.end());
std::cout << input << "\n";
}
So you first move all extra spaces to the end of the string and then truncate it.
The great advantage of C++ is that is universal enough to port your code to plain-c-static strings with only few modifications:
void erase(char * p) {
// note that this ony works good when initial array is allocated in the static array
// so we do not need to rearrange memory
*p = 0;
}
int main()
{
char input [] {"asfa sas f f dgdgd dg ggg"};
bool prev_is_space = true;
erase(std::remove_if(std::begin(input), std::end(input), [&prev_is_space](unsigned char curr) {
bool r = std::isspace(curr) && prev_is_space;
prev_is_space = std::isspace(curr);
return r;
}));
std::cout << input << "\n";
}
Interesting enough remove step here is string-representation independent. It will work with std::string without modifications at all.
I have the sinking feeling that good ol' scanf will do (in fact, this is the C school equivalent to Anatoly's C++ solution):
void remove_extra_whitespaces(char* input, char* output)
{
int srcOffs = 0, destOffs = 0, numRead = 0;
while(sscanf(input + srcOffs, "%s%n", output + destOffs, &numRead) > 0)
{
srcOffs += numRead;
destOffs += strlen(output + destOffs);
output[destOffs++] = ' '; // overwrite 0, advance past that
}
output[destOffs > 0 ? destOffs-1 : 0] = '\0';
}
We exploit the fact that scanf has magical built-in space skipping capabilities. We then use the perhaps less known %n "conversion" specification which gives us the amount of chars consumed by scanf. This feature frequently comes in handy when reading from strings, like here. The bitter drop which makes this solution less-than-perfect is the strlen call on the output (there is no "how many bytes have I actually just written" conversion specifier, unfortunately).
Last not least use of scanf is easy here because sufficient memory is guaranteed to exist at output; if that were not the case, the code would become more complex due to buffering and overflow handling.
Since you are writing c-style, here's a way to do what you want.
Note that you can remove '\r' and '\n' which are line breaks (but of course that's up to you if you consider those whitespaces or not).
This function should be as fast or faster than any other alternative and no memory allocation takes place even when it's called with std::strings (I've overloaded it).
char temp[] = " alsdasdl gasdasd ee";
remove_whitesaces(temp);
printf("%s\n", temp);
int remove_whitesaces(char *p)
{
int len = strlen(p);
int new_len = 0;
bool space = false;
for (int i = 0; i < len; i++)
{
switch (p[i])
{
case ' ': space = true; break;
case '\t': space = true; break;
case '\n': break; // you could set space true for \r and \n
case '\r': break; // if you consider them spaces, I just ignore them.
default:
if (space && new_len > 0)
p[new_len++] = ' ';
p[new_len++] = p[i];
space = false;
}
}
p[new_len] = '\0';
return new_len;
}
// and you can use it with strings too,
inline int remove_whitesaces(std::string &str)
{
int len = remove_whitesaces(&str[0]);
str.resize(len);
return len; // returning len for consistency with the primary function
// but u can return std::string instead.
}
// again no memory allocation is gonna take place,
// since resize does not not free memory because the length is either equal or lower
If you take a brief look at the C++ Standard library, you will notice that a lot C++ functions that return std::string, or other std::objects are basically a wrapper to a well written extern "C" function. So don't be afraid to use C functions in C++ applications, if they are well written and you can overload them to support std::strings and such.
For example, in Visual Studio 2015, std::to_string is written exactly like this:
inline string to_string(int _Val)
{ // convert int to string
return (_Integral_to_string("%d", _Val));
}
inline string to_string(unsigned int _Val)
{ // convert unsigned int to string
return (_Integral_to_string("%u", _Val));
}
and _Integral_to_string is a wrapper to a C function sprintf_s
template<class _Ty> inline
string _Integral_to_string(const char *_Fmt, _Ty _Val)
{ // convert _Ty to string
static_assert(is_integral<_Ty>::value,
"_Ty must be integral");
char _Buf[_TO_STRING_BUF_SIZE];
int _Len = _CSTD sprintf_s(_Buf, _TO_STRING_BUF_SIZE, _Fmt, _Val);
return (string(_Buf, _Len));
}
Well here is a longish(but easy) solution that does not use pointers.
It can be optimized further but hey it works.
#include <iostream>
#include <string>
using namespace std;
void removeExtraSpace(string str);
int main(){
string s;
cout << "Enter a string with extra spaces: ";
getline(cin, s);
removeExtraSpace(s);
return 0;
}
void removeExtraSpace(string str){
int len = str.size();
if(len==0){
cout << "Simplified String: " << endl;
cout << "I would appreciate it if you could enter more than 0 characters. " << endl;
return;
}
char ch1[len];
char ch2[len];
//Placing characters of str in ch1[]
for(int i=0; i<len; i++){
ch1[i]=str[i];
}
//Computing index of 1st non-space character
int pos=0;
for(int i=0; i<len; i++){
if(ch1[i] != ' '){
pos = i;
break;
}
}
int cons_arr = 1;
ch2[0] = ch1[pos];
for(int i=(pos+1); i<len; i++){
char x = ch1[i];
if(x==char(32)){
//Checking whether character at ch2[i]==' '
if(ch2[cons_arr-1] == ' '){
continue;
}
else{
ch2[cons_arr] = ' ';
cons_arr++;
continue;
}
}
ch2[cons_arr] = x;
cons_arr++;
}
//Printing the char array
cout << "Simplified string: " << endl;
for(int i=0; i<cons_arr; i++){
cout << ch2[i];
}
cout << endl;
}
I don't know if this helps but this is how I did it on my homework. The only case where it might break a bit is when there is spaces at the beginning of the string EX " wor ds " In that case, it will change it to " wor ds"
void ShortenSpace(string &usrStr){
char cha1;
char cha2;
for (int i = 0; i < usrStr.size() - 1; ++i) {
cha1 = usrStr.at(i);
cha2 = usrStr.at(i + 1);
if ((cha1 == ' ') && (cha2 == ' ')) {
usrStr.erase(usrStr.begin() + 1 + i);
--i;//edit: was ++i instead of --i, made code not work properly
}
}
}
I ended up here for a slighly different problem. Since I don't know where else to put it, and I found out what was wrong, I share it here. Don't be cross with me, please.
I had some strings that would print additional spaces at their ends, while showing up without spaces in debugging. The strings where formed in windows calls like VerQueryValue(), which besides other stuff outputs a string length, as e.g. iProductNameLen in the following line converting the result to a string named strProductName:
strProductName = string((LPCSTR)pvProductName, iProductNameLen)
then produced a string with a \0 byte at the end, which did not show easily in de debugger, but printed on screen as a space. I'll leave the solution of this as an excercise, since it is not hard at all, once you are aware of this.
I don't know whether or not this is possible but I have used this technique in different languages but am struggling to use it in C++. I have 10 images that I am trying to load into an array using a loop as so:
for (int i = 0; i < 10; i++)
{
Sprite[i] = IMG_LoadTexture(renderer, "Graphics/Player" + i + ".png");
}
This however does not seem to work in C++ so I was wondering what I am doing wrong, or what can I do to get the same result without having to load each image individually like so:
Sprite[0] = IMG_LoadTexture(renderer, "Graphics/Player0.png");
My error is: "Expression must have integral or unscoped enum type"
Thanks for any help =)
You cannot do this:
"This is my number: " + (int)4 + "!";
This is illegal. It will give you an error for trying to operator+ a const char* and a const char[SOME_INT_GOES_HERE] or another error for trying to use operator+ to add an int onto a string. Things just don't work that way.
You'd either have to use C (i.e. snprintf()) or a string stream. Here's my test code for isolating your problem:
#include <iostream>
#include <string>
int main()
{
int a = 1;
std::string str = "blah";
std::string end = "!";
//std::string hello = str + a + end;// GIVES AN ERROR for operator+
std::string hello = "blah" + a + "!";
//const char* c_str = "blah" + a + "end";
//std::cout << c_str << std::endl;
std::cout << hello << std::endl;
return 0;
}
Here's an alternative solution using string streams.
#include <iostream>
#include <string>
#include <sstream>
int main()
{
int i = 0;
std::string str;
std::stringstream ss;
while (i < 10)
{
//Send text to string stream.
ss << "text" << i;
//Set string to the text inside string stream
str = ss.str();
//Print out the string
std::cout << str << std::endl;
//ss.clear() doesn't work. Calling a constructor
//for std::string() and setting ss.str(std::string())
//will set the string stream to an empty string.
ss.str(std::string());
//Remember to increment the variable inside of while{}
++i;
}
}
Alternatively, you can also use std::to_string() if you're using C++11 (which just requires -std=c++11) but std::to_string() is broken on some sets of compilers (i.e. regular MinGW). Either switch to another flavor where it works (i.e. MinGW-w64) or just write your own to_string() function using string streams behind the scenes.
snprintf() may be the fastest way of doing such a thing, but for safer C++ and better style, it is recommended you use a non-C way of doing things.
I had a similar problem and I solwed it this way:
#include <iostream>
using namespace std;
int main() {
string line;
for (int i = 0; i < 10; i++) {
line = "Graphics/Player" + inttostr(i) + ".png"; //I wrote inttostr function because built in inttostr functions messed up my program (see below)
char charger[line.length()]; //creating char array
for (int i = 0; i < sizeof(line); i++) {
charger[i] = line[i]; // copying string to char arry
}
Sprite[i] = IMG_LoadTexture(renderer, charger);
}
}
string inttostr(int integer) { //I know it isn't the best way to convert integer to string, but it works
string charakter;
int swap;
bool negativ = false;
if (integer < 0) {
integer = -integer;
negativ = true;
}
if (integer == 0) {
charakter = "0";
}
while (integer >= 1) {
swap = integer % 10;
integer = integer / 10;
charakter = char(swap + 48) + charakter;
}
if (negativ) {
charakter = "-" + charakter;
}
return charakter;
}
Let say you have a string:
std::string s = "ABCD\t1234";
I can use std::string::find to get an offset to the '\t' character, but as far as I know there are no functions with the following signature:
int atoi_n(char *, int len);
An I missing anything? strtok replaces the \t with \0, and I don't want to touch the original buffer. I find it hard to believe that there aren't instances of the atoi, atof, etc which take a length parameter, but I can't find anything.
Anyone know if there is something I'm missing? I know boost has some tokenizers but I'd like avoid having to add the dependency of boost.
Looking the comments so far I'd like to clarify. Let's change the scenario:
char buffer[1024];
char *pStartPos;
char *pEndPost;
pStartPos = buffer + 5;
pEndPos = buffer + 10;
Let's also say you can't make any assumptions about the memory outside pStartPos and pEndPos. How do you convert the charaters between pStartPos and pEndPos to an int without adding a '\0' to buffer or copying using substr?
If you want to parse only the end of the string (from the character after \t to the end) you just need to pass a pointer to the first character to parse to atoi...
int n = atoi(s.c_str()+s.find('\t')+1);
(error checking omitted for brevity - in particular, we are always assuming that a \t is actually present)
If, instead, you want to parse from the beginning of the string up to \t you can just do
int n = atoi(s.c_str());
since atoi stops at the first non-numeric character anyway.
By the way, you should consider using more robust solutions for parsing the number, like strtol, sscanf or the C++ streams - they all can report a parsing error in some way, while atoi just returns 0 (which isn't distinguishable from a 0 that comes from parsing the string).
Incidentally, atoi is not in the "STL" by any means - it's just part of the C standard library.
I know that atoi is not in the STL. I was wondering if there was anything in STL like it where you can specify the last character which you want to include in the conversion. Basically I have a buffer which may be partially filled with garbage. I know the start of possible valid data and the end of possible valid data. I don't want to depend on whitespace to end the conversion, I want to be explicit about the length of the "field" because it also may not be /0 terminated.
If you are sure that the garbage doesn't start with digits you can use atoi/strtol/istringstream as is - they automatically stop just when they see the garbage. Otherwise, use the substr method to extract the exact substring you need:
std::string mayContainGarbage="alcak123456amaclmò";
std::string onlyTheDigits=mayContainGarbage.substr(5, 6);
// now parse onlyTheDigits as you prefer
To my knowledge, there is no such function out-of-box, but it shouldn't be difficult to implement.
For example:
template <typename ForwardIterator>
int range_to_int(ForwardIterator begin, ForwardIterator past_end) {
if (begin != past_end) {
bool negative = false;
auto ch = *begin;
if (ch == '-') {
negative = true;
++begin;
}
else if (ch == '+')
++begin;
if (begin != past_end) {
int result = 0;
do {
auto ch = *begin;
if (ch < '0' || ch > '9')
throw std::invalid_argument("Invalid digit.");
result = result * 10 + (ch - '0');
++begin;
} while (begin != past_end);
if (negative)
result = -result;
return result;
}
throw std::invalid_argument("+ or - must be followed by at least one digit.");
}
throw std::invalid_argument("Empty range.");
}
And you can use it like this:
int main() {
const char* buffer = "abc-123def";
int i = range_to_int(buffer + 4, buffer + 7);
assert(i == 123);
i = range_to_int(buffer + 3, buffer + 7);
assert(i == -123);
try {
i = range_to_int(buffer + 3, buffer + 8);
assert(false);
}
catch (const std::exception& ex) {
std::cout << ex.what() << std::endl;
}
try {
i = range_to_int(buffer + 3, buffer + 4);
assert(false);
}
catch (const std::exception& ex) {
std::cout << ex.what() << std::endl;
}
try {
i = range_to_int(buffer + 4, buffer + 4);
assert(false);
}
catch (const std::exception& ex) {
std::cout << ex.what() << std::endl;
}
// You can use it on std::string as well....
const std::string str = buffer;
i = range_to_int(str.begin() + 4, str.begin() + 7);
assert(i == 123);
// Etc...
return EXIT_SUCCESS;
}
I'm reading a string from a file so it's in the form of a char array. I need to tokenize the string and save each char array token as a uint8_t hex value in an array.
char* starting = "001122AABBCC";
// ...
uint8_t[] ending = {0x00,0x11,0x22,0xAA,0xBB,0xCC}
How can I convert from starting to ending? Thanks.
Here is a complete working program. It is based on Rob I's solution, but fixes several problems has been tested to work.
#include <string>
#include <stdio.h>
#include <stdlib.h>
#include <vector>
#include <iostream>
const char* starting = "001122AABBCC";
int main()
{
std::string starting_str = starting;
std::vector<unsigned char> ending;
ending.reserve( starting_str.size());
for (int i = 0 ; i < starting_str.length() ; i+=2) {
std::string pair = starting_str.substr( i, 2 );
ending.push_back(::strtol( pair.c_str(), 0, 16 ));
}
for(int i=0; i<ending.size(); ++i) {
printf("0x%X\n", ending[i]);
}
}
strtoul will convert text in any base you choose into bytes. You have to do a little work to chop the input string into individual digits, or you can convert 32 or 64bits at a time.
ps uint8_t[] ending = {0x00,0x11,0x22,0xAA,0xBB,0xCC}
Doesn't mean anything, you aren't storing the data in a uint8 as 'hex', you are storing bytes, it's upto how you (or your debugger) interpretes the binary data
With C++11, you may use std::stoi for that :
std::vector<uint8_t> convert(const std::string& s)
{
if (s.size() % 2 != 0) {
throw std::runtime_error("Bad size argument");
}
std::vector<uint8_t> res;
res.reserve(s.size() / 2);
for (std::size_t i = 0, size = s.size(); i != size; i += 2) {
std::size_t pos = 0;
res.push_back(std::stoi(s.substr(i, 2), &pos, 16));
if (pos != 2) {
throw std::runtime_error("bad character in argument");
}
}
return res;
}
Live example.
I think any canonical answer (w.r.t. the bounty notes) would involve some distinct phases in the solution:
Error checking for valid input
Length check and
Data content check
Element conversion
Output creation
Given the usefulness of such conversions, the solution should probably include some flexibility w.r.t. the types being used and the locale required.
From the outset, given the date of the request for a "more canonical answer" (circa August 2014) liberal use of C++11 will be applied.
An annotated version of the code, with types corresponding to the OP:
std::vector<std::uint8_t> convert(std::string const& src)
{
// error check on the length
if ((src.length() % 2) != 0) {
throw std::invalid_argument("conversion error: input is not even length");
}
auto ishex = [] (decltype(*src.begin()) c) {
return std::isxdigit(c, std::locale()); };
// error check on the data contents
if (!std::all_of(std::begin(src), std::end(src), ishex)) {
throw std::invalid_argument("conversion error: input values are not not all xdigits");
}
// allocate the result, initialised to 0 and size it to the correct length
std::vector<std::uint8_t> result(src.length() / 2, 0);
// run the actual conversion
auto str = src.begin(); // track the location in the string
std::for_each(result.begin(), result.end(), [&str](decltype(*result.begin())& element) {
element = static_cast<std::uint8_t>(std::stoul(std::string(str, str + 2), nullptr, 16));
std::advance(str, 2); // next two elements
});
return result;
}
The template version of the code adds flexibility;
template <typename Int /*= std::uint8_t*/,
typename Char = char,
typename Traits = std::char_traits<Char>,
typename Allocate = std::allocator<Char>,
typename Locale = std::locale>
std::vector<Int> basic_convert(std::basic_string<Char, Traits, Allocate> const& src, Locale locale = Locale())
{
using string_type = std::basic_string<Char, Traits, Allocate>;
auto ishex = [&locale] (decltype(*src.begin()) c) {
return std::isxdigit(c, locale); };
if ((src.length() % 2) != 0) {
throw std::invalid_argument("conversion error: input is not even length");
}
if (!std::all_of(std::begin(src), std::end(src), ishex)) {
throw std::invalid_argument("conversion error: input values are not not all xdigits");
}
std::vector<Int> result(src.length() / 2, 0);
auto str = std::begin(src);
std::for_each(std::begin(result), std::end(result), [&str](decltype(*std::begin(result))& element) {
element = static_cast<Int>(std::stoul(string_type(str, str + 2), nullptr, 16));
std::advance(str, 2);
});
return result;
}
The convert() function can then be based on the basic_convert() as follows:
std::vector<std::uint8_t> convert(std::string const& src)
{
return basic_convert<std::uint8_t>(src, std::locale());
}
Live sample.
uint8_t is typically no more than a typedef of an unsigned char. If you're reading characters from a file, you should be able to read them into an unsigned char array just as easily as a signed char array, and an unsigned char array is a uint8_t array.
I'd try something like this:
std::string starting_str = starting;
uint8_t[] ending = new uint8_t[starting_str.length()/2];
for (int i = 0 ; i < starting_str.length() ; i+=2) {
std::string pair = starting_str.substr( i, i+2 );
ending[i/2] = ::strtol( pair.c_str(), 0, 16 );
}
Didn't test it but it looks good to me...
You may add your own conversion from set of char { '0','1',...'E','F' } to uint8_t:
uint8_t ctoa(char c)
{
if( c >= '0' && c <= '9' ) return c - '0';
else if( c >= 'a' && c <= 'f' ) return 0xA + c - 'a';
else if( c >= 'A' && c <= 'F' ) return 0xA + c - 'A';
else return 0;
}
Then it will be easy to convert a string in to array:
uint32_t endingSize = strlen(starting)/2;
uint8_t* ending = new uint8_t[endingSize];
for( uint32_t i=0; i<endingSize; i++ )
{
ending[i] = ( ctoa( starting[i*2] ) << 4 ) + ctoa( starting[i*2+1] );
}
This simple solution should work for your problem
char* starting = "001122AABBCC";
uint8_t ending[12];
// This algo will work for any size of starting
// However, you have to make sure that the ending have enough space.
int i=0;
while (i<strlen(starting))
{
// convert the character to string
char str[2] = "\0";
str[0] = starting[i];
// convert string to int base 16
ending[i]= (uint8_t)atoi(str,16);
i++;
}
uint8_t* ending = static_cast<uint8_t*>(starting);