Example
For example:
string MyString = "Normal\tString";
cout << MyString << endl;
produces the following: "Normal String"
Appending the raw string modifier to the string like so:
string MyString = R"(Normal\tString)";
cout << MyString << endl;
produces the following: "Normal\tString"
The Question
Is there a way to append the raw string modifier to a variable containing a string in order to print the raw form of the string contained within the variable?
string TestString = "Test\tString";
cout << R(TestString) << endl;
So you get: "Test\tString"
Is there a way to append the raw string modifier to a variable containing a string in order to print the raw form of the string contained within the variable?
No.
However, you can write a function that substitutes the characters that are defined by escape sequences by an appropriate string, i.e. replace the character '\t' by the string "\\t".
Sample program:
#include <iostream>
#include <string>
// Performs only one substitution of \t.
// Needs to be updated to do it for all occurrences of \t and
// all other escape sequences that can be found in raw strings.
std::string toRawString(std::string const& in)
{
std::string ret = in;
auto p = ret.find('\t');
if ( p != ret.npos )
{
ret.replace(p, 1, "\\t");
}
return ret;
}
int main()
{
std::string TestString = "Test\tString";
std::cout << toRawString(TestString) << std::endl;
}
Output:
Test\tString
This question is tagged as C++11, in which case rolling your own conversion function is probably the best call.
However, if you have a C++14 compiler, you can use the std::quoted stream manipulator:
#include <iostream>
#include <iomanip>
int main() {
string s = "Hello\tWorld!";
std::cout << std::quoted(s) << std::endl; // Prints "Hello\tWorld!"
}
Kind of. This works but it's not a pretty solution:
std::string strToInsert = " insert this "
std::string myRawString = R"(
raw string raw string raw string
raw string raw string raw string
)"; myRawString += strToInsert; myRawString += R"(raw string
raw string raw string raw string
)";
Related
In a small utility I'm writing, I want to read a file record and:
look for the presence of an XOR checksum in the form *XX, where XX are hex digits
replace it if it's incorrect
add one if it's not present
So far, I'm only to the point of reading the file and looking for the checksum. The problem I'm up against is that std::string::find is not finding the * I know to be present; it returns npos every time.
The find() is on line 37. The first line read into strInput is:
$GPGGA,14240.99,2732.581,S,15301.947,E,1,06,3,65,M,37,M,-1.0,0006*6E\n
Here's the code:
#include <iostream>
#include <string>
int main()
{
std::cout << "Enter input file name:\n";
std::string strFileName = "";
std::getline(std::cin, strFileName);
std::cout << "Filename entered: " << strFileName << '\n';
FILE* fp;
int nErrCode = EXIT_FAILURE;
fopen_s(&fp, strFileName.c_str(), "r+");
if (!fp)
{
std::string strErr = "Failed to open " + strFileName;
perror(strErr.c_str());
return nErrCode;
}
rewind(fp);
std::string strInput;
strInput.reserve(100);
std::string::size_type n;
do
{
fgets(&strInput[0], 99, fp);
std::cout << strInput.c_str();
//n = 0;
n = strInput.find('*');
if (n != std::string::npos)
std::cout << "Found checksum at position " << n;
else
std::cout << "Did not find checksum";
} while (!feof(fp));
}
Thanks in advance.
I've debugged this and all is well up to the find(). At that point, I can see that the return value is npos, even though I can see the value of strInput and verify that it contains an asterisk.
Is it possible to edit a .mm file before it gets compiled in AppCenter?
In an attempt to fix a build error, I want to find and replace a string in ../node_modules/react-native/React/CxxBridge/RCTCxxBridge.mm.
I tried using sed -i 's/oldString/newString/g' ../node_modules/react-native/React/CxxBridge/RCTCxxBridge.mm inside appcenter-pre-build.sh but it does not work.
Any help will be appreciated,
Thanks.
Not sure if this is your case, but I needed to update a version number on a complex project. To replace the counter of the current version with a new one, I considered updating the file with each build. After some versions of bash scripts, I realized that it's easier for me to write a console application in C with a couple of parameters to solve this problem. Works almost perfect for me. If you need I can share the simple code of this program.
Here is the C code that looks for a string in the file passed as a parameter and replaces the version number in it.
int main(int argc, char* argv[])
{
cout << "Version changer start\n";
if (argc < 2) {
cout << "There is no path argument. Version was no changed.";
return 1;
}
string sourcePath = argv[1];
string targetPath = sourcePath + ".tmp";
bool firstLine = true;
cout << sourcePath << endl;
ifstream sourceFile(sourcePath); // open file for input
ofstream targetFile(targetPath); // open file for output
string line;
while (getline(sourceFile, line)) // for each line read from the file
{
// looking for the desired line
if (line.find("public static String VER") != std::string::npos) { // replace "public static String VER" to your string definition code
line.replace(0, 32, "");
line.replace(line.length() - 2, 2, "");
int number = atoi(line.c_str());
number++; // In my case, I get an integer and add one to it
string v = to_string(number);
line = " public static String VER = \"" + v + "\";";
cout << v;
}
if (firstLine) {
targetFile << line;
firstLine = false;
}
else
targetFile << endl << line;
}
sourceFile.close();
targetFile.close();
remove(sourcePath.c_str());
if (rename(targetPath.c_str(), sourcePath.c_str()) != 0)
perror("Error renaming file");
else
cout << endl << "----------- done -----------\n";
}
I'm new to coding in C and C++, and I have a program with
an issue. When I (try) to run it, it gives me this error:
"No suitable constructor exists to convert from "char" to "std::string".
I'm not sure what it means. My code is an example of a simple
substitution cipher covered in the book "Cracking Codes with Python" by Al Sweigart.
I just want to replicate it in C++. Here's my code:
#include <iostream> // for communicating with user
#include <string>
using namespace std;
string symbols = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"; // all symbols
string encrypt(string msg, string key, string mode) {
string ca = symbols;
string cb = key;
string translated;
if (mode == "decrypt") {
ca, cb = cb, ca;
}
int index = 0;
for (index = 0; index < msg.length(); index++) {
cout << "index is " << index << endl;
int sindex = ca.find(msg[index]); // it works here
cout << "sindex is " << sindex << endl;
string cl = cb[sindex]; // the problem
translated += cl;
}
return translated;
}
int main() {
string msg = "";
string key = "";
string mode = "";
string ciphertext = ""; // our variables
cout << "Enter message: (no spaces please)\n";
cin >> msg;
cout << "Enter key (or \"none\" for using default):\n";
cin >> key;
if (key == "none") {
key = "QWERTYUIOPASDFGHJKLZXCVBNM";
}
cout << "Enter mode: (\"encrypt\" or \"decrypt\")\n";
cin >> mode;
ciphertext = encrypt(msg, key, mode);
cout << "The ciphertext is\n" << ciphertext;
}
For some reason it works with msg on line 17 but not with cb on line 19, even though
they're both std::string. The actual error is on line 19 with string cl = cb[sindex];.
Not even sure what's wrong. It works on line 17 int sindex = ca.find(/*The thing here*/msg[index]);.
(Maybe my Visual Studio 2019 has gone nuts.) If I replace cb with msg it still gives me the
same error. Maybe line 17 is a lucky line? Who knows? But please help, I'm so
confused!
This code is giving error on 7th line?
why is *it=*rit not working?
string temp;
string::reverse_iterator rit;
rit = demo.rbegin();
string::iterator it;
it = temp.begin();
for (; rit != demo.rend(); rit++, it++)
{
*it = *rit;
}
To answer why your version doesn't work.
( I don't see a definition for demo - I assume its a non-empty string. )
temp is defined though, as an empty string.
Your loop checks "rit" but not "it" this could be OK if the 2 strings had the same length.
"it" is trying to iterate along an empty string, and that's why dereferencing it fails.
If temp was already a string with the correct length, then it could be iterated and edited successfully.
You could achieve this with
temp.resize(demo.length());
or just
temp = demo;
before the loop.
Try it like this:
#include <iostream>
#include <string>
#include <iterator>
int main()
{
std::string s = "Hello, world";
std::reverse_iterator<std::string::iterator> r = s.rbegin();
std::string rev(r, s.rend());
std::cout << rev << '\n';
}
I have problem converting String^ containing 8 bytes as characters (as ascii) to double.
I want to take those 8 characters and convert them binary to double.
What would you recommend to do this conversion in C++/cli?
I was trying to use Marshal::Copy, Double::TryParse, etc.
Maybe I use wrong specifications of parameters, but I really lost my last hopes.
There must be something easy to do this conversion.
Thanks.
Well, the bad news is that the System.String class uses only Unicode encoding internally.
So if you give it bytes it will map them to its internal encoding hiding the original value.
The good news is that you can play with the System.Text.Encoding class to retrieve 8bits values corresponding to the unicode characters.
Here is a sample :
#include <iostream>
using namespace System;
using namespace System::Text;
int main()
{
int n = 123456;
double d = 123.456;
std::cout << n << std::endl;
std::cout << d << std::endl;
char* n_as_bytes = (char*)&n;
char* d_as_bytes = (char*)&d;
String^ n_as_string = gcnew String(n_as_bytes, 0, sizeof(n));
String^ d_as_string = gcnew String(d_as_bytes, 0, sizeof(d));
Encoding^ ascii = Encoding::GetEncoding("iso-8859-1");
array<Byte>^ n_as_array = ascii->GetBytes(n_as_string);
array<Byte>^ d_as_array = ascii->GetBytes(d_as_string);
cli::pin_ptr<unsigned char> pin_ptr_n = &n_as_array[0];
cli::pin_ptr<unsigned char> pin_ptr_d = &d_as_array[0];
unsigned char* ptr_n = pin_ptr_n;
unsigned char* ptr_d = pin_ptr_d;
int n_out = *(int*)ptr_n;
double d_out = *(double*)ptr_d;
std::cout << n_out << std::endl;
std::cout << d_out << std::endl;
return 0;
}
This should give you :
123456
123.456
123456
123.456
Not sure it is completely safe, but trying it in your context should be a good start to ensure it is viable. :)