This seems like it should be a fairly simple thing to do, but I'm having a hard time figuring out how to do it. Basically I just have a QString object that will always be one character (specifically a letter) long, and I need to convert it to a char to set it to an object in my class of that type. I know how to make it char*, with QString("myTextHere").toStdString.c_str(), but the object that the value needs to be set to isn't char*, it's char. Is there a way to do this? Thanks!
I'm pretty sure I found a way to convert the value without getting the error about QCharRef, I found if I use the QString::at() function to get the first index instead of QString("myText")[0], and then use toAscii() on that value, it seems to work and produce the correct value I want. In other words, this is what I did:
char c = QString("A").at(0).toAscii();
If you claim that this returns a char*:
QString("myTextHere").toStdString.c_str();
Then obviously, this will get the first character, which would be char:
QString("myTextHere").toStdString.c_str()[0];
It may not look pretty, and probably there are better ways of getting the first character, but by definition, this code should (probably must) work.
Use the index operator to get the first character of the string:
QString("myTextHere").toStdString()[0]
However, there is no need to convert to std::string, so the following is a better idea:
The main problem is that QString is in UTF-16 and contains QChars, two-byte characters. If your character can be represented using latin-1 charset, you should use
char ch = QString("myTextHere")[0].latin1();
Otherwise you need a bigger type (NOTE: even this cuts non-BMP characters in half):
int i = QString("myTextHere")[0].unicode();
Like others have said, use QString::operator[]. Using this on a QString will return a QCharRef, a helper class that is equivalent to a QChar. Since QChar is meant to support unicode characters there is only one further specification. From QChar in qt documentation (4.7):
QChar provides constructors and cast operators that make it easy to
convert to and from traditional 8-bit chars. If you defined
QT_NO_CAST_FROM_ASCII and QT_NO_CAST_TO_ASCII, as explained in the
QString documentation, you will need to explicitly call fromAscii() or
fromLatin1(), or use QLatin1Char, to construct a QChar from an 8-bit
char, and you will need to call toAscii() or toLatin1() to get the
8-bit value back.
My Qt build will not allow
char c = QString("myTextHere")[0];
and instead I would use
char c = QString("myTextHere")[0].toAscii();
this worked for me in case you are getting error
error: reference to non-static member function must be called; did you
mean to call it with no arguments?
QString mystr = "mystring";
char myvar = mystr.toStdString().c_str()[0];
Related
So I've looked around for how to convert a string to a short and found a lot on how to convert a string to an integer. I would leave a question as a comment on those threads, but I don't have enough reputation. So, what I want to do is convert a string to a short, because the number should never go above three or below zero and shorts save memory (as far as I'm aware).
To be clear, I'm not referring to ASCII codes.
Another thing I want to be able to do is to check if the conversion of the string to the short fails, because I'll be using a string which consists of a users input.
I know I can do this with a while loop, but if there's a built in function to do this in C++ that would be just as, or more, efficient than a while loop, I would love to hear about it.
Basically, an std::stos function is missing for unknown reasons, but you can easily roll your own. Use std::stoi to convert to int, check value against short boundaries given by e.g. std::numeric_limits<short>, throw std::range_error if it's not in range, otherwise return that value. There.
If you already have the Boost library installed you might use boost::lexical_cast for convenience, but otherwise I would avoid it (mainly for the verbosity and library dependency, and it's also a little inefficient).
Earlier boost::lexical_cast was known for not being very efficient, I believe because it was based internally on stringstreams, but as reported in comments here the modern version is faster than conversion via stringstream, and for that matter than via scanf.
An efficient way is to use boost::lexical_cast:
short myShort = boost::lexical_cast<short>(myString);
You will need to install boost library and the following include: #include <boost/lexical_cast.hpp>
You should catch bad_lexical_cast in case the cast fails:
try
{
short myShort = boost::lexical_cast<short>(myString);
}
catch(bad_lexical_cast &)
{
// Do something
}
You can also use ssprintf with the %hi format specifier.
Example:
short port;
char szPort[] = "80";
sscanf(szPort, "%hi", &port);
the number should never go above three or below zero
If you really really need to save memory, then this will also fit in a char (regardless whether char is signed or unsigned).
Another 'extreme' trick: if you can trust there are no weird things like "002" then what you have is a single character string. If that is the case, and you really really need performance, try:
char result = (char)( *ptr_c_string - '0' );
m working on some homework but don't even know where to start on this one. If you could can you throw me in the right direction. This is what i'm suppose to do
Write your own version of the str_c function that takes a C++ string as an argument (with the parameter set as a constant reference variable) and returns a pointer to the equivalent C-string. Be sure to test it with an appropriate driver.
There are different possibilities to write such a function.
First, take a look at the C++ reference for std::string, which is the starting point for your problem.
In the Iterator section on that page, you might find some methods which can help you to get the string character by character.
It can also help to read the documentation for the std::string::c_str method, you'd like to imitate: string::c_string. It's important to understand, how the system works with normal C-strings (char*):
Due to the fact, that a C-string has now length- or size-attribute, a trick is used to determine the end of the string: The last character in the string has to be a '\0'.
Make sure you understand, that a char* string can also be seen as array of characters (char[]). This might help you, when understanding and solving your problem.
as we know, C-string is null-terminated array of char. you can put char by char from std::string to an array of char, and then closed with '\0'. and remember a pointer to a char (char*) is also representation of array of char. you can use this concept
Entire string needs to be made of integers which as we know are 0123456789 I am trying with following function but it doesnt seem to work
bool isNumeric( const char* pszInput, int nNumberBase )
{
string base = "0123456789";
string input = pszInput;
return (::strspn(input.substr(0, nNumberBase).c_str(), base.c_str()) == input.length());
}
and the example of using it in code...
isdigit = (isNumeric((char*)text, 11));
It returns true even with text in the string
Presumably the issue is that text is actually LPCWSTR which is const wchar_t*. We have to infer this fact from the question title and the cast that you made.
Now, that cast is a problem. The compiler objected to you passing text. It said that text is not const char*. By casting you have not changed what text is, you simply lied to the compiler. And the compiler took its revenge.
What happens next is that you reinterpret the wide char buffer as being a narrow 8 bit buffer. If your wide char buffer has latin text, encoded as UTF-16, then every other byte will be zero. Hence the reinterpret cast that you do results in isNumeric thinking that the string is only 1 character long.
What you need to do is either:
Start using UTF-16 encoded wchar_t buffers in isNumeric.
Convert from UTF-16 to ANSI before calling isNumeric.
You should think about this carefully. It seems that at present you have a rather unholy mix of ANSI and UTF-16 in your program. You really ought to settle on a standard character encoding an use it consistently throughout. That is tenable internal to your program, but you will encounter external text that could use different encodings. Deal with that by converting at the boundary between your program and the outside world.
Personally I don't understand why you are using C strings at all. Surely you should be using std::wstring or std::string.
what is the best way to call a function with the following declaration
string Extract(const char* pattern,const char* input);
i use
string str=Extract("something","input text");
is there a problem with this usage
should i use the following
char pattern[]="something";
char input[]="input";
//or use pointers with new operator and copy then free?
the both works but i like the first one but i want to know the best practice.
A literal string (e.g. "something") works just fine as a const char* argument to a function call.
The first method, i.e. passing them literally in, is usually preferable.
There are occasions though where you don't want your strings hard-coded into the text. In some ways you can say that, a bit like magic numbers, they are magic words / phrases. So you prefer to use constant identifier to store the values and pass those in instead.
This would happen often when:
1. a word has a special meaning, and is passed in many times in the code to have that meaning.
or
2. the word may be cryptic in some way and a constant identifier may be more descriptive
Unless you plain to have duplicates of the same strings, or alter those strings, I'm a fan of the first way (passing the literals directly), it means less dotting about code to find what the parameters actually are, it also means less work in passing parameters.
Seeing as this is tagged for C++, passing the literals directly allows you to easily switch the function parameters to std::string with little effort.
I've found that
NetUserChangePassword(0, 0, L"ab", L"cd");
changes the user password from ab to cd. However,
NetUserChangePassword(0, 0, (LPCWSTR) "ab", (LPCWSTR) "cd");
doesn't work. The returned value indicates invalid password.
I need to pass const char* as last two parameters for this function call. How can I do that? For example,
NetUserChangePassword(0, 0, (LPCWSTR) vs[0].c_str(), (LPCWSTR) vs[1].c_str());
Where vs is std::vector<std::string>.
Those are two totally different L's. The first is a part of the C++ language syntax. Prefix a string literal with L and it becomes a wide string literal; instead of an array of char, you get an array of wchar_t.
The L in LPCWSTR doesn't describe the width of the characters, though. Instead, it describes the size of the pointer. Or, at least, it used to. The L abbreviation on type names is a relic of 16-bit Windows, when there were two kinds of pointers. There were near pointers, where the address was somewhere within the current 64 KB segment, and there were far, or long pointers, which could point beyond the current segment. The OS required callers to provide the latter to its APIs, so all the pointer-type names use LP. Nowadays, there's only one type of pointer; Microsoft keeps the same type names so that old code continues to compile.
The part of LPCWSTR that specifies wide characters is the W. But merely type-casting a char string literal to LPCWSTR is not sufficient to transform those characters into wide characters. Instead, what happens is the type-cast tells the compiler that what you wrote really is a pointer to a wide string, even though it really isn't. The compiler trusts you. Don't type-cast unless you really know better than the compiler what the real types are.
If you really need to pass a const char*, then you don't need to type-cast anything, and you don't need any L prefix. A plain old string literal is sufficient. (If you really want to cast to a Windows type, use LPCSTR — no W.) But it looks like what you really need to pass in is a const wchar_t*. As we learned above, you can get that with the L prefix on the string literal.
In a real program, you probably don't have a string literal. The user will provide a password, or you'll read a password from some other external source. Ideally, you would store that password in a std::wstring, which is like std::string but for wchar_t instead of char. The c_str() method of that type returns a const wchar_t*. If you don't have a wstring, a plain array of wchar_t might be sufficient.
But if you're storing the password in a std::string, then you'll need to convert it into wide characters some other way. To do a conversion, you need to know what code page the std::string characters use. The "current ANSI code page" is usually a safe bet; it's represented by the constant CP_ACP. You'll use that when calling MultiByteToWideString to have the OS convert from the password's code page into Unicode.
int required_size = MultiByteToWideChar(CP_ACP, 0, vs[0].c_str(), vs[0].size(), NULL, 0);
if (required_size == 0)
ERROR;
// We'll be storing the Unicode password in this vector. Reserve at
// least enough space for all the characters plus a null character
// at the end.
std::vector<wchar_t> wv(required_size);
int result = MultiByteToWideChar(CP_ACP, 0, vs[0].c_str(), vs[0].size(), &wv[0], required_size);
if (result != required_size - 1)
ERROR;
Now, when you need a wchar_t*, just use a pointer to the first element of that vector: &wv[0]. If you need it in a wstring, you can construct it from the vector in a few ways:
// The vector is null-terminated, so use "const wchar_t*" constructor
std::wstring ws1 = &wv[0];
// Use iterator constructor. The vector is null-terminated, so omit
// the final character from the iterator range.
std::wstring ws2(wv.begin(), wv.end() - 1);
// Use pointer/length constructor.
std::wstring ws3(&wv[0], wv.size() - 1);
You have two problems.
The first is the practical problem - how to do this. You are confusing wide and narrow strings and casting from one to the other. A string with an L prefix is a wide string, where each character is two bytes (a wchar_t). A string without the L is a single byte (a char). You cannot cast from one to the other using the C-style cast (LPCWSTR) "ab" because you have an array of chars, and are casting it to a pointer to wide chars. It is simply changing the pointer type, not the underlying data.
To convert from a narrow string to a wide string, you would normally use MultiByteToWideChar. You don't mention what code page your narrow strings are in; you would probably pass in CP_ACP for the first parameter. However, since you are converting between a string and a wstring, you might be interested in other ways to convert (one, two). This will give you a wstring with your characters, not a string, and a wstring's .c_str() method returns a pointer to wchar_ts.
The second is the following misunderstanding:
I need to pass const char* as last two parameters for this function call. How can I do that?
No you don't. You need to pass a wide string, which you got above. Your approach to this (casting the pointer) indicates you probably don't know about different string types and character encodings, and this is something every software developer should know. So, on the assumption you're interested, hopefully you'll find the following references handy:
The Absolute Minimum Every Software Developer Absolutely, Positively Must Know About Unicode and Character Sets (No Excuses!)
The Unicode FAQ (covers things like 'What is Unicode?')
MSDN introduction to wide characters.
I'd recommend you investigate recompiling your application with UNICODE and using wide strings. Many APIs are defined in both narrow and wide versions, and normally this would mean you access the narrow version by default (you can access either the ANSI (narrow) or Wide versions of these APIs by directly calling the A or W version - they have A or W appended to their name, such as CreateWindowW - see the bottom of that page for the two names. You normally don't need to worry about this.) As far as I can tell, this API is always available as-is regardless of UNICODE, it's just it's only prototyped as wide.
C style casts as you've used here are a very blunt instrument. They assume you know exactly what you're doing.
You'll need to convert your ASCII or multi-byte strings into Unicode strings for the API. There might be a NetUserChangePasswordA function that takes the char * types you're trying to pass, try that first.
LPWSTR is defined wchar_t* (whcar_T is 2 byte char-type) which are interpreted differently then normal 1 byte chars.
LPCWSTR means you need to pass a wchar_t*.
if you change vs to std::vector<std::wstring> that will give you a wide char when you pass vs[0].c_str()
if you look at the example at http://msdn.microsoft.com/en-us/library/aa370650(v=vs.85).aspx you can see that they define UNICODE which is why they use the wchar_t.