I want to overload the + operator for my mainwindow class(or any other class) for Qstring.
this is what I have done so far :
void operator+(QString a,QString b)
{
qDebug()<<"works";
}
but the thing is, the QString + operator is already overloaded(to concatenate, I guess).
so, if I use the above code, it results in ambiguity(both the signatures are same).
how do I override the actual function to my own function without making a new class to hold QString?
Try QString operator+(QString a,QString b). The return type of the overloaded function is QString, not void. The compiler cannot differ between two overloads only differ by return type.
Related
Documentation: http://doc.qt.io/qt-5/qflags.html#operator-Int
The question. I want to know what flags are set withouth testing one by one so I want the int number. Can anyone provide an example of how to use that operator in one of the many qt methods that rerturn a QFlags?
By referring to QFlags.h source code (https://android.googlesource.com/platform/prebuilts/android-emulator-build/qt/+/master/common/include/QtCore/qflags.h)
This is the definition in QFlags for "Int" operator.
Q_DECL_CONSTEXPR inline operator Int() const Q_DECL_NOTHROW { return i; }
And the "i" in return statement is declared as
Int i;
And the "Int" is declared as
typedef int Int
Notice the below two constructors of QFlags. The first constructor takes Enum as parameter and the second constructor takes QFlag as parameter.
Q_DECL_CONSTEXPR inline QFlags(Enum f) Q_DECL_NOTHROW : i(Int(f)) {}
Q_DECL_CONSTEXPR inline QFlags(QFlag f) Q_DECL_NOTHROW : i(f) {}
After noticing the above constructors, if Enum is passed to the constructor, the Enum can be a signed one or unsigned one. QFlags internally type casts it to int using Int.
Consider below example now.
//Qt::CursorShape is an Enum
Qt::CursorShape shape = Qt::ArrowCursor;
//Create QFlags object by passing "ENUM" as parameter
QFlags<Qt::CursorShape> qF(shape);
//Create QFlags object by just passing FLAG as a parameter
QFlags<Qt::CursorShape> q(Qt::ArrowCursor);
Now the situation where "Int" operator is called: In the below piece of code the first statement invokes Int operator and not in the second statement.
//Now try getting the values.
int test = qF; //In this case the "Int" operator is called.
int test1 = q;
I am making a simple scheduler that executes functions contained in a FIFO queue.
Those functions have a same return type int, but have different number of int arguments.
I tried to implement it this way, but it does not seem to work. The compiler forbids conversion between int(*)() , int(*)(int), int(*)(int, int), or to any of those sort. (Arduino Sketch compiler)
Is there a way to solve this problem, or could you recommend a better way around? Thanks!
My code:
typedef int (*fnptr)(); // Tried this!
int foo(int var) {
return 0;
}
int main() {
fnptr fp = &foo; // error: invalid conversion from
// 'int (*)(int)' to 'int (*)()'
// [-fpermissive]
return 0;
}
You can cast:
fnptr fp = reinterpret_cast<fnptr>(foo);
The ()s are the "function call operator", adding them makes no sense at all in this situation, it changes the expression from "take the address of this function" to "take the address of this function's return value".
Note that aboev I don't even include the &, this is because the name of a function acts pretty much like a function pointer so it's already an address.
I get an illegal indirection error at generateCSVHeader(*file4);.
function declaration:
void generateCSVHeader(QFile * file);
function use:
str="MyData.csv";
QFile file4(str);
generateCSVHeader(*file4);
when I drop the dereference designator, it gives me a cannot convert QFIle to QFile * error.
You should pass pointer to your QFile object (which is an address) instead of passing the object itself (dereferencing is irrelevant, because it is used only with pointers, not objects). To get an address of your object, you should use the & operator. So you have to invoke your function like this:
generateCSVHeader(&file4)
Also, you may consider using references instead of pointers.
I want to add a new line in this. This is my sample code:
ui->button->setText(" Tips " + "\n" + TipsCount );
This is the error it shows:
invalid operands of types 'const char [7]' and 'const char [2]' to binary 'operator+'
But when I add to label it gets appended!
ui->label->setText(name + "\n" + City );
Can someone please help me?
This is a very common problem in C++ (in general, not just QT).
Thanks to the magic of operator overloading, name + "\n" gets turned into a method call (couldn't say which one since you don't list the type). In other words, because one of the two things is an object with + overloaded it works.
However when you try to do "abc" + "de", it blows up. The reason is because the compiler attempts to add two arrays together. It doesn't understand that you mean concatenation, and tries to treat it as an arithmetic operation.
To correct this, wrap your string literals in the appropriate string object type (std::string or QString most likely).
Here is a little case study:
QString h = "Hello"; // works
QString w = "World"; // works too, of course
QString a = h + "World"; // works
QString b = "Hello" + w; // also works
QString c = "Hello" + "World"; // does not work
String literals in C++ (text in quotes) are not objects and don't have methods...just like numeric values aren't objects. To make a string start acting "object-like" it has to get wrapped up into an object. QString is one of those wrapping objects, as is the std::string in C++.
Yet the behavior you see in a and b show we're somehow able to add a string literal to an object. That comes from the fact that Qt has defined global operator overloads for both the case where the left operand is a QString with the right a const char*:
http://doc.qt.nokia.com/latest/qstring.html#operator-2b-24
...as well as the other case where the left is a const char* and the right is a QString:
http://doc.qt.nokia.com/latest/qstring.html#operator-2b-27
If those did not exist then you would have had to write:
QString a = h + QString("World");
QString b = QString("Hello") + w;
You could still do that if you want. In that case what you'll cause to run will be the addition overload for both operands as QString:
http://doc.qt.nokia.com/latest/qstring.html#operator-2b-24
But if even that didn't exist, you'd have to call a member function. For instance, append():
http://doc.qt.nokia.com/latest/qstring.html#append
In fact, you might notice that there's no overload for appending an integer to a string. (There's one for a char, however.) So if your TipsCount is an integer, you'll have to find some way of turning it into a QString. The static number() methods are one way.
http://doc.qt.nokia.com/latest/qstring.html#number
So you might find you need:
ui->button->setText(QString(" Tips ") + "\n" + QString::number(TipsCount));
like: vector<void *(*func)(void *)>...
You can declare a vector of pointers to functions taking a single void * argument and returning void * like this:
#include <vector>
std::vector<void *(*)(void *)> v;
If you want to store pointers to functions with varying prototypes, it becomes more difficult/dangerous. Then you must cast the functions to the right type when adding them to the vector and cast them back to the original prototype when calling. Just an example how ugly this gets:
#include <vector>
int mult(int a) { return 2*a; }
int main()
{
int b;
std::vector<void *(*)(void *)> v;
v.push_back((void *(*)(void *))mult);
b = ((int (*)(int)) v[0])(2); // The value of b is 4.
return 0;
}
You can use typedef's to partially hide the function casting syntax, but there is still the danger of calling a function as the wrong type, leading to crashes or other undefined behaviour. So don't do this.
// shorter
std::vector<int (*)(int)> v;
v.push_back(mult);
b = v[0](2); // The value of b is 4.
Storing a function in vector might be a difficult task as illustrated above. In that case if u want to dynamically use a function u can also store a function in pointer which is much easier. Main advantage of this is u can store any type of function either it is a normal function or a paramatrized one(having some input as parametrs). Complete process is described in the link given below with examples...just have a look...!!!
how can we store Function in pointer