I guess the title is quite confusing, I'll explain my case with some code.
template<uint16_t Len>
void add(const int8_t (&i_array)[Len])
{
// Do something
}
class Test
{
public:
int8_t* GetName()
{
return name;
}
private:
int8_t name[10] = "myname";
}
int main()
{
Test mytest;
add(mytest.GetName()); // Compilation error
}
This code does not compile. The following error is generated :
"Error#304 : no instance of function template add matches the argument list"
It seems that the compilator is not able to determine that GetName() return an array of size 10. Is that right ?
How could I call "add" with a pointer on an array ?
Thanks,
Nicolas
Test::GetName returns a pointer, not an array. You cannot bind its result to a function that expects an array reference. However, you could change the signature of GetName to make it return the array (by reference, of course):
int8_t (&GetName())[10] { return name; }
Alternatively you could use a cast, but that would defeat the purpose of the type system.
Related
Currently learning some c++ and unsure why this is giving me "expression must have pointer type".
Mapp.hpp
class RouteMap
{
public:
RouteMap();
string getCurrent_();
void StoreCity(string b);
private:
std::vector<string>* cities();
string current_;
};
mapp.cpp
RouteMap::RouteMap(){}
string RouteMap::getCurrent_()
{
return current_;
}
void RouteMap::StoreCity(string b)
{
cities->push_back(b); //Error
}
std::vector<string> RouteMap::cities()
{
return std::vector<string>();
}
I am attempting to have a vector Cities as a private member so that when I run the member function StoreCity(string x), it would push_back the specific string into Cities.
I'm going to take a stab and say that the problem is cities() is a function and requires parenthesis:
cities()->push_back(b);
Edit Just found the implementation of cities() (silly me). You have another problem, and that is that your declaration and implementation don't match.
// declaration
std::vector<string>* cities();
// implementation. Notice the lack of a pointer type return
std::vector<string> RouteMap::cities()
{
return std::vector<string>();
}
It's also weird that you're returning a new vector each time. You probably want a member variable:
class RouteMap
{
//...
private:
std::vector<string> my_cities;
//...
};
and then return the member variable from there:
std::vector<string>* RouteMap::cities()
{
return &my_cities;
}
Edit2: It has come to my attention that you probably, while you could fix these things like this and get it working, the truth is that you probably don't mean for cities() to be a function at all. You probably mean for it to be a member variable instead:
class RouteMap
{
//...
private:
std::vector<string> cities;
//...
};
This requires no implementation, (aka RouteMap::cities(){}), and you can just use it inside any member function (because it's a private member) like current_.
I have two functions with a little different functionality, so I can't make them as template functions.
int func64(__int64 a) {
return (int) a/2;
}
int func32(int a) {
return a--;
}
Depending on variable b64, I would like to call func64 or func32. I don't want check if b64 is true many times in my code, so I use pointers to functions.
void do_func(bool b64) {
typedef int (*pfunc32)(int);
typedef int (*pfunc64)(__int64);
pfunc32 call_func;
if (b64)
call_func = func64; //error C2440: '=' : cannot convert from 'int (__cdecl *)(__int64)' to 'pfunc32'
else
call_func = func32;
//...
call_func(6);
}
How can I avoid this error and cast call_func to pfunc32 or pfunc64?
The language requires all functions called through the same function pointer to have the same prototype.
Depending on what you want to achieve, you could use the pointer/cast aproach already mentioned (which satisfies this requirement at the loss of type safety) or pass a union instead:
union u32_64
{
__int64 i64;
int i32;
};
int func64(union u32_64 a) {
return (int) a.i64/2;
}
int func32(union u32_64 a) {
return --a.i32;
}
void do_func(bool b64) {
typedef int (*pfunc)(union u32_64);
pfunc call_func;
if (b64)
call_func = func64;
else
call_func = func32;
//...
union u32_64 u = { .i64 = 6 };
call_func(u);
}
Pass a void pointer and cast it in the function body.
Of course this means less compiler control if you use the wrong type; if you call func64 and pass an int to it the program will compile and produce wrong results without giving you any tip of what is going wrong.
int func64(void *a) {
__int64 b = *((__int64*) a);
return (int) b/2;
}
int func32(void *a) {
int b = *((int *) a)
return b-1;
}
I need to call func32() or func64() depending on flag b64
So do that:
void do_func(bool b64) {
if (b64)
func64(6);
else
func32(6);
}
Well, first of all, please note that function func32 is returning the input argument as is.
This is because with return a--, you are returning the value of a before decrementing it.
Perhaps you meant to return a-1 instead?
In any case, you can simply declare this function as int func32(__int64 a).
This way, it will have the same prototype as function func64, but will work exactly as before.
BTW, calling a function through a pointer might be more "expensive" than a simple branch operation, so depending on your application, you might be better off with a simple if/else conditional statement...
Make a wrapper for func64:
int func64_as_32(int a) {
return func64(a);
}
Now you can assign either func32 or func64_as_32 to call_func since they have the same signature. The value you pass in, 6, has type int so passing it to func64_as_32 has the same effect as passing it directly to func64.
If you have call sites where you pass in a value of type __int64 then you'd do it the other way around, wrap func32.
As bool in C++ converts to int ( true => 1, false => 0 ) you can use b64 as array index. So take SJuan76's advice, convert your functions prototype to int f(void*) and put them into array int (*array_fun[2])(void* x); . You can call these functions then like that :
int p = 6;
array_fun[b64](&p);
I need to find some way to mock an overload of a function return type in C++.
I know that there isn't a way to do that directly, but I'm hoping there's some out-of-the-box way around it.
We're creating an API for users to work under, and they'll be passing in a data string that retrieves a value based on the string information. Those values are different types. In essence, we would like to let them do:
int = RetrieveValue(dataString1);
double = RetrieveValue(dataString2);
// Obviously, since they don't know the type, they wouldn't use int =.... It would be:
AnotherFunction(RetrieveValue(dataString1)); // param of type int
AnotherFunction(RetrieveValue(dataString2)); // param of type double
But that doesn't work in C++ (obviously).
Right now, we're having it set up so that they call:
int = RetrieveValueInt(dataString1);
double = RetrieveValueDouble(dataString2);
However, we don't want them to need to know what the type of their data string is.
Unfortunately, we're not allowed to use external libraries, so no using Boost.
Are there any ways we can get around this?
Just to clarify, I understand that C++ can't natively do it. But there must be some way to get around it. For example, I thought about doing RetrieveValue(dataString1, GetType(dataString1)). That doesn't really fix anything, because GetType also can only have one return type. But I need something like that.
I understand that this question has been asked before, but in a different sense. I can't use any of the obvious answers. I need something completely out-of-the-box for it to be useful to me, which was not the case with any of the answers in the other question asked.
You've to start with this:
template<typename T>
T RetrieveValue(std::string key)
{
//get value and convert into T and return it
}
To support this function, you've to work a bit more, in order to convert the value into the type T. One easy way to convert value could be this:
template<typename T>
T RetrieveValue(std::string key)
{
//get value
std::string value = get_value(key, etc);
std::stringstream ss(value);
T convertedValue;
if ( ss >> convertedValue ) return convertedValue;
else throw std::runtime_error("conversion failed");
}
Note that you still have to call this function as:
int x = RetrieveValue<int>(key);
You could avoid mentioning int twice, if you could do this instead:
Value RetrieveValue(std::string key)
{
//get value
std::string value = get_value(key, etc);
return { value };
}
where Value is implemented as:
struct Value
{
std::string _value;
template<typename T>
operator T() const //implicitly convert into T
{
std::stringstream ss(_value);
T convertedValue;
if ( ss >> convertedValue ) return convertedValue;
else throw std::runtime_error("conversion failed");
}
}
Then you could write this:
int x = RetrieveValue(key1);
double y = RetrieveValue(key2);
which is which you want, right?
The only sane way to do this is to move the return value to the parameters.
void retrieve_value(std::string s, double& p);
void retrieve_value(std::string s, int& p);
<...>
double x;
retrieve_value(data_string1, x);
int y;
retrieve_value(data_string2, y);
Whether it is an overload or a specialization, you'll need the information to be in the function signature. You could pass the variable in as an unused 2nd argument:
int RetrieveValue(const std::string& s, const int&) {
return atoi(s.c_str());
}
double RetrieveValue(const std::string& s, const double&) {
return atof(s.c_str());
}
int i = RetrieveValue(dataString1, i);
double d = RetrieveValue(dataString2, d);
If you know your value can never be something like zero or negative, just return a struct holding int and double and zero out the one you don't need...
It's a cheap and dirty, but easy way...
struct MyStruct{
int myInt;
double myDouble;
};
MyStruct MyFunction(){
}
If the datastrings are compile-time constants (as said in answering my comment), you could use some template magic to do the job. An even simpler option is to not use strings at all but some data types which allow you then to overload on argument.
struct retrieve_int {} as_int;
struct retrieve_double {} as_double;
int RetrieveValue(retrieve_int) { return 3; }
double RetrieveValue(retrieve_double) { return 7.0; }
auto x = RetrieveValue(as_int); // x is int
auto y = RetrieveValue(as_double); // y is double
Unfortunately there is no way to overload the function return type see this answer
Overloading by return type
int a=itoa(retrieveValue(dataString));
double a=ftoa(retrieveValue(dataString));
both return a string.
As an alternative to the template solution, you can have the function return a reference or a pointer to a class, then create subclasses of that class to contain the different data types that you'd like to return. RetrieveValue would then return a reference to the appropriate subclass.
That would then let the user pass the returned object to other functions without knowing which subclass it belonged to.
The problem in this case would then become one of memory management -- choosing which function allocates the returned object and which function deletes it, and when, in such a way that we avoid memory leaks.
The answer is simple just declare the function returning void* type and in the definition return a reference to the variable of different types. For instance in the header (.h) declare
void* RetrieveValue(string dataString1);
And in the definition (.cpp) just write
void* RetrieveValue(string dataString1)
{
if(dataString1.size()<9)
{
static double value1=(double)dataString1.size();
return &value1;
}
else
{
static string value2=dataString1+"some string";
return &value2;
}
}
Then in the code calling RetrieveValue just cast to the right value
string str;
string str_value;
double dbl_value;
if(is_string)
{
str_value=*static_cast<*string>(RetrieveValue(str));
}
else
{
dbl_value=*static_cast<*double>(RetrieveValue(str));
}
Since you used an example that wasn't really what you wanted, you threw everyone off a bit.
The setup you really have (calling a function with the return value of this function whose return type is unknowable) will not work because function calls are resolved at compile time.
You are then restricted to a runtime solution. I recommend the visitor pattern, and you'll have to change your design substantially to allow for this change. There isn't really another way to do it that I can see.
This question is similar to what I'm trying to do Calling C++ member function pointer from a struct .
However my structure contains a member function pointer that is defined in a different class then the one the structure is defined and used in. Here is some example code of how my classes, structures and function pointers are laid out.
// Alpha.h:
class Alpha{
public:
void function1(char name[], int number);
void function2(char name[], int number);
void function3(char name[], int number);
typedef void (Alpha::*My_func_ptr)(char name[], int number);
static My_func_ptr functionTable[];
};
// Alpha.cpp:
#include "Alpha.h"
Alpha::My_func_ptr Alpha::functionTable[] = {
&Alpha::function1,
&Alpha::function2,
&Alpha::function3
};
void Alpha::function1(char name[], int number)
{
//some stuff
}
void Alpha::function2(char name[], int number)
{
//some stuff
}
void Alpha::function3(char name[], int number)
{
//some stuff
}
// Beta.h:
#include "Alpha.h"
typdef struct{
char bName[10];
Alpha::My_func_ptr fptr;
}ptr_structure;
class Beta{
public:
void betafunction();
Alpha alphaobject;
ptr_structure str_array[3];
};
// Beta.cpp:
#include "Beta.h"
void betafunction()
{
str_array[0].fptr = alphaobject.functionTable[0];
str_array[1].fptr = alphaobject.functionTable[1];
str_array[2].fptr = alphaobject.functionTable[2];
(str_array[0].fptr)("name", 1); //gives error expression must have
//(pointer-to-) function type
(this->*str_array[0].fptr)("name", 1);
//error pointer-to-member selection class types are incompatible "Beta" and "Alpha"
//sample function pointer call using function table from other class,
//this syntax compiles and runs without error.
(alphaobject.*Alpha::functionTable[0]("name", 1);
}
As you can see I can call the function pointer from an array, but can't seem to figure out how to call a function pointer from inside an array of structures.
When calling a through member function pointer, you need to have an instance of the object associated with that pointer:
(alphaobject.*(str_array[0].fptr))("name", 1)
^^^^^^^^^^^
I would think:
(object.*functionTable[0])(args, ...);
(objptr->*functionTable[0])(args, ....);
IIRC, the combination of object and the .* operator is like a big unary operator. So that has lower precedence to the [0] postfix. However, it also has lower prededence than the function call postfix operator (args, ...)
Analogy:
(*foo)(); /* classic C */
Of course the * operator is not required when calling a regular function. But if you do write it, you need the parens, because *foo() means something else.
You can go to one of two solutions, depending on how readable you want the code. The unreadable version (which might even be wrong, and I won't even try to compile):
void Beta::betafunction() {
Alpha a;
(a.*(strArray[0].fptr))("name",1);
}
But I would actually try to make things a bit simpler:
void Beta::betafunction() {
Alpha a;
Alpha::My_func_ptr mptr = strArray[0].fptr;
(a.*mptr)("name",1);
}
I believe the second to be much more readable, and the compiler can optimize away mptr pretty easily, so there is no point in trying to play guru with the syntax.
Why do I get this error in the code below?
class ST : public Instruction{
public:
ST (string _name, int _value):Instruction(_name,_value){}
void execute(int[]& anArr, int aVal){
//not implemented yet
cout << "im an st" <<endl;
anArr[value] = aVal;
}
virtual Instruction* Clone(){
return new ST(*this);
}
};
classes.h:81: error: ‘anArr’ was not declared in this scope
classes.h:81: error: ‘aVal’ was not declared in this scope
You have a problem with the type of the first parameter of your execute function. Read this up to know more about how to pass arrays around.
Because the type of anArr is invalid.
Also you may be interested in using a covariant return type on your clone method. I.e. it can return a pointer to ST instead of Instruction.
Try this out :
void execute(int anArr[] , int aVal)
since You cant use array of reference .
If execute() is supposed to be taking an array of integers, you should probably declare it like this:
void execute(int* anArr, int anArrLength, int aVal)
{
// ...
}
Note that there are several differences to your method:
anArr is passed in as a pointer to the start of the array. The client code can simply pass in the array variable name, as by definition this is equivalent to "a pointer to the start of the array".
anArrLength is passed in to indicate the length of the array. This is required to ensure that the execute() method doesn't access memory which is out of the bounds of the array (or what has been allocated for the array). Doing so could result in memory corruption.
You could improve the method signature above by adding a return value to indicate success or failure. This would allow the client code to detect if there have been any problems. For example:
// Returns true on success, false on failure
bool execute(int* anArr, int anArrLength, int aVal)
{
// Get "value" through whatever means necessary
// ...
if (value >= anArrLength)
{
// Out of bounds of array!
return false;
}
anArr[value] = aVal;
// Do whatever else you need to do
// ...
return true;
}