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Pattern to avoid nested try catch blocks?
(16 answers)
Closed 6 years ago.
I have a yaml-cpp which always converts into a std::string, and sometimes also into something else. For example, if the string actually is "3.14", it would also convert into double. I'd first like to try int, then double, then bool, and if that doesn't work, convert to a std::string. Alright, so let's nest those try-catches:
try {
const int a = node.as<int>();
std::cout << "int!" << a << std::endl;
} catch (YAML::BadConversion) {
try {
const double a = node.as<double>();
std::cout << "double!" << a << std::endl;
} catch (YAML::BadConversion) {
try {
const bool a = node.as<bool>();
std::cout << "bool!" << a << std::endl;
} catch (YAML::BadConversion) {
const std::string a = node.as<std::string>();
std::cout << "string!" << a << std::endl;
}
}
}
Hm, the deeper and deeper nesting tells me that this isn't the best way to write that code.
Any suggestions on how to improve the design here? Flat nesting would certainly be advised.
You may put it in a function like:
template<typename N, typename T>
bool tryParseNode(N& node, T& val) {
try {
val = node.as<T>();
return true;
} catch (YAML::BadConversion) {
return false;
}
}
then:
int a;
double d;
bool b;
std::string s;
if (tryParseNode(node, a) {
std::cout << "int!" << a << std::endl;
}
else if (tryParseNode(node, d) {
std::cout << "double!" << d << std::endl;
}
else if (tryParseNode(node, b) {
std::cout << "bool!" << b << std::endl;
}
else if (tryParseNode(node, s) {
std::cout << "string!" << s << std::endl;
}
Try the other way round:
Convert into to a string, then try bool, etc.
Everything within a single try-catch and ignore the exception.
Using exceptions for normal control flow is considered bad practice. In this case, the as method uses the `YAML::convert::decode' method to attempt to convert the node into the requested type returning a false if it fails instead of throwing an exception.
int anInt;
double aDouble;
bool aBool;
if (YAML::convert <int>::decode (node, anInt))
std::cout << "int!" << anInt << std::endl;
else
if (YAML::convert <double>::decode (node, aDouble))
std::cout << "double!" << aDouble << std::endl;
else
if (YAML::convert <bool>::decode (node, aBool))
std::cout << "double!" << aBool << std::endl;
else
std::cout << "string!" << node.as <std::string> () << std::endl;
which could be further simplified to
template <typename value_type>
std::optional <value_type> decode (YAML::Node const & Node)
{
value_type Value;
if (YAML::convert <value_type>::decode (node, Value))
return { Value };
else
return {};
}
if (auto anInt = decode <int> (node))
std::cout << "int!" << *anInt << std::endl;
else
if (auto aDouble = decode <double> (node))
std::cout << "double!" << *aDouble << std::endl;
else
if (auto aBool = decode <bool> (node))
std::cout << "double!" << *aBool << std::endl;
else
std::cout << "string!" << node.as <std::string> () << std::endl;
Related
I have multiple functions that have long and similar implementations. The only difference is they call different calls, which is basically based on the function name like below.
// A.h
class A : public ParentA
{
public:
explicit A(B& b);
~A() override = default;
// accessors
C does_same_thing1() const;
C does_same_thing2() const;
C does_same_thing3() const;
// ...
}
// A.cpp
C A::does_same_thing1() const
{
...
return xyz.values().thing1();
}
C A::does_same_thing2() const
{
...
return xyz.values().thing2();
}
C A::does_same_thing3() const
{
...
return xyz.values().thing3();
}
I wonder if there's a way to dynamically fill out the functions that are almost the same except the accessors they call (thing1(), thing2(), and thing3(), and this actually happens more than once, not just on the return line) based on their function names. Would this be possible in C++?
Thanks!
You can write one function template and let the caller choose what is to be returned:
template <typename F>
auto foo(F f) {
...
return f(xyz.values());
}
Details depend on details you left out from the question. For example, is the type of xyz.values() available to the caller? Also, it is up to you to let the caller pick f or write wrappers:
auto does_same_thing1() {
foo([](auto& x) { return x.thing1(); }
}
// ... and same for the others
Some options are:
Using an abstract and overriding the parts you require.
Using lambas and passing in the functions your require.
Using template functions is kind of a mix of the two above, but I'll let someone else explain that one.
Create your base class
class Base
{
protected:
int value;
public:
virtual void differentFunction(int mathThing) = 0;
void longFunction()
{
value = 0;
std::cout << "I do a lot of steps" << std::endl;
std::cout << "Step 1" << std::endl;
value++;
std::cout << "Step 2" << std::endl;
value++;
std::cout << "Step 3" << std::endl;
value++;
std::cout << "Step 4" << std::endl;
value++;
std::cout << "Step 5" << std::endl;
value++;
std::cout << "Step 6" << std::endl;
//And finally I do a unique thing
differentFunction(3);
std::cout << "Resulting value: " << value << std::endl;
}
void longLamdaFunction(std::function<void(int& value, int mathThing)> mathFunction)
{
value = 0;
std::cout << "I do a lot of steps" << std::endl;
std::cout << "Step 1" << std::endl;
value++;
std::cout << "Step 2" << std::endl;
value++;
std::cout << "Step 3" << std::endl;
value++;
std::cout << "Step 4" << std::endl;
value++;
std::cout << "Step 5" << std::endl;
value++;
std::cout << "Step 6" << std::endl;
//And finally I do a unique thing
mathFunction(value, 3);
std::cout << "Resulting value: " << value << std::endl;
}
};
Create an overriding class
class Derived1 : public Base
{
public:
void differentFunction(int mathThing) override
{
std::cout << "I multiply the value" << std::endl;
value *= mathThing;
}
};
Create a different overriding class
class Derived2 : public Base
{
public:
void differentFunction(int mathThing) override
{
std::cout << "I divide the value" << std::endl;
value /= mathThing;
}
};
Example on use, you can see the Lambda example here too
int main()
{
Derived1 d1;
Derived2 d2;
std::cout << "\nUsing multiple interface\n";
d1.longFunction();
std::cout << "\nUsing divide interface\n";
d2.longFunction();
std::cout << "\nUsing add lamda\n";
//I now add them
auto addFunction = [](int& x, int y) -> void { x += y; };
d1.longLamdaFunction(addFunction);
std::cout << "\nUsing subtract lamda\n";
//I now subtract them
auto subtractFunction = [](int& x, int y) -> void { x -= y; };
d1.longLamdaFunction(subtractFunction);
}
I have a boost::any object, and I would like to check its type.
typedef boost::any Value;
Value a = 12;
if(a.type() == typeid(int)) {
std::cout << boost::any_cast<int>(a) << std::endl;
}
This is easy enough when the type is defined, however how would I achieve a the same result when the type is not defined (i.e. because its value has not been set yet).
Value b;
if(b is undefined) {
std::cout << "b is not defined" << std::endl;
}
boost::any::empty will return true if there is no value.
Demo
#include "boost/any.hpp"
#include <iostream>
int main()
{
boost::any a = 42;
if (!a.empty())
std::cout << "a has a value\n";
boost::any b;
if (b.empty())
std::cout << "b does not have a value\n";
}
Alternatively, you can use boost::any::type like you did in the first example and, if there's no value, it will return typeid(void):
Demo 2
boost::any a = 42;
std::cout << std::boolalpha << (a.type() == typeid(int)) << std::endl; // true
boost::any b;
std::cout << std::boolalpha << (b.type() == typeid(void)) << std::endl; // true
I want to make a function that can wrap any lambda to log start/end calls on it.
The code below works except for:
any lambda that has captures
any lambda that returns void (although this can easily be fixed by writing a second function)
#include <iostream>
#include <functional>
template <class T, class... Inputs>
auto logLambda(T lambda) {
return [&lambda](Inputs...inputs) {
std::cout << "STARTING " << std::endl;
auto result = lambda(inputs...);
std::cout << "END " << std::endl;
return result;
};
}
int main() {
int a = 1;
int b = 2;
// works
auto simple = []() -> int {
std::cout << "Hello" << std::endl; return 1;
};
logLambda(simple)();
// works so long as explicit type is declared
auto with_args = [](int a, int b) -> int {
std::cout << "A: " << a << " B: " << b << std::endl;
return 1;
};
logLambda<int(int, int), int, int>(with_args)(a, b);
// Does not work
// error: no matching function for call to ‘logLambda<int(int), int>(main()::<lambda(int)>&)’
auto with_captures = [&a](int b) -> int {
std::cout << "A: " << a << " B: " << b << std::endl;
return 1;
};
logLambda<int(int), int>(with_captures)(b);
}
Is there any way to do this? Macros are also acceptable
Use Raii to handle both void and non-void return type,
and capture functor by value to avoid dangling reference,
and use generic lambda to avoid to have to specify argument your self
It results something like:
template <class F>
auto logLambda(F f) {
return [f](auto... args) -> decltype(f(args...)) {
struct RAII {
RAII() { std::cout << "STARTING " << std::endl; }
~RAII() { std::cout << "END " << std::endl; }
} raii;
return f(args...);
};
}
Call look like:
const char* hello = "Hello";
logLambda([=](const char* s){ std::cout << hello << " " << s << std::endl; })("world");
Demo
That code has undefined behavior.
auto logLambda(T lambda) {
return [&lambda]
You are capturing local parameter by reference.
Suppose I have a variable, double x, as a result of some calculations, which can have any value, including zero, and I need it passed to std::cout. How can I avoid printing x if its value is zero?
As an example, this will print 1+<value_of_x> if x, else just 1:
std::cout << (x ? "1+" : "1") << x << '\n';
Is there a way to make the same but for x? Something like the following nonsense:
std::cout << (x ? ("1+" << x) : "1") << '\n';
I should probably add that I am not advanced in C++.
You could say
std::cout << (x ? "1+" + std::to_string(x) : "1") << '\n';
but
if (x)
std::cout << "1+" << x << '\n';
else
std::cout << "1" << '\n';
is perhaps more readable.
(I consider this largely a matter of personal preference.)
If x is 0, don't print it:
if (x != 0)
std::cout << x << '\n';
Any further variations should be self-evident.
Using an if statement would be a simple and readable approach:
if (x)
std::cout << "1+" << x;
else
std::cout << "1";
std::cout << '\n';
Or even:
std::cout << "1";
if (x) std::cout << "+" << x;
std::cout << '\n';
But, if you really want to print out the value inline, you can define a custom operator<< to format the value however you want:
struct to_coefficient_str
{
double m_value;
to_coefficient_str(double value) : m_value(value) {}
void print(std::ostream &out) const
{
out << "1";
if (m_value)
out << "+" << m_value;
}
};
std::ostream& operator<<(std::ostream &out, const to_coefficient_str &ce)
{
ce.print(out);
return out;
}
Then you can use it like this:
std::cout << to_coefficient_str(x) << '\n';
In this program, I am using template class, I have a header file and this is my main file. I am having trouble displaying the (".....") IndexOutOfBounds and displaying it on the screen.
#include "XArray.h"
#include <iomanip>
#include <string>
using namespace std;
template<class T>
void afriend ( XArray<T> );
int main()
{
XArray<double> myAD(18);
myAD.randGen(15, 100);
cout << myAD.getType() << endl;
cout << setprecision(1) << fixed << "\n\n Unsorted: " << myAD;
myAD.sort();
cout << "\n Now Sorted: " << myAD;
cout << "\n\n";
**try
{
cout << "A[-5] = " << setw(6) << myAD[-5] << endl;
}
catch(XArray<double>::IndexOutOfBound e)
{
e.print();
}
try
{
cout << "A[8] = " << setw(6) << myAD[8] << endl;
}
catch(XArray<double>::IndexOutOfBound e)
{
e.print();
}**
cout << "\n\n" << setprecision(2) << fixed;
cout << "Size = " << setw(6) << myAD.getSize() << endl;
cout << "Mean = " << setw(6) << myAD.mean() << endl;
cout << "Median = " << setw(6) << myAD.median() << endl;
cout << "STD = " << setw(6) << myAD.std() << endl;
cout << "Min # = " << setw(6) << myAD.min() << endl;
cout << "Max # = " << setw(6) << myAD.max() << endl;
return 0;
}
There is the Array.h file posted as a dropbox link
Array.h
The code for operator[] in Array.h is:
template <class T>
T XArray<T>::operator[] (int idx)
{
if( (idx = 0) && (idx < size) )
{
return Array[idx];
}
else
{
throw IndexOutOfBound();
return numeric_limits<T>::epsilon();
}
}
Although the question is somewhat obscure, give a try to these suggestions.
Firstly, it can happen that XArray<>::IndexOutOfBounds have no proper copy ctor. You can try catching by const reference to workaround that:
try
{
...
}
catch(const XArray<double>::IndexOutOfBound& e)
{
e.print();
}
Index operator in standard library containers does not check for bounds, there is a special getter that does the check called at(). If the XArray class is designed with standard library in mind, it could behave similarly.
However to get more adequate response you need to be more specific describing the trouble you are having.
I'm still wondering what exact question is.
However, I'm understanding the question is that how I can use 'catch' by using 'IndexOutOfBound'.
#include <exception>
#include <iostream>
using namespace std;
template <typename T>
class Array
{
private:
int m_nLength;
T *m_ptData;
public:
...
...
T& operator[](int nIndex)
{
//assert(nIndex >= 0 && nIndex < m_nLength);
if(nIndex < 0 || nIndex > m_nLength)
{
throw myex;
}
else
{
return m_ptData[nIndex];
}
}
//class definition for 'IndexOutOfBound'
class IndexOutOfBound: public exception
{
public:
virtual const char* print() const throw()
{
return "Exception occured 'Index Out Of Bound'";
}
}myex;
};
int main()
{
Array<double> arr(3);
try
{
arr[0] = 1;
//exception will occur here.
arr[4] = 2;
}
catch(Array<double>::IndexOutOfBound &e)
{
cout << e.print() << '\n';
}
return 0;
}
Here is no 'XArray.h', so I've written a sample array class for example.
The problem is in the operator[] function. The code idx = 0 sets idx to 0. So all of your calls to operator[] will return the first element, and therefore there is no out-of-bounds error unless the array is empty.
You probably meant to write if ( idx >= 0 && idx < size ).
BTW the throw aborts the function, it makes no sense to return after throw.