I wonder if there is a way of constructing a temporary map to be passed so that the following implementation is possible:
void func(map<string,int> & input) {
cout << input["m1"] << endl;
cout << input["m2"] << endl;
}
func ( map<string,int>{{"m1",1},{"m2",2}} ; // causing error when compiled
CodePudding user response:
The problem is that you're trying to bind an rvalue expression to an lvalue reference to non-const std::map
.
To solve this you can add a low-level const
in the parameter and use std::map::find
as shown below:
void func(const std::map<string, int>& input) {
auto it1 = input.find("m1");
if(it1!=input.end())
{
cout << it1->second << endl;
}
else
{
std::cout<<"element cannot be found"<<std::endl;
}
//do the same for other key "m2"
}
int main()
{
func({{"m1", 1}, {"m2", 2}});
return 0;
}
Note if you want to just print all the elements of the map you can use structure binding with C 17 as shown below:
void func(const std::map<string, int>& input) {
for(const auto&[key, value]: input)
{
std::cout<<key<<" --> "<<value<<std::endl;
}
}
int main()
{
func({{"m1", 1}, {"m2", 2}});
return 0;
}
Note structure bindings are available from C 17, so if you're using C 11 or C 14 you can use the range-based for loop :
void func(const std::map<string, int>& input) {
for(const auto&element: input)
{
std::cout<<element.first<<" --> "<<element.second<<std::endl;
}
}
CodePudding user response:
Generally, yes, that's easily possible, you just need to make your func
accept temporary (rvalue) values.
You don't need to modify the map, so
void func(const std::map<string, int>& input) {
cout << input.at("m1") << endl;
cout << input.at("m2") << endl;
}
func({{"m1", 1}, {"m2", 2}});
should do.
Note that, as comfortable as they are, []
of map modifies the map (if there key is not in there before, it's inserted and a value is value-initialized). So, you can't use it on const references.
The way around that is passing in the map as special rval ref, &&
:
void func(map<string,int>&& input) {
cout << input["m1"] << endl;
cout << input["m2"] << endl;
}
However, that is not always the right thing to do, and I do prefer the const & approach when the function really has no need to modify the parameter, as it actually makes compile-time guarantees that this won't happen, even on complex objects.