如何从std :: map返回值 <int, std::string> 使用uint64_t键类型？

Question

Novice question, but I searched for this and couldn't find something clearly solving my issue - apologies if this is obvious.

I have defined a map which looks like this:

map<int, string> testmap = {
    { 0, "a" },
    { 1, "b" },
    { 2, "c" }
}

However, I need to retrieve a value from testmap using a uint64_t value provided by another function.

When I do testmap[my_uint64_t_value] it returns an empty string, so I think this is because it's adding my_uint64_t_value as a key and setting the value to NULL .

This is the same if I set the map type to <uint64_t, string> , at least the way I'm currently defining my keys.

However, is there a way that I either:

1. convert the uint64_t value to a regular int
2. define the map as <uint64_t, string> , and be able to define my keys as the 'correct' type?

It seems like int type conversion isn't that common, is this something that should be avoided?

Answer 1

The reason why you get an empty string is std::map::operator[] returns a reference to the value if and only if it exists , otherwise it performs an insertion . I suspect you have the latter case.

You need to use std::map::find for search.

uint64_t keyToFind =  1;
if (auto iter = testmap.find(keyToFind); iter != testmap.cend()) 
{
   // do something
   std::cout << "Found!\n";
}
else { std::cout << "Not Found!\n"; }

Like @Rene mentioned in the comments, casting from uint64_t to int can cause overflow. Therefore, making the key to larger type(as per requirement) would be a good idea.

std::map<uint64_t, std::string> testmap;

Answer 2

As said in another answer, the [] operator of the map class will perform an insertion with a default-constructed value if the key is not present in the map.

You can first use the count method to determine if the key is present in the map before accessing it.

if(testmap.count(keyToFind))
    return testmap[keyToFind];
else
    report_key_not_found();

An alternative solution is to use the at method to access the value. It will throw an std::out_of_range exception if the key is not present instead of inserting a new key.