C++ unordered map with uint8_t

Question

I'm having a strange problem using an unordered_map with key type uint64_t and value type uint8_t. (I have a few potentially large integers that I want to count with the uint8_t values.) Here's some code:

std::unordered_map<uint64_t,uint8_t> mymap;
mymap.emplace(2,1);
mymap.emplace(20,3);
mymap.emplace(200,34);

for (auto it = mymap.begin(); it != mymap.end(); ++it) {
    std::cout << it->first << ":" << it->second << std::endl;
    printf("%lld : %d\n", it->first, it->second);
}

It appears that the keys and values are saved perfectly fine in the unordered_map. But, the std::cout << it->second << std::endl; code is garbling the output of the uint8_t values. That is, cout (the first line in the for loop) garbles the output, while the old school cstdio printf (the second line in the for loop) prints out the values just fine. Weird. Any idea as to what is happening here?

If I change the unordered_map to <uint64_t,uint16_t> or <uint64_t,int>, it works perfectly fine. I've tried various terminal and pseudo terminal environments plus text files, all with garbled or missing outputs for the uint8_ts. I'm stumped.

Here is the output:

20:
20 : 3
200:"
200 : 34
2:
2 : 1

Care to add a garbled pair or two? – Deduplicator Apr 14 '14 at 18:55 — Deduplicator, Apr 14 '14 at 18:55
hint: what is `uint8_t` a typedef of? – Mooing Duck Apr 14 '14 at 18:56 — Mooing Duck, Apr 14 '14 at 18:56

score 2 · Answer 1 · answered Apr 14 '14 at 19:03

Note: There are no formatted input functions taking a signed or unsigned char in the iostream library. Hence the the uint8_t is converted and interpreted as character.

std::cout << it->first << ":" << ùnsigned(it->second) << std::endl; will fix it.

Yifu Wang · Answer 2 · 2014-04-14T19:03:33.383

0

You could do a cast:

std::cout << int(it->second);

edited Apr 14 '14 at 19:03

answered Apr 14 '14 at 18:57

Yifu Wang

313
2
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Vlad from Moscow · Answer 3 · 2014-04-14T19:12:51.223

It seems that type uint8_t is an alias for type unsigned char. If you will cast the value to unsigned int you will get the same result as with printf.

for (auto it = mymap.begin(); it != mymap.end(); ++it) {
    std::cout << it->first << ":" << ( unsigned int )it->second << std::endl;
    printf("%lld : %d\n", it->first, it->second);
}

As for your original code there is used the following operator for uint8_t

template<class traits>
basic_ostream<char,traits>& operator<<(basic_ostream<char,traits>&,
unsigned char);

C++ unordered map with uint8_t

3 Answers3