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edited Oct 3, 2013 at 11:32

You can utilize std::transform here (I'll assume C++11 for the lambda):

int main()
{
    std::vector<uint8_t> first_list = {0x01, 0x10, 0x32, 0x1A};
    std::vector<uint8_t> second; 
    std::transform(first_list.begin(), first_list.end(), std::back_inserter(second),
                   [](uint8_t u) { return ~u; });
    for(uint8_t it : second) {}

}

You also don't need to make a copy of the first vector to do this; better to simply use push_back (or back_inserter here, since we're working with iterators).

You can utilize std::transform here (I'll assume C++11 for the lambda):

int main()
{
    std::vector<uint8_t> first_list = {0x01, 0x10, 0x32, 0x1A};
    std::vector<uint8_t> second; 
    std::transform(first_list.begin(), first_list.end(), std::back_inserter(second),
                   [](uint8_t u) { return ~u; });
    for(uint8_t it : second) {

}

You also don't need to make a copy of the first vector to do this; better to simply use push_back (or back_inserter here, since we're working with iterators).

You can utilize std::transform here (I'll assume C++11 for the lambda):

int main()
{
    std::vector<uint8_t> first_list = {0x01, 0x10, 0x32, 0x1A};
    std::vector<uint8_t> second; 
    std::transform(first_list.begin(), first_list.end(), std::back_inserter(second),
                   [](uint8_t u) { return ~u; });
}

You also don't need to make a copy of the first vector to do this; better to simply use push_back (or back_inserter here, since we're working with iterators).

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created Oct 3, 2013 at 8:37

Yuushi

You can utilize std::transform here (I'll assume C++11 for the lambda):

int main()
{
    std::vector<uint8_t> first_list = {0x01, 0x10, 0x32, 0x1A};
    std::vector<uint8_t> second; 
    std::transform(first_list.begin(), first_list.end(), std::back_inserter(second),
                   [](uint8_t u) { return ~u; });
    for(uint8_t it : second) {

}

You also don't need to make a copy of the first vector to do this; better to simply use push_back (or back_inserter here, since we're working with iterators).