Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
197 views
in Technique[技术] by (71.8m points)

c++ - how to prevent numbers from showing up in scientific notations

We have a StreamBuffer class in which we haven't implemented std::fixed operations and I am trying to prevent number showing up in scientific notations. With my below code some numbers are getting shown in scientific notations. We want to avoid doing any allocations so that's why we implemented StreamBuffer class because of performance reason.

Below is the code:

T value = 0;

template<typename U> void process(U& buf, DataOption holder) const {
    if (holder == DataOption::TYPES) {
        switch (type_) {
        case teck::PROC_FLOAT:
            buf << "{"float":" << value << "}";
            break;
        case teck::PROC_DOUBLE:
            buf << "{"double":" << value << "}";
            break;
        default:
            buf << "{"" << type_ << "":" << value << "}";
        }
    }
}

And this is the way it is being called:

void HolderProcess::dump(std::ostream& os, DataOption holder) const 
{
    process<std::ostream>(os, holder);
}

void HolderProcess::dump(StreamBuffer& buffer, DataOption holder) const
{
    process<StreamBuffer>(buffer, holder);
}

I tried using like as shown below and I got an error by which I understood we cannot use std::fixed on my StreamBuffer class.

case teck::PROC_DOUBLE:
    buf << "{"double":" << std::fixed << value << "}";

What is the alternative to std::fixed I can use here which doesn't do any allocations at all. I was thinking of converting number to string and then apply std::fixed on it but that will do some allocations as well which I want to avoid that.

What is the best way to do this which is performance efficient and doesn't do any allocations? I have a below solution but it will do some allocations as it uses string. I can call below method from my above code.

template <typename T> string str(T number)
   {
       std::ostringstream ss;
       ss << std::fixed << number;
       return ss.str();
   }

Is there any other optimized and efficient way?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

StreamBuffer class must inherit from std::ios_base (or some of it's derivatives such as std::ostream) for your expected behaviour. std::fixed can only work with derivative implementations of what is available as part of the STL.

Additionally, if you have access to an std::ios_base, you can also play around with std::ios_base::precision.

If you are stuck in a situation where you cannot update the class then the most commonly used and traditional way is by way of scaling floats. In the interest of reducing duplication please have a look at the already answered question here. For example, for a 3rd degree of precision, I'd replace all 'value' instances with:

// case teck::PROC_FLOAT:
static_cast<float>( static_cast<int>(value*1000) ) / 1000
// case techk::PROC_DOUBLE:
static_cast<double>( static_cast<long long>(value*1000) ) / 1000

Having better understood the questioner's requirements. I have realised that the above would not work with exponents. In order to get around this I propose doing the following:

case teck::PROC_FLOAT:
        std::stringstream ss;

        ss << std::fixed << value;
        buf << "{"float":" << ss.str() << "}";
        break;

This, however, will most certainly allocate more memory.



与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

1.4m articles

1.4m replys

5 comments

57.0k users

...