如何在一个大数字插入空格，使其更加可读？ [英] How to insert spaces in a big number to make it more readable?

问题描述

我想出了这个，因为stackoverflow提供的其他例子在C＃

I came up with this, since other examples provided on stackoverflow were in C#

string number_fmt(ulong n)
{
    // cout << "(" << n << ")" << endl;
    char s[128];
    sprintf(s, "%lu", n);
    string r(s);
    reverse(r.begin(), r.end());
    int space_inserted = 0;
    size_t how_many_spaces = r.length() / 3;

    if(r.length() % 3 != 0)
        how_many_spaces += 1;

    for(int i = 1; i < how_many_spaces; ++i)
    {
        r.insert(3 * i + space_inserted, " ");
        space_inserted += 1;
    }
    reverse(r.begin(), r.end());

    return r;
}

你知道更好的解决方案吗？

Do you know any better solution ?

推荐答案

这一个是不同的，但更好的是主观的。我认为它非常简洁明了它在做什么：

This one is different, but better is subjective. I think it's very succinct and clear what it's doing though:

string number_fmt(unsigned long long n, char sep = ',') {
    stringstream fmt;
    fmt << n;
    string s = fmt.str();
    s.reserve(s.length() + s.length() / 3);

    // loop until the end of the string and use j to keep track of every
    // third loop starting taking into account the leading x digits (this probably
    // can be rewritten in terms of just i, but it seems more clear when you use
    // a seperate variable)
    for (int i = 0, j = 3 - s.length() % 3; i < s.length(); ++i, ++j)
        if (i != 0 && j % 3 == 0)
            s.insert(i++, 1, sep);

    return s;
}

使用

cout << number_fmt(43615091387465) << endl;

print

43,615,091,387,465

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