Convert a number to a string with specified length

问题:

I have some numbers of different length (like 1, 999, 76492, so on) and I want to convert them all to strings with a common length (for example, if the length is 6, then those strings will be: '000001', '000999', '076492').

In other words, I need to add correct amount of leading zeros to the number.

int n = 999;
string str = some_function(n,6);
//str = '000999'

Is there a function like this in C++?

回答1:

or using the stringstreams:

#include <sstream>
#include <iomanip>

std::stringstream ss;
ss << std::setw(10) << std::setfill('0') << i;
std::string s = ss.str();

I compiled the information I found on arachnoid.com because I like the type-safe way of iostreams more. Besides, you can equally use this code on any other output stream.

回答2:

char str[7];
snprintf (str, 7, "%06d", n);

See snprintf

回答3:

One thing that you may want to be aware of is the potential locking that may go on when you use the stringstream approach. In the STL that ships with Visual Studio 2008, at least, there are many locks taken out and released as various locale information is used during formatting. This may, or may not, be an issue for you depending on how many threads you have that might be concurrently converting numbers to strings...

The sprintf version doesn't take any locks (at least according to the lock monitoring tool that I'm developing at the moment...) and so might be 'better' for use in concurrent situations.

I only noticed this because my tool recently spat out the 'locale' locks as being amongst the most contended for locks in my server system; it came as a bit of a surprise and may cause me to revise the approach that I've been taking (i.e. move back towards sprintf from stringstream)...

回答4:

stringstream will do (as xtofl pointed out). Boost format is a more convenient replacement for snprintf.

回答5:

This method doesn't use streams nor sprintf. Other than having locking problems, streams incur a performance overhead and is really an overkill. For streams the overhead comes from the need to construct the steam and stream buffer. For sprintf, the overhead comes from needing to interpret the format string. This works even when n is negative or when the string representation of n is longer than len. This is the FASTEST solution.

inline string some_function(int n, int len)
{
    string result(len--, '0');
    for (int val=(n<0)?-n:n; len>=0&&val!=0; --len,val/=10)
       result[len]='0'+val%10;
    if (len>=0&&n<0) result[0]='-';
    return result;
}

回答6:

There are many ways of doing this. The simplest would be:

int n = 999;
char buffer[256]; sprintf(buffer, "%06d", n);
string str(buffer);

回答7:

sprintf is the C-like way of doing this, which also works in C++.

In C++, a combination of a stringstream and stream output formatting (see http://www.arachnoid.com/cpptutor/student3.html ) will do the job.

回答8:

This is an old thread, but as fmt might make it into the standard, here is an additional solution:

#include <fmt/format.h>

int n = 999;

const auto str = fmt::format("{:0>{}}", n, 6);

Note that the fmt::format("{:0>6}", n) works equally well when the desired width is known at compile time. Another option is abseil:

#include <absl/strings/str_format.h>

int n = 999;

const auto str = absl::StrFormat("%0*d", 6, n);

Again, abs::StrFormat("%06d", n) is possible. boost format is another tool for this problem:

#include <boost/format.hpp>

int n = 999;

const auto str = boost::str(boost::format("%06d") % n);

Unfortunately, variable width specifier as arguments chained with the % operator are unsupported, this requires a format string setup (e.g. const std::string fmt = "%0" + std::to_string(6) + "d";).

In terms of performance, abseil and fmt claim to be very attractive and faster than boost. In any case, all three solutions should be more efficient than std::stringstream approaches, and other than the std::*printf family, they do not sacrifice type safety.