Convert uint64_t to std::string

Question

How can I transfer uint64_t value to std::string? I need to construct the std::string containing this value For example something like this:

void genString(uint64_t val)
{
      std::string str;
      //.....some code for str 
      str+=(unsigned int)val;//????
}

Thank you

On Windows (MSVC) you can use the `%I64d` format specifier when using `std::sprintf` — Jonas Engström, Sep 08 '11 at 12:38
@Jonas - given this is clearly C++ I'd strongly advise keeping well clear of `sprintf`, especially with non-portable extensions. — Flexo, Sep 08 '11 at 12:39

score 35 · Answer 1 · answered Nov 06 '14 at 14:11

35

In C++ 11 you may just use:

std::to_string()

it's defined in header

answered Nov 06 '14 at 14:11

dau_sama

score 12 · Accepted Answer · answered Sep 08 '11 at 12:33

12

use either boost::lexical_cast or std::ostringstream

e.g.:

str += boost::lexical_cast<std::string>(val);

or

std::ostringstream o;
o << val;
str += o.str();

answered Sep 08 '11 at 12:33

Flexo

Can you give me the code example how to use std::ostringstream? – YAKOVM Sep 08 '11 at 12:36
@yakov - added examples for both. You'll need either `#include ` or `#include ` depending on which solution you use. – Flexo Sep 08 '11 at 12:38
1

Downvoted since by now, the standard C++11 solution should be used - std::to_string. – Liosan Aug 29 '16 at 21:15

score 5 · Answer 3 · answered Sep 08 '11 at 13:08

I use something like this code below. Because it's a template it will work with any type the supports operator<< to a stream.

#include <sstream>

template <typename T>
std::string tostring(const T& t)
{
    std::ostringstream ss;
    ss << t;
    return ss.str();
}

for example

uint64_t data = 123;
std::string mystring = tostring(data);

score 2 · Answer 4 · answered Sep 08 '11 at 12:43

2

string genString(uint64_t val)
{
   char temp[21];
   sprintf(temp, "%z", val);
   return temp;
}

answered Sep 08 '11 at 12:43

NMI

To whomever may use this in the future, make sure to make your buffer big enough. Also you could make that buffer `static` if you wanted to add a little more efficiency. – Seth Carnegie Sep 08 '11 at 12:44
1

@Seth - is static (and hence not reentrant) really faster than changing the stack pointer increase at entering the function from `n` to `n + 21`? – Flexo Sep 08 '11 at 12:56
@Seth - gcc 4.4 with -O3 on ia32 gives me the exact opposite behaviour. `static` makes it consistently (slightly) slower (more than the error) on 100 repetitions of a 10,000,000 call loop. Seems like premature optimisation to me. The main chance in the local version is `subl $72, %esp` vs `subl $52, %esp`, an O(1) change. – Flexo Sep 08 '11 at 13:42
@nmi - I don't think `%z` is correct - I get `test.cc:10:29: warning: conversion lacks type at end of format` and `test.cc:10:29: warning: too many arguments for format` with gcc – Flexo Sep 08 '11 at 13:43

4 Answers4