CODEWITHSUNDEEP
c++c++11printfstring-formatting
ar2015
ar2015

Reputation: 6150

snprintf c++ alternative

How can I convert this code from C into C++ ?

char out[61]; //null terminator
for (i = 0; i < 20; i++) {
    snprintf(out+i*3, 4, "%02x ", obuf[i])
}

I can't find any alternative for snprintf.

Upvotes: 1

Views: 13594

Answers (4)

mrts
mrts

Reputation: 18963

You can convert this code from C to C++ easily with standard library's std::stringstream and iomanip I/O stream manipulators:

#include <sstream>
#include <iomanip>
...

std::ostringstream stream;
stream << std::setfill('0') << std::hex;

for (const auto byte : obuf)
  stream << std::setw(2) << byte;

const auto out = stream.str();

Upvotes: 0

M.M
M.M

Reputation: 141628

This code is valid C++11, if you have #include <cstdio> and type std::snprintf (or using namespace std;).

No need to "fix" what isn't broken.

Upvotes: 3

jyw
jyw

Reputation: 135

You can use Boost.Format.

#include <boost/format.hpp>
#include <string>

std::string out;
for (size_t i=0; i<20; ++i)
    out += (boost::format("%02x") % int(obuf[i])).str();

Upvotes: 0

VolAnd
VolAnd

Reputation: 6407

Use stringstream class from <sstream>.

E.g.:

#include <iostream>
#include <iomanip>
#include <sstream>
#include <string>
using namespace std;

int main()
{
    stringstream ss;
    for (int i = 0; i < 20; i++) {
        ss << setw(3) << i;
    }
    cout << "Resulting string: " << endl;
    cout << ss.str() << endl;
    printf("Resulting char*: \n%s\n", ss.str().c_str() );
    return 0;
}

Upvotes: 8

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