compiler opions for linux

Question

2.00/5 (1 vote)

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C++

#include <iostream>
#include <fstream>
#include <cstdlib>
#include <random>

int main ( int argc , char ** argv ) {

  std::random_device rd;
  std::mt19937_64 gen(rd());

 std::uniform_real_distribution<> dis(0, 1);

 std::cout << dis(gen) << std::endl;

  return 0;
}

I am using the random library in c++ on linux, but this compiler options

"g++ code.cpp -o code.exe -std=c++0x"

doesn't work because of the generator of random variables "mt19937_64"

this is the error:

erreur: ‘mt19937_64’ is not a member of ‘std’

Posted 9-Sep-13 3:54am

Member 10258053

Updated 9-Sep-13 4:44am

Nelek

v3

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Comments

Sergey Alexandrovich Kryukov 9-Sep-13 9:59am

Not a question. Not clear. Incomplete. Not clear at all. Please click on Improve question above and clarify.
—SA

Richard MacCutchan 9-Sep-13 10:17am

This class is only available with C++11. You need to get a copy of the latest compiler.

Member 10258053 9-Sep-13 10:25am

Ok Thank you.

Sergey Alexandrovich Kryukov 9-Sep-13 17:22pm

Right. Just a note: C++0x is an obsolete name for what is now called C++11.
—SA

Richard MacCutchan 10-Sep-13 2:59am

And what did I call it in my reply?

Sergey Alexandrovich Kryukov 10-Sep-13 10:53am

You called it "C++11", but OP mentioned "C++0x" in the parameters.
I only added that "C++11" is the same thing.

By the way, the fact that the compiler uses obsolete name "C++0x" may really indicate it's obsolete. In this documentation, C++11 is mentioned: http://gcc.gnu.org/onlinedocs/gcc/C_002b_002b-Dialect-Options.html#C_002b_002b-Dialect-Options.

—SA

Member 10258053 10-Sep-13 16:46pm

when I use c + +11 instead of c + +0 x it doesn't work but this :
g+ +-std = c + +0 x-o code code.cpp with gcc/7.4.2 it works

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Member 10258053 · Accepted Answer · 2013-09-10T04:05:00

Solution 1

it compiles without problems when I use the version compiler gcc/4.7.2

Thank you.

Posted 10-Sep-13 4:05am

Member 10258053