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Home/ Questions/Q 977207
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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T03:52:03+00:00 2026-05-16T03:52:03+00:00

here is code #include <iostream> #include <fstream> #include <cstring> using namespace std; int main()

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here is code

#include <iostream>
#include <fstream>
#include <cstring>
using namespace std;

int main()
{
  ofstream out("test", ios::out | ios::binary);

  if(!out) {
    cout << "Cannot open output file.\n";
    return 1;
  }

  double num = 100.45;
  char str[] = "www.java2s.com";

  out.write((char *) &num, sizeof(double));
  out.write(str, strlen(str));

  out.close();

  return 0;
}

i dont understand only this

out.write((char *) &num, sizeof(double));

why we need (char *)&num?or sizeof(double)?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-16T03:52:03+00:00Added an answer on May 16, 2026 at 3:52 am

    write takes two parameters, a char* and a length.

    &num is actually a double*. It’s the value we want, but it’s the wrong type, and the compiler would complain. The (char*) tells the compiler to treat this as a char*. Basically, it says to the compiler “Shutup. I know what I’m doing”.

    sizeof(double) is the size of the double in characters. (Usually, this is 8).

    Together, they say to write the 8 bytes starting at the address given by &num. (or in other words, write out the bytes that make up num)

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