What does "double + 1e-6" mean?

August 13, 2022

The result of this cpp is 72.740, but the answer should be like 72.741

mx = 72.74050000;
printf("%.3lf \n", mx);

So I found the solution on website, and it told me to add "+1e-7" and it works

mx = 72.74050000;
printf("%.3lf \n", mx + 1e-7);

but I dont know the reason in this method, can anyone explain how it works?

And I also try to print it but nothing special happens…, and it turn out to be 72.7405

mx = 72.74050003;
cout << mx + 1e-10;

>Solution :

If you will output the value like

printf( "mx = %.16f\n", mx );

you will see

mx = 72.7404999999999973

So to make the result like 72.741 due to rounding in outputting with a call of printf you need to make the next digit equal to 5 instead of 4. It is enough to add 0.00001.

Here is a demonstration program.

#include <iostream>
#include <iomanip>
#include <cstdio>

int main( void ) 
{
    double mx = 72.74050000;

    printf( "mx = %.3f\n", mx + 0.00001);

    std::cout << "mx = " << std::setprecision( 5 ) <<  mx + 0.00001 << '\n';
}

The program output is