C++中printf中的点

Question

I encountered this snippet but couldn't understand how it works, especially the printf statements. Can someone explain

void remove_trailing_zeroes()
{
    int a,b; 
    bool f1,f2;
    f1=a%2;
    f2=b%2;
    if (f1==f2) {
        printf("%.0lf\n",(a*1.+b)/2.);
    }
    else {
        printf("%.1lf\n",(a*1.+b)/2.);
    }

}

EDIT: I have rephrased my question, help me improve it

Answer 1

If you are puzzled about the dots here is what they are:

• %.1lf is the format specification for precision. This is requesting one digit after the decimal point in the printf output.
• The 1. and 2. in (a*1.+b)/2. mean that those literals are double (as opposed to 1 that would be int and 1.f that would be float ). Whoever wrote that snippet was probably trying to avoid truncation in computing that average (given a and b are int ).

Answer 2

Sounds like:

printf("%g", (a+b)/2.);

emulation.

Answer 3

It looks like it prints average value between a and b .

This if decides when the result will need decimal point .5 :

bool f1,f2;
f1=a%2;
f2=b%2;
if (f1==f2)

It would be better to just write:

// get a and b from somewhere
if ((a+b)%2) // check if sum can be divided by 2
    printf("%.1lf\n",(a+b)/2.); // %.1lf will print value with 1 decimal ("xX.X")
else
    printf("%.0lf\n",(a+b)/2.); // %.0lf will print value without decimals ("xX")