Thread: Printing float number with incremented precision

Hi

The following code gives the corresponding print:

Code:

floatValue = 12345.67890;

printf("%.1f\n", floatValue);
printf("%.2f\n", floatValue);
printf("%.3f\n", floatValue);
printf("%.4f\n", floatValue);
printf("%.5f\n", floatValue);

Print:
12345.7
12345.68
12345.679
12345.6789
12345.67890

Is there a way to make this precision increment in a loop instead?
i.e.

Code:

for(int i=1; i < 6; i++){
    printf("%.if\n", floatValue);
    i++;
}

BR,
Flanderso

It's actually in the manual if you look.

printf("%.*f\n", i, floatValue);

Thanks a lot whiteflags!

Btw, what manual?

Originally Posted by iMalc

Your example is misleading by the way.
For it to output that, the variable floatValue must actually be declared as a double, not a float.

What? I thought for sure that a float is promoted to double when passed to a variadic function (or when there is no function declaration) like printf.

Originally Posted by christop

What? I thought for sure that a float is promoted to double when passed to a variadic function (or when there is no function declaration) like printf.

I think what iMalc means is that, usually, float values do not have the kind of precision indicated by the example.
For an 32-bit IEEE floating point number, with 23+1 binary digits of precision, the precision of the decimal expansion (~7 digits) is not enough for the result shown.