C中的舍入浮点

Question

While testing the float type and printing it with it's format specifier %f I was testing it's rounding methods.

I've declared the variable as float and gave it the value 5.123456. As you know float must represent at least 6 significant figures.

I then changed it's value to 5.1234567 and printed the value with the %f . It baffles me why it prints out as 5.123456. But if I change the variable value to 5.1234568, it prints out as 5.123457. It rounds properly.

If I haven't made myself clear or the explanation is very confusing:

float a = 5.1234567
printf("%d", a); 
// prints out as 5.123456

float a = 5.1234568
printf("%d", a); 
// prints out as 5.123457

I've compiled using CodeBlocks and MinGW, same result.

Answer 1

OP is experiencing the effects of double rounding

First, the values 5.123456, 5.1234567, etc. are rounded by the compiler to the closest representable float . Then printf() is rounding the float value to the closest 0.000001 decimal textual representation.

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I've declared the variable as float and gave it the value 5.123456. As you know float must represent at least 6 significant figures.

A float can represent about 2^32 different values. 5.123456 is not one of them. The closest value a typical float can represent is 5.12345600128173828125 and that is correct for 6 significant digits: 5.12345...

float x = 5.123456f;
//  5.123455524444580078125 representable float just smaller than 5.123456
//  5.123456                OP's code
//  5.12345600128173828125  representable float just larger  than 5.123456 (best)

// The following prints 7 significant digits
// %f prints 6 places after the decimal point.
printf("%f", 5.123456f); // --> 5.123456

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With 5.1234567, the closest float has an exact value of 5.123456478118896484375. When using "%f" , this is expected print rounded to the closest 0.000001 or 5.123456

float x = 5.1234567f;
//  5.123456478118896484375 representable float just smaller than 5.1234567 (best)
//  5.1234567                OP's code
//  5.1234569549560546875   representable float just larger  than 5.1234567

// %f prints 6 places after the decimal point.
printf("%f", 5.1234567f); // --> 5.123456

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Significant digits is not the number of digit after the decimal point. It is the number of digits starting with the left-most (most significant) digit.

To print a float to 6 significant figures, use "%.*e" .
See Printf width specifier to maintain precision of floating-point value for more details.

float x = 5.1234567;         
printf("%.*e\n", 6 - 1, x);  // 5.12346e+00
                             // x xxxxx   6 significant digits

Answer 2

There is no exact float representation for the number 5.1234567 you intend to show here.

If you check here: https://www.h-schmidt.net/FloatConverter/IEEE754.html

You can see that this number is converted into 5.1234565, or the double 5.1234564781188965 and this rounds down,

While the number 5.1234568 is representable in float, and has a double representation of 5.123456954956055, and this rounds up.

Answer 3

There are two levels of rounding going on:

1. Your constant of 5.1234567 gets rounded to the nearest value which can be represented by a float (5.123456478...).
2. The float gets rounded to 6 digits when printed.

It will become obvious if you print the value with more digits.

What it comes down to is that the mantissa of a float has 23 bits and this is not the same as 6 decimal digits (or any number of digits really). Even some apparently simple values like 0.1 don't have an exact float representation.