Home Backend Development Golang Why Does `fmt.Printf` with `%g` and Width/Precision Fields Behave Unexpectedly?

Why Does `fmt.Printf` with `%g` and Width/Precision Fields Behave Unexpectedly?

Oct 26, 2024 pm 05:02 PM

Why Does `fmt.Printf` with `%g` and Width/Precision Fields Behave Unexpectedly?

fmt.Printf with Width and Precision Fields in %g Behaves Unexpectedly

When trying to format floats with a consistent width using fmt.Printf(), it may not behave as expected when using the %g format specifier with both width and precision fields.

According to the documentation, the precision field for %g sets the number of total digits, excluding any leading zeros. However, the %g format specifier has some unique behaviors that can be confusing.

Leading Zeros and Digit Counting:

  • Leading zeros in the fractional part are not counted as digits.
  • Trailing zeros after non-zero digits are counted as digits.

Examples:

  • fmt.Printf("%.4g", 0.12345) gives 0.1235.
  • fmt.Printf("%.4g", 0.012345) gives 0.01235.
  • fmt.Printf("%.4g", 0.0012345) gives 0.001234.

Precision and Width:

  • The width field specifies the minimum width, including any decimal points or exponents.
  • If the number of digits exceeds the specified width, the width field will be ignored.

Your Case:

In your example, you used .9g, which specifies a minimum width of 10 and a total of 9 digits. However, the following behaviors occur:

  • 0.0606060606060606: Requires 12 digits, so the minimum width is ignored.
  • 0.3333333333333333: Requires 11 digits, so the minimum width is ignored.
  • 0.05 and 0.4: Require 9 digits, which are less than the minimum width, so they are padded with spaces.
  • 0.1818181818181818: Requires 9 digits after rounding, so the minimum width is ignored.

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