# Test that integers format to exact values. for b in [13, 123, 457, 23456]: for r in range(1, 10): e_fmt = "{:." + str(r) + "e}" f_fmt = "{:." + str(r) + "f}" g_fmt = "{:." + str(r) + "g}" for e in range(0, 5): f = b * (10**e) title = str(b) + " x 10^" + str(e) print(title, "with format", e_fmt, "gives", e_fmt.format(f)) print(title, "with format", f_fmt, "gives", f_fmt.format(f)) print(title, "with format", g_fmt, "gives", g_fmt.format(f)) # The tests below check border cases involving all mantissa bits. # In case of REPR_C, where the mantissa is missing two bits, the # the string representation for such numbers might not always be exactly # the same but nevertheless be correct, so we must allow a few exceptions. is_REPR_C = float("1.0000001") == float("1.0") # 16777215 is 2^24 - 1, the largest integer that can be completely held # in a float32. val_str = "{:f}".format(16777215) # When using REPR_C, 16777215.0 is the same as 16777212.0 or 16777214.4 # (depending on the implementation of pow() function, the result may differ) if is_REPR_C and (val_str == "16777212.000000" or val_str == "16777214.400000"): val_str = "16777215.000000" print(val_str) # 4294967040 = 16777215 * 128 is the largest integer that is exactly # represented by a float32 and that will also fit within a (signed) int32. # The upper bound of our integer-handling code is actually double this, # but that constant might cause trouble on systems using 32 bit ints. val_str = "{:f}".format(2147483520) # When using FLOAT_IMPL_FLOAT, 2147483520.0 == 2147483500.0 # Both representations are valid, the second being "simpler" is_float32 = float("1e300") == float("inf") if is_float32 and val_str == "2147483500.000000": val_str = "2147483520.000000" # When using REPR_C, 2147483520.0 is the same as 2147483200.0 # Both representations are valid, the second being "simpler" if is_REPR_C and val_str == "2147483200.000000": val_str = "2147483520.000000" print(val_str) # Very large positive integers can be a test for precision and resolution. # This is a weird way to represent 1e38 (largest power of 10 for float32). print("{:.6e}".format(float("9" * 30 + "e8")))