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Jeff Epler
10b624751e
Clang -m32 rounds some floating point reprs differently, likely due to x87 temporary excess precision. Accept this value in addition to the other values that are accepted. Closes: #19120 Signed-off-by: Jeff Epler <jepler@unpythonic.net>
61 lines
2.5 KiB
Python
61 lines
2.5 KiB
Python
# Test that integers format to exact values.
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for b in [13, 123, 457, 23456]:
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for r in range(1, 10):
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e_fmt = "{:." + str(r) + "e}"
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f_fmt = "{:." + str(r) + "f}"
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g_fmt = "{:." + str(r) + "g}"
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for e in range(0, 5):
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f = b * (10**e)
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title = str(b) + " x 10^" + str(e)
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print(title, "with format", e_fmt, "gives", e_fmt.format(f))
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print(title, "with format", f_fmt, "gives", f_fmt.format(f))
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print(title, "with format", g_fmt, "gives", g_fmt.format(f))
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# The tests below check border cases involving all mantissa bits.
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# In case of REPR_C, where the mantissa is missing two bits, the
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# the string representation for such numbers might not always be exactly
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# the same but nevertheless be correct, so we must allow a few exceptions.
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is_REPR_C = float("1.0000001") == float("1.0")
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# 16777215 is 2^24 - 1, the largest integer that can be completely held
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# in a float32.
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val_str = "{:f}".format(16777215)
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# When using REPR_C, 16777215.0 is the same as 16777212.0 or 16777214.4
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# (depending on the implementation of pow() function, the result may differ)
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if is_REPR_C and (val_str == "16777212.000000" or val_str == "16777214.400000"):
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val_str = "16777215.000000"
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print(val_str)
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# 4294967040 = 16777215 * 128 is the largest integer that is exactly
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# represented by a float32 and that will also fit within a (signed) int32.
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# The upper bound of our integer-handling code is actually double this,
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# but that constant might cause trouble on systems using 32 bit ints.
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val_str = "{:f}".format(2147483520)
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# When using FLOAT_IMPL_FLOAT, 2147483520.0 == 2147483500.0
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# Both representations are valid, the second being "simpler"
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is_float32 = float("1e300") == float("inf")
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if is_float32 and val_str == "2147483500.000000":
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val_str = "2147483520.000000"
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# When using REPR_C, 2147483520.0 is the same as 2147483200.0
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# Both representations are valid, the second being "simpler"
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if is_REPR_C and val_str == "2147483200.000000":
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val_str = "2147483520.000000"
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# When using REPR_C, x86 and clang, 2147483520.0 is the same
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# as 2147483100.0, the second being "simple" but rounded differently
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# due to x87 extra precision on intermediates.
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# Both representations are valid.
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if is_REPR_C and val_str == "2147483100.000000":
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val_str = "2147483520.000000"
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print(val_str)
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# Very large positive integers can be a test for precision and resolution.
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# This is a weird way to represent 1e38 (largest power of 10 for float32).
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print("{:.6e}".format(float("9" * 30 + "e8")))
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