Hi everyone.

New to Julia, I try to reproduce a tricky case study of floating-point arithmetic with the BigFloat module: the “Rump’s example” described in

https://tel.archives-ouvertes.fr/tel-00657843v2/document

page 27. I got completely different results between the output of the C+MPFR code provided in the document, and my Julia code:

```
Precision | C + MPFR | Julia + BigFloat
23 | 1.1726038455963134 | 1.284406e+09
24 | -6.3382530011411471e29 | -6.338253e+29
53 | -1.1805916207174114e21 | -1.1805916207161269e+21
54 | 1.1726039400531785 | 1.28440611600000000e+09
55 | 1.1726039400531786 | 1.28440611600000000e+09
56 | 1.1726039400531786 | 1.28440611600000000e+09
57 | 1.1726039400531786 | 1.284406116000000000e+09
58 | 3.6893488147419103e19 | 3.689348814870350925e+19
59 | -1.8446744073709552e19 | -1.84467440724251455e+19
60 | -1.8446744073709552e19 | -1.8446744072425145504e+19
62 | 1.1726039400531786 | -2.305843007929287836e+18
63 | 1.1529215046068469e18 | 1.152921505891253092e+18
65 | 1.1726039400531786 | 1.28440611600000000000e+09
70 | 1.1726039400531786 | 1.2844061160000000000000e+09
80 | 1.1726039400531786 | 1.2844061160000000000000000e+09
85 | 2.7487790694517260e11 | 2.76162313060000000000000000e+11
90 | 1.7179869185172603e10 | 9.8743407080000000000000000000e+09
100 | 1.1726039400531786 | 1.2844061160000000000000000000000e+09
110 | 1.1726039400531786 | 1.2844061160000000000000000000000000e+09
120 | 1.1726039400531786 | 1.2844061160000000000000000000000000000e+09
122 | -8.2739605994682137e-1 | 1.2844061140000000000000000000000000000e+09
```

To my understanding, BigFloat uses MPFR as a back-end so the results should be the same. But results above are completely different (except for single (24) and double (53) precision).

Can someone help me on understanding this result. As I am new to Julia, I may have coded this wrong, so here is the Julia code I used:

```
function rump(T)
a::T = parse(T, "77617.0");
b::T = parse(T, "33096.0");
b2 = b*b
b4 = b2*b2
b6 = b4*b2
b8 = b4*b4
a2 = a*a
firstexpr = 11*a2*b2 - b6 - 121*b4 - 2
f = parse(T, "333.75")*b6 + a2 * firstexpr + parse(T, "5.5")*b8 + (a/(parse(T, "2.0")/b))
println(" Precision is ", f.prec , " Result is ", f)
end
function main()
prec = eval(parse(ARGS[1]))
Base.MPFR.setprecision(prec)
rump(BigFloat)
end
main()
```

Thanks for your help!