C vs Java primitive casting and expressions

Question

Why does this equation produce two different values for e in C (32-bit) and Java (64-bit)? They are off by an unexpectedly large amount.

int a, b, c, d;
double e;

a = -12806;
b = 1;
c = 800;
d = 19209;

double f = (32768 / 10.24);

e = (a + (double) b / c * d) / f;

C produces -3.9940624237060547. Java produces -3.9943714843750002.

UPDATE:

Sorry folks, this error appears to be something else than I expected. I reduced my code to just this equation and the numbers it produces are much closer.

Any examples of that "unexpectedly large amount" ? – dee-see Feb 24 '12 at 01:45 — dee-see, Feb 24 '12 at 01:45
Which inputs do you use and which results do you get? – Andres F. Feb 24 '12 at 01:45 — Andres F., Feb 24 '12 at 01:45
what is `sizeof(double)` in C and Java on your system? – Bingo Feb 24 '12 at 01:57 — Bingo, Feb 24 '12 at 01:57

score 1 · Accepted Answer · answered Feb 24 '12 at 01:57

In Java the implicit braces are a bit different:

    int a, b, c, d;
    double e;

    a = 3;
    b = 4;
    c = 5;
    d = 6;
    e = a + (double) b / c * d;
    System.out.println("e=" + e);
    e = a + (((double) b) / c) * d; // Java: 7.8
    System.out.println("e=" + e);

If you run this in C you will see the difference.

score 0 · Answer 2 · edited May 23 '17 at 12:20

0

This has been answered here:

Retain precision with double in Java

Your best bet would be to use BigDecimals as these retain this precision.

edited May 23 '17 at 12:20

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answered Feb 24 '12 at 01:58

C vs Java primitive casting and expressions

2 Answers2