4 ms·

Here gcc (9.4.0) throws a warning if the first declaration is commented and the second one uncommented, but it prints "0". Care to explain why?

• WalterBright 4y ago
  Because foo() gets implicitly declared as accepting int arguments, when it actually accepts a double. The double bits interpreted as an int are 0.
  • edflsafoiewq 4y ago
    On x64, foo(double) expects its argument in xmm0, but foo(int) expects it in rdi. So the 3 passed in rdi is ignored and it operates on whatever happened to be in xmm0.
    • WalterBright 4y ago
      Yes, you're right. Thanks for the correction. I was thinking of the Windows C ABI.