Because foo() gets implicitly declared as accepting int arguments, when it actua... | Hacker News

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		WalterBright on Sept 11, 2022 \| parent \| context \| favorite \| on: Teaching C (2016) 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 on Sept 11, 2022 [–]

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 on Sept 11, 2022 | [–]

Yes, you're right. Thanks for the correction. I was thinking of the Windows C ABI.

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