InvInt 的 xi2 和 xi3 数组应该预计算为 1/I² 和 1/I³, 与 Fortran INITIA 中的初始化逻辑一致。 Co-Authored-By: Claude Opus 4.6 <noreply@anthropic.com> |
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| .. | ||
| fortran-analyzer | ||
| fortran-extractor | ||
| fortran-to-rust | ||
| pua | ||
| self-improving-agent | ||
InvInt 的 xi2 和 xi3 数组应该预计算为 1/I² 和 1/I³, 与 Fortran INITIA 中的初始化逻辑一致。 Co-Authored-By: Claude Opus 4.6 <noreply@anthropic.com> |
||
|---|---|---|
| .. | ||
| fortran-analyzer | ||
| fortran-extractor | ||
| fortran-to-rust | ||
| pua | ||
| self-improving-agent | ||