error c4996: 'fopen': this function or variable may be unsafe. consider using fopen_s instead. to di
亲，您这应该是在 C 或 C++编程中遇到的错误提示呢。它说‘fopen’这个函数或变量可能不太安全，建议您使用‘fopen_s’来替代。您能跟我多说一说出现这个错误的具体情况不？比如您是在什么样的代码环境里碰到的，或者您的代码大致是咋写的？ [2024-12-11 14:10:18 | AI写代码神器 | 82点数解答]
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