localtime vs localtime_s和适当的输入参数 [英] localtime vs localtime_s and appropriate input arguments

问题描述

time_t rawtime;
struct tm * timeinfo;
time ( &rawtime );
timeinfo = localtime ( &rawtime );

这会返回：warning C4996：'localtime'：此函数或变量可能不安全。请考虑使用localtime_s。

This returns: warning C4996: 'localtime': This function or variable may be unsafe. Consider using localtime_s instead.

time_t rawtime;
struct tm * timeinfo;
time ( &rawtime );
timeinfo = localtime_s ( &rawtime );

当我将localtime更改为localtime_s时，我得到：error C2660：'localtime_s'：function does not take 1参数

When I change localtime to localtime_s I get: error C2660: 'localtime_s' : function does not take 1 arguments

这是我认为在第一个代码块：

Here is what I think is going on in the first block of code:

• 创建一个空的time_t变量


• 创建一个指向在ctime中定义的timeinfo的指针


• 将rawtime写入rawtime参考
• 
  

• 将原始时间转换为对行人有意义的东西

• create an empty time_t variable.
• create a pointer to timeinfo which is defined in ctime
• write the rawtime into a rawtime reference

• convert the rawtime into something meaningful to pedestrians

1. 
   
2. localtime_s需要什么第二个输入参数？


3. 如果我忽略整个本地时间安全问题，最糟糕的情况是什么。

推荐答案

localtime 返回指向静态分配的指针 struct tm 。

localtime returns a pointer to a statically allocated struct tm.

通过localtime_s， tm和 localtime_s 将其结果数据写入其中，因此您的代码将更改为：

With localtime_s, you pass in a pointer to a struct tm, and localtime_s writes its result data into that, so your code would change from:

struct tm *timeinfo;
timeinfo = localtime(&rawtime);

：

struct tm timeinfo;
localtime_s(&timeinfo, &rawtime);

这样，它正在写入 缓冲区，它自己的。

This way, it's writing to your buffer instead of having a buffer of its own.

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