SBRM/RAII 用于 std::va_list/va_start()/va_end 使用 [英] SBRM/RAII for std::va_list/va_start()/va_end use

问题描述

我的代码包含如下片段:

My code contains snippets like these:

    std::va_list ap;
    va_start(ap, msgfmt);
    snprintf_buf buf;
    const tchar * msg = buf.print_va_list(msgfmt, ap);
    va_end(ap);

它们很短，va_start() 和 va_end() 靠得很近，所以它们不是什么大问题.两者之间调用的异常可能是一个问题(或不是?).

These are short and va_start() and va_end() are close together so they are not much of a problem. Exceptions from calls in between the two could be a problem (or not?).

简单测试表明，不允许从没有省略号的函数中调用 va_start().调用 va_end() 是否从不同于 va_start() 的函数调用?

Simple test shows that calling va_start() from a function without ellipsis is not allowed. Is calling va_end() from a different function than va_start() was called from allowed or not?

基本上，我很好奇是否可以对这些调用使用 SBRM/RAII 习惯用法，即使有必要手动调用 va_start() 然后传递 的实例std::va_list 到我的 RAII/SBRM 守卫实例?

Basically, I am curious if it is possible to use the SBRM/RAII idiom for these calls, even if it were necessary to call va_start() manually and then to pass instance of std::va_list into my RAII/SBRM guard instance?

推荐答案

很遗憾，没有.va_start 和 va_end 的规范要求:

Unfortunately, no. The specification of va_start and va_end requires that:

va_start 和 va_copy 宏的每次调用应与 va_end 宏 的相应调用相匹配功能.

Each invocation of the va_start and va_copy macros shall be matched by a corresponding invocation of the va_end macro in the same function.

因此，va_end 必须在可变参数函数本身中，而不是类析构函数.

Therefore, va_end must be in the variadic function itself, not a class destructor.

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