Question

我从另一个零碎问题中借了这个问题,但我不赞成这个问题。我更担心我根本不理解这一职能。类型和数据寿命:

同一数据由以下几个方面构成:探测器(带有内部元数据的动态射线型)、该数据所载示意图器(回归参数)和申报的回归类型,后者是一个 st脚石。

QUESTIONS: How does the data get safely out of the function when the std::vector is going to be destroyed? Is it implicitly copied? Is the dynamic array of char in the vector detached from the vector and returned as a std::string type so that no bulk copy is required? Sometimes, I think that C++ and the std library is trying to get me...

I ve been using C++ for some while but stuff like this does my ed in.

std::string TestFragmentation()
{
    std::vector<char> buffer(500);
    SomeCApiFunction( &buffer[0], buffer.size() ); // Sets buffer to null-terminated string data
    return &buffer[0];
}

Answer 1

The data stored by a std::vector is guaranteed to be contiguous, so &buffer[0] gets you a raw pointer to the beginning of that data.¹

<代码>std:string has a Constructionor which take a const char *, which texts the data. 收益报表中含蓄地提到这一点(允许汇编者在大多数暗中转换作业中打电话,以避免汇编时间错误)。

In both cases (the vector and the string), the memory for the corresponding backing buffer is managed by the container class, so there is no possibility of a memory leak or similar (so long as your raw C function creates a valid null-terminated C-style string and doesn t trample beyond the buffer bounds).

页: 1 然而,注意到没有保证它会停留在一个地方。当你生长或缩小病媒时,它有可能在其他地方复制,以备记忆,使在病媒中发现的所有原始点失效。如果是<条码>, 本身被拆解,当然,背后的数据不再有效。

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