Question

鉴于原始记忆的“博彩”是个空白点,因此,有两种方式撰写材料。

首先是使用新安置办法。这种方法的好处是,当我们处理阶级类型时,就能够把机器人称为汽车。然而,当我处理非阶级类型时,更不用说 cast? 我认为,这样做可能更快。

(pLocation is a void pointer to a blob of memory)

// ----- Is this better -----
*reinterpret_cast<char*>(pLocation) = pattern;

// ----- Or is this better -----
::new(pLocation) char(pattern);

Answer 1

我利用以下方案审视了这些技术的生成组合:

#include <new>

char blob[128];

int main() {
    void *pLocation = blob;
    char pattern =  x ;
#ifdef CAST
    *reinterpret_cast<char*>(pLocation) = pattern;
#else
    ::new(pLocation) char(pattern);
#endif
}

I m 采用4.4.3g++,在有违约汇编旗的六氯环己烷64轨道上使用。

The relevant part of the asm for placement new:

    movb    $120, -1(%rbp)
    movq    -16(%rbp), %rax
    movq    %rax, %rsi
    movl    $1, %edi
    call    _ZnwmPv
    movq    %rax, %rdx
    testq   %rdx, %rdx
    je  .L5
    movzbl  -1(%rbp), %edx
    movb    %dl, (%rax)
.L5:

从我收集的情况来看,这实际上称作安置新经营者,并检查其返回价值,尽管它总是成功。然后,将<代码>x的数值写入交还的记忆中。

<<><>>t>reinterpret_cast:

    movb    $120, -1(%rbp)
    movq    -16(%rbp), %rax
    movzbl  -1(%rbp), %edx
    movb    %dl, (%rax)

请注意,这些指示与前两个和最后两个位置:<编码>新。

使用<代码>-O1,两种代码生成i Equal。集会:

movb    $120, blob(%rip)

因此,如果你重新担心业绩,就没有。任何其他电解机也可能将两者都减至相同的代码。

Answer 2

虽然将原始记忆注入物体在实践中可能发挥作用,但它正式援引未经界定的行为,因此根据C++标准,你的代码可能做任何事情。

Placement new, OTOH, is a technique to invoke a constructor at a particular address, and construction is what officially turns raw memory into valid objects. That s why I would prefer placement new.
Just to make sure, I would also have the destructor for such objects is called. While you say that you only need this for PODs and PODs destruction is a no-op, many bugs I have seen in my carrier were in code that was written with a set of restrictions in mind, but had later some of the restrictions lifted and suddenly found itself in an environment with which it was unable to cope.

还指出,可能有一些平台,因为并非所有可能的比照模式都是有效的价值观,哪怕是内在的。这种平台还可能妨碍人们获得这种模式的价值。例如,全方位轨道型号不能成为浮动型的有效数值,因此即使事先使记忆变得零也不能防止硬件例外。

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