setTimeout/setImmediate 返回結(jié)果的順序不一定。 那么兩種可能性的分別是怎樣的？

那當(dāng)然，你都沒法保證是不是同一次輪詢（你可能是問這個？）。這就是為什么后面要加fs.readFile。

cut -b 試試

試試ENTRYPOINT ["gunicorn"] 然后 docker run -it name --config python:config web:app

在C++里使用C庫的函數(shù)時，你加了 extern "C" { } 嗎？ 否則就會鏈接找不到。

看這里

加多一個if判斷，或者在sql 語句添加多個where的條件判斷即可。

支持的，可能您的一些設(shè)置之間沖突了，所以會這樣提醒

vis.js

你的表格是v-for生成的嗎

 <button @click="addnewTr"></button>
    <tr v-for="(item,index) in trs" :key="index">
        <td><input type="text"></td>
        <td><input type="text"></td>
    </tr>
    tr=[...]

 addnewTr(){
       this.trs.push(..)
   }

比如這樣

default: ()=>{this.$t('notice')}

rsync -azW --exclude='signup.php' --exclude='apache'  "/var/www/html/." "/var/www/release/" --delete

如果 release 下可以有其他文件 就不要 --delete

先把防火墻關(guān)掉試一下能不能訪問到，能的話多半是端口的問題，不能的話就是服務(wù)器的問題了.

new is not an operator!

In c++, new and operator are both keywords. new int(x) is a(n) (new-)expression. operator new is a function. new operator in your title is new-expression indeed. new-expression will invoke oeprator new function.

placement new的作用就是在operator new分配好的內(nèi)存上執(zhí)行對象的構(gòu)造，
Yes, that's true. To help you understand, here is a sample:

char* ptr = new char[sizeof(T)]; // ptr use a new-expression(newchar [sizeof(T)] to allocate memory, char is guaranteed to be sizeof 1
T* tptr = new(ptr) T;            // Here is placement new

In a nutshell, placement new will use already allocated memory to construct the object. Where is the already allocated memory from? Either from new expression(free store) or allocated from activation record like int buffer[10], both ok.

那么new operator使用了operator new來申請內(nèi)存之后是使用了placement new嗎？如果沒有話是怎么構(gòu)造的呢？和placement new有關(guān)系嗎？

Above is my answer to your questions

BTW, from the case int buffer[10], we can see pre-new-expression is not a must for placement new(however, note that placement new itself is a new-expression, which will invoke operator new function because all it does is just construct here). If your question is "will placement new always be after operator new/new-expression", it will be a good question.

Update

One year ago, I was confused about how to combine operator new with the constructor then asked a question, FrankHB answered my question: https://tieba.baidu.com/p/508... Now, look back to this question, it is a X-Y question, what really confused me was how does new expression invoke constructor, so it is not related to placement new. Hope it will also inspire you.

Update again

所以我認(rèn)為或許自己和您一年前的疑問相似，內(nèi)存申請和構(gòu)造函數(shù)這兩個過程是如何結(jié)合的呢？
the word combination(結(jié)合) is not properly now(I also make such mistake as said above), let me re-organize my wording:

new expression does two things:

1. allocate memory
2. initialization of the object

You and I(one year ago) are both confused about how does compiler initialize the object(via constructor or else) after allocating. Yes, I mentioned compiler, because C++ standard guarantee new will do the two things, but didn't how to, so, its compiler's work. So, it is not c++'s work, just compiler's. Now, we can see the assembly:

struct Foo
{
    Foo(int i) {}
};
int main()
{
    auto *a = new Foo(1);
}  

-O0:

Foo::Foo(int):
        push    rbp
        mov     rbp, rsp
        mov     QWORD PTR [rbp-8], rdi
        mov     DWORD PTR [rbp-12], esi
        pop     rbp
        ret
main:
        push    rbp
        mov     rbp, rsp
        push    rbx
        sub     rsp, 24
        mov     edi, 1
        call    operator new(unsigned long)
        mov     rbx, rax
        mov     esi, 1
        mov     rdi, rbx
        call    Foo::Foo(int)
        mov     QWORD PTR [rbp-24], rbx
        mov     eax, 0
        add     rsp, 24
        pop     rbx
        pop     rbp
        ret

codes related the new expression is follows: a

    mov     edi, 1
    call    operator new(unsigned long)
    mov     rbx, rax
    mov     esi, 1
    mov     rdi, rbx
    call    Foo::Foo(int)
    mov     QWORD PTR [rbp-24], rbx
    

Now, it is clear enough, right? assemble calls two procedures, oeprator new and Foo::Foo(int)

That's all, cheers!

So, your question is how the two combined?

有的，利用php的gd函數(shù)庫完全可以實現(xiàn)這個功能。主要用到的函數(shù)有創(chuàng)建畫布imagecreate類函數(shù)，復(fù)制函數(shù)imagecopy和imagecopyresampled，上色函數(shù)imagecolorallocate，保持透明imagecolortransparent

https://segmentfault.com/q/10...
可以參考這個問題

可以考慮使用背景圖片，設(shè)置 background-size: cover，位置一般習(xí)慣性會設(shè)置成 background-position: center center，也可以根據(jù)圖片的焦點來仔細(xì)調(diào)整位置。

然后我仔細(xì)觀察了一下你的圖，我猜你還是想全部顯示出來，那么由于比例問題，一定會有空白區(qū)域，這種情況下，把背景色設(shè)置為接近圖片背景的顏色，或者就當(dāng)是寬幕電影，設(shè)置成黑色或深灰色，然后 background-size: contain, background-repeat: no-repeat，再把位置設(shè)置成居中。

找到辦法了。

在計算機圖標(biāo)右鍵選擇管理。然后在新出來的窗口選擇設(shè)備管理器。在右側(cè)選擇網(wǎng)絡(luò)適配器。就會出來所有的網(wǎng)卡，然后對指定網(wǎng)卡進行卸載即可。

我記得localCompare好像也不是總是準(zhǔn)確，如果可以的話，數(shù)據(jù)結(jié)構(gòu)改下，專門增加一個用來排序的字段，比如字母之類的。不能改數(shù)據(jù)的話，可以找個node的拼音庫試下。

webpack吧，用的人多呀

@blur="fun()"