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鍍金池/ 問答/ 網(wǎng)絡安全問答
久不遇 回答

'Access-Control-Allow-Origin' 是跨域錯誤:

協(xié)議://域名:端口 要保持一致才算是同一域的訪問。

一些高版本瀏覽器中,已經(jīng)支持跨域請求,不過需要后端進行配置,后端允許某域的請求。具體如何配置,又是另外的一個問題了。

不懂的話,下面評論我再詳細解答

笑忘初 回答

檢查ckplayer.js的路徑是否正確,檢查ckplayer.js里面定義的名稱是否是ckplayer,否則檢查版本是否合適

若相惜 回答

react不知道有沒有,應該是有的,vue用的axios,可以利用返回的http狀態(tài)碼去進行相關(guān)的跳轉(zhuǎn)

話寡 回答

初步找到原因,因為先裝的laragon,并且安裝的時候勾選了auto virtual hosts,所以創(chuàng)建的所有{name}.dev域名,在chrome下都被強制轉(zhuǎn)到https,而IE下則是直接訪問http沒問題。暫時的解決辦法是,停止自建虛擬主機,手動添加。

問題已解決,https://forum.laragon.org/top...

痞性 回答

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?

檸檬藍 回答

把你的項目結(jié)構(gòu)貼出來以及相關(guān)代碼貼出來

舊顏 回答

await后面是有異步請求的話,不加await獲取不到result
async,await是generator的語法糖,await是阻塞進程,執(zhí)行完await后面的才會執(zhí)行下一步

貓小柒 回答

既然你modalVisiable是通過data算出來的,那么可以通過@action改變data從而間接改變modalVisiable為false

大濕胸 回答

可以翻墻不一定就可以通過terminal翻墻了, 你可以在終端里面輸入curl https://google.com來試試。
如果不能的話, 在終端里執(zhí)行執(zhí)行export http_proxy=127.0.0.1:<proxy_port; export https_proxy=127.0.0.1:<proxy_port>, 然后就可以通過終端使用代理了。
另外, 使用nvm的話, 可以使用淘寶鏡像,
export NVM_NODEJS_ORG_MIRROR=https://npm.taobao.org/dist;
把這個加入~/.bashrc里面就可以了。

病癮 回答

1.要理解這里的this指向哪里,只需要理解函數(shù)作為構(gòu)造函數(shù)之后,也就是放在new 關(guān)鍵字之后發(fā)生了什么
2.其實這里的this是指向AppViewModel new出來的對象,可以參考https://developer.mozilla.org...
3.匿名函數(shù)就是一個函數(shù),只不過這個函數(shù)不需要在其他地方調(diào)用,所以不需要有一個名字,在JavaScript里函數(shù)是一等公民,也是對象,可以作為參數(shù)傳遞

笑忘初 回答

https://clipboardjs.com/
該插件兼容絕大多數(shù)瀏覽器;
非常方便;

執(zhí)念 回答

done是回調(diào)函數(shù),意思是告訴調(diào)用你的人(passport)你已經(jīng)執(zhí)行完你要執(zhí)行的內(nèi)容了,讓它繼續(xù)跑它的邏輯。
一般是以done(err,data)的形式回調(diào)。

假灑脫 回答

clipboard.png

修改AndroidManifest.xml 文件這個

法克魷 回答

多進程,是并發(fā)執(zhí)行的操作,這就沒什么次序可言了。主程序的for循環(huán)是非??斓?,幾乎瞬間5個進程就全開啟了,而先啟動的先輸出的概率大,但并不是絕對。這就跟五個運動員在同一起跑線,即使他們起跑的時間可能會有差異,有人快一點起跑,有人慢一點起跑,但是最先到終點的就不一定是最先起跑的那個人了。

朽鹿 回答

不用管它,因為模擬器在壓縮1個像素的時候出現(xiàn)的四舍五入而已

瘋浪 回答

單詞拼錯了大哥

'./src/js/mian.js' main