小编asd*_*sds的帖子

我正在阅读Beej\xe2\x80\x99s 网络编程指南，在第 45 页上编写了以下代码；

struct pollfd *pfds = malloc(sizeof *pfds * fd_size);\n

我想知道是否应该

struct pollfd *pfds = malloc(sizeof(struct pollfd) * fd_size);\n

由于malloc返回一个指向指定大小的内存块的指针。

由于 *pfds 是一个指针，它的大小将为 4 或 8 字节，所以我无法理解为什么在创建数组时要考虑它的大小。

import React, {useState, useEffect, Component} from 'react';
import {Grid, Paper, TextField} from '@material-ui/core'
import DataManager from './../data_manager/data_manager'

const dataManager = new DataManager();

const Tile = (props)=>{
    // Initializing State Variables
    const [state, setState] = useState({
        data : {}
    })

    const { status, data, error, isFetching } = useQuery("data",async()=>{
        const res = await fetch("localhost:8000");
        return res.json()
    }

    if(status==="success"){
        setState({data})
    }else{
        return(<p>Doing</p>)
    }
}

此代码会导致无限循环，渲染不断循环进行。

我认为这是因为setState导致useQuery再次执行再次设置状态等等。

任何帮助表示赞赏。我想将从 useQuery 获得的数据存储在状态变量中。

TIA。

#include <stdio.h>
#include <stdlib.h>

typedef struct StupidAssignment{
    long length;
    char* destination_ip;
    char* destination_port;
    long timestamp;
    long uid;
    char* message;
}packet;

void main(){
    int number_of_packets=10;int i;
    packet* all_packets[number_of_packets];
    for(i=0;i<number_of_packets;i+=1)all_packets[i]=malloc(sizeof packet);
}

上面的代码片段不会编译，并出现以下错误：-

reciever.c: In function ‘main’:
reciever.c:16:64: error: expected expression before ‘packet’
  for(i=0;i<number_of_packets;i+=1)all_packets[i]=malloc(sizeof packet);

但是，以下代码确实可以编译：-

#include <stdio.h>
#include <stdlib.h>

typedef struct StupidAssignment{
    long length;
    char* destination_ip;
    char* destination_port;
    long timestamp;
    long uid;
    char* message;
}packet;

void main(){
    int number_of_packets=10;int i;
    packet* all_packets[number_of_packets];
    for(i=0;i<number_of_packets;i+=1)all_packets[i]=malloc(sizeof(packet));
}

唯一的区别是sizeof(packet)和sizeof packet。

在上一个答案中，我了解到这 …

我编写了以下控制器来生成 PreSigned S3 上传链接。

case class S3Controller(private val s3Config: S3Config, private val awsConfig: AwsConfig) {
  val URL_TIMEOUT_IN_MILLIS: Long = 60 * 5 * 1000

  def getPreSignedURLForUpload(keyName: String): String = {
    val preSigner: S3Presigner = DefaultS3Presigner.builder()
      .serviceConfiguration(s3Config.s3Configuration)
      .credentialsProvider(awsConfig.awsCredentialsProvider).build()

    val objectRequest: PutObjectRequest = PutObjectRequest.builder()
      .bucket(s3Config.bucketName)
      .key(keyName)
      .contentType("text/plain")
      .build()

    val preSignRequest: PutObjectPresignRequest = PutObjectPresignRequest.builder()
      .signatureDuration(Duration.ofMinutes(10))
      .putObjectRequest(objectRequest)
      .build()

    val preSignedRequest: PresignedPutObjectRequest = preSigner.presignPutObject(preSignRequest)


    val myURL: String = preSignedRequest.url().toString

    myURL
  }
}

我使用的配置对象

case class S3Config (
                      bucketName: String,
                      s3Configuration: S3Configuration
                    )

case class AwsConfig …

def merge[X](list1: Option[List[X]], list2: Option[List[X]]): Option[List[X]]  
def merge[X](list1: Option[List[X]], elem: Option[X]): Option[List[X]]

编译器说这两个函数在擦除后具有相同的类型

def merge[X](list1: Option[List[X]],list2: Option[List[X]]): Option[List[X]] at line 122 and
def merge[X](list1: Option[List[X]],elem: Option[X]): Option[List[X]] at line 131
have same type after erasure: (list1: Option, list2: Option)Option

为什么擦除后Option[List[X]]和Option[X]变得相同？我怎样才能让它们与众不同？