我是C ++的新手。我正在阅读Bjarne Stroustrup撰写的C ++编程语言。
我的问题是:
另一个问题是,为什么在以下声明中没有Node* end(lst)
struct Node { Node* next; int data; };
Node* operator++(Node* p) { return p−>next; }
int operator*(Node* p) { return p−>data; }
Node* end(lst) { return nullptr; }
void test(Node* lst)
{
int s = sum<int*>(lst,end(lst));
}
与<?php
$keyword1 = $_POST["keyword1"]; // Warning! SQL Injection Security Issue!
$keyword2 = $_POST["keyword2"]; // Warning! SQL Injection Security Issue!
$keyword3 = $_POST["keyword3"]; // Warning! SQL Injection Security Issue!
$keyword4 = $_POST["keyword4"]; // Warning! SQL Injection Security Issue!
$keyword5 = $_POST["keyword5"]; // Warning! SQL Injection Security Issue!
$keyword6 = $_POST["keyword6"]; // Warning! SQL Injection Security Issue!
$keyword7 = $_POST["keyword7"]; // Warning! SQL Injection Security Issue!
$keyword8 = $_POST["keyword8"]; // Warning! SQL Injection Security Issue!
$where = NULL;
if( $keyword1 != NULL ) { $where .= "and hospital='$keyword1'"; }
if( $keyword2 != NULL ) { $where .= "and room='$keyword2'"; }
if( $keyword3 != NULL ) { $where .= "and equipmentNAME='$keyword3'"; }
if( $keyword4 != NULL ) { $where .= "and supplier='$keyword4'"; }
if( $keyword5 != NULL ) { $where .= "and brand='$keyword5'"; }
if( $keyword6 != NULL ) { $where .= "and quantity='$keyword6'"; }
if( $keyword7 != NULL ) { $where .= "and yearpurchased='$keyword7'"; }
if( $keyword8 != NULL ) { $where .= "and nxtyrPlan='$keyword8'"; }
$sql =" SELECT * FROM equipmentTable WHERE 1=1 $where";
?>
相关联:
import boto3
ecs_client = boto3.client('ecs')
logger = logging.getLogger()
def lambda_handler(event, context):
response = ecs_client.run_task(
cluster='your-cluster',
taskDefinition='your-task-name',
count=1,
launchType='EC2', # or fargate
)
logger.info(response)
response = {
"isBase64Encoded": "false",
"statusCode": 200,
"headers": { "COntent-Type": "application/json"},
"body": "hello"
}
return response
答案 0 :(得分:2)
operator++
和operator*
是重载运算符,它们只是特殊功能。当您实现自定义数据类型并决定支持增量(++
)和取消引用(*
)操作时,它们会派上用场。在发布这些微不足道的问题之前,cppreference应该是一个值得寻找的好地方。
问题的第二部分 - 您发布的内容会产生编译错误,实际签名应为Node* end(Node* lst)
。我不确定你从哪里得到它,也许作者不关心lst
的类型(假设从上下文推断),因为列表标记的末尾由null
表示(约定) )。