Question

我在C ++中执行以下程序，但是没有编译。请帮助找到问题

**Here is my gradle code:**

import org.jooq.util.jaxb.*
import org.jooq.util.*

buildscript {
  repositories {
    jcenter()
  }
  dependencies {
    classpath 'com.h2database:h2:1.4.186'   
    classpath 'org.jooq:jooq-codegen:3.9.1' 
    classpath 'com.microsoft.sqlserver:sqljdbc4:4.0' 
   classpath "io.ratpack:ratpack-gradle:0.9.7-SNAPSHOT"
    classpath 'com.github.jengelman.gradle.plugins:shadow:1.0.2'

  }
}
plugins {
  id 'io.ratpack.ratpack-java' version '1.3.3'
}

repositories {
  jcenter()
}

dependencies {
  compile ratpack.dependency('hikari') 
  compile 'com.h2database:h2:1.4.186' 
  compile 'org.jooq:jooq:3.9.1' 
  compile 'com.microsoft.sqlserver:sqljdbc4:4.0' 
  compile "io.ratpack:ratpack-gradle:0.9.7-SNAPSHOT"
  compile 'com.github.jengelman.gradle.plugins:shadow:1.0.2'

}
mainClassName = 'App'

task jooqCodegen {
  doLast {
    String init = "<directory>/init.sql"
    Configuration configuration = new Configuration()
      .withJdbc(new Jdbc()
      .withDriver("com.microsoft.sqlserver.jdbc.SQLServerDriver") 
      .withUrl("jdbc:sqlserver://<server address>;DatabaseName=JOOQ; INIT=RUNSCRIPT FROM '$init'") 
      .withUser("****")
      .withPassword("****")
    )
      .withGenerator(new Generator()
        .withDatabase(new Database()
         .withName("org.jooq.util.sqlserver.SQLServerDatabase")
          .withIncludes(".*")
          .withExcludes("")
          .withInputSchema("dbo")
        )
      .withTarget(new Target()
        .withDirectory("$projectDir/src/main/java") 
        .withPackageName("jooq"))) 
    GenerationTool.generate(configuration)
  }
}

**Note :** Added one create table statement in init.sql file

它给出了以下编译错误：

#include<iostream>
class A;
void showA(A& x) {
    // Since showA() is a friend, it can access
    // private members of A
    std::cout << "A::a=" << x.a;
}
class A
{
    int a;
public:
    A() {a = 0;}
    friend void showA(A& x); // global friend function
};



int main()
{
    A a;
    showA(a);
    return 0;
}

Answer 1

class A;
void showA(A& x) {
    // Since showA() is a friend, it can access
    // private members of A
    std::cout << "A::a=" << x.a;
}

我们不知道A会成为朋友。我们甚至不知道A会有会员。您无法访问不完整类型的成员。尝试这样做是无效使用不完整类型。

解决方案：在访问成员之前完成类型（即定义类）。

功能顺序在C ++中不明确？

1 个答案: