使用super向父类传递必填参数失败的问题排查

问题描述 #

子类调用super方法后无法将参数值传递给父类，编译时出现错误。

原代码 #

void main() {
  MobilePhone mp = new MobilePhone(modelName: "7T", brandName: "OnePlus");
}

class Electronics {
  // 父类
  double height = 56;
  double width = 56;
  double thickness = 56;

  Electronics({required String brandName}) {
    print("The Brand Name of this Mobile Phone is - $brandName");
  }
}

class MobilePhone extends Electronics {
  // 子类
  MobilePhone({required String modelName, required String brandName})
      : super(brandName) {
    print("The model of this mobile phone is - $modelName");
  }
}

编译错误信息 #

main.dart:19:14: Error: Too many positional arguments: 0 allowed, but 1 found.
Try removing the extra positional arguments.
      : super(brandName) {

错误原因 #

父类Electronics的构造函数使用了命名参数（参数被{}包裹），但子类调用super时，却以位置参数的形式传递brandName，违反了Dart的语法规则，因此触发报错。

解决方案 #

子类调用super时，必须以命名参数的形式传递参数，也就是在参数前加上参数名brandName:。

修正后的代码 #

void main() {
  MobilePhone mp = new MobilePhone(modelName: "7T", brandName: "OnePlus");
}

class Electronics {
  double height = 56;
  double width = 56;
  double thickness = 56;

  Electronics({required String brandName}) {
    print("The Brand Name of this Mobile Phone is - $brandName");
  }
}

class MobilePhone extends Electronics {
  MobilePhone({required String modelName, required String brandName})
      : super(brandName: brandName) { // 修改为命名参数传递方式
    print("The model of this mobile phone is - $modelName");
  }
}

运行修正后的代码，会正常输出：

The Brand Name of this Mobile Phone is - OnePlus
The model of this mobile phone is - 7T

内容的提问来源于stack exchange，提问作者Rajiv Iyer