Question

我遇到一种情况，我想用不是重复的元素（设置相同）来满足数组，因此我写了一个简单的函数，以便在插入之前比较地址。

功能非常简单，但无法正常工作，因此我不得不寻求帮助...谢谢。

// function to prevent insertion of duplicates in array
function isDuplicate() public view returns (bool){
    for(uint i = 0; i < addressIndices.length; i++){
        if(addressIndices[i] == msg.sender) return true;
        else return false;
  }
}

为了使问题更完整，我将为您提供更多信息：

第一：主要结构：

struct Match{
    bytes32 teamHome;
    bytes32 teamAway;
    uint teamHomeGoals;
    uint teamAwayGoals;
    uint  oddsTeamHome;
    uint  oddsTeamAway;
    uint  oddsDraw;
    uint outcome; // outcome of a match ( possible values: '1', 'X', '2', mapped to, because of uint, as: '1', '2', '3');
}
struct Bet{
    address bettor; // address a of creator of a bet;
    bytes32 name; // name in a relation with ^ a provided address;
    uint amount; // deposit on bet;
    uint bet; // match index being bet on;
    uint outcome; // bet placed on a outcome; defined as: '1', 'X', '2', mapped to, because of uint, as: '1', '2', '3';

第二：下注的功能。

function placeBet(bytes32 name, uint _outcome, uint desiredMatchIndex, uint _amount) public payable{
// find a way to store a bid in order to be easily searchable, in order to easily send money to winners;
    //   require(!roundEnd, "Interactions with contract are locked, be careful next time!");
    //   require(state == State.Active, "Betting is over, game have already started!");
      require(msg.value > 0, "It isn't possible to place a bet without a money ");
    if(!isDuplicate()){
        addressIndices.push(msg.sender);
    }
      existingBets[msg.sender].push(Bet({
          bettor: msg.sender,
          name: name,
          amount: _amount,
          bet: desiredMatchIndex,
          outcome: _outcome
      }));
    // emit event, finally;
}

这个问题真的很奇怪：在用3个不同的地址进行3次下注之后，我要用作集合的数组长度为7。如果函数起作用，则应为3，否则应为9。

有什么主意吗？

编辑：isDuplicate（）函数仅适用于第一个插入的地址

Answer 1

isDuplicated（）问题已解决->

函数isDuplicate（address _address）公共视图返回（布尔）{

for(uint i = 0; i < addressIndices.length; i++){
    if(addressIndices[i] == _address) return true;
}
return false;

}

Answer 2

当前函数的外观永远不会超出数组的第一个元素。它立即返回true或false。请改用此方法，但是正如我在评论中指出的那样，mapping可以避免迭代：

// function to prevent insertion of duplicates in array
function isDuplicate() public view returns (bool) {
    for(uint i = 0; i < addressIndices.length; i++){
        if(addressIndices[i] == msg.sender) return true;
    }

    return false;
}

实体地址比较

2 个答案: