为什么最后bankaccounttest函数中使用bankaccount account而不是bankaccount *account呢？ #include <stdio.h> typedef struct { double balance; } bankaccount; //开户 void openaccount(bankaccount *account) { account->balance = 0; printf("开户成功"); } void deposit(bankaccount *account, double amount) { account->balance += amount; printf("存款 %.2f 成功", amount); } void withdraw(bankaccount *account, double amount) { if (account->balance >= amount) { account->balance -= amount; printf("取款 %.2f 成功", amount); } else { printf("取款
在 `bankaccountTest` 函数中使用 `bankaccount account` 而不是 `bankaccount *account` 可能取决于具体的设计和需求。 如果使用 `bankaccount account` ，这意味着在函数内部操作的是 `bankaccount` 类型的对象本身，而不是通过指针来间接操作。 这样做的优点可能是代码更直观，不需要解引用指针来访问成员。但如果需要在函数内部修改传入对象的状态并在函数外部反映这些修改，通常会使用指针（`bankaccount *account`），因为通过值传递对象时，对其的修改不会影响到函数外部的原始对象。 如果在这个特定的函数中不需要修改传入对象的状态或者只是读取其数据，或者这个对象本身规模较小，值传递（使用 `bankaccount account` ）可能是一个合适的选择。 要确切地理解为什么在这个特定情况下使用了值传递而不是指针传递，还需要查看这个函数的具体实现和功能需求。 [2024-08-08 15:48:26 | AI写代码神器 | 218点数解答]
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