具有指針的C ++程序不起作用

Question

我是一個初學者，被困住了。 我已經寫了這篇文章，到目前為止還行不通。 在“添加或刪除交易者”之后，它什么也不做。 任何幫助或花樣如何使這一功能將不勝感激。 謝謝。

#include<iostream>
#include<iomanip>
#include<string>
using namespace std;

struct Department{
string deptName;
int numTraders;     
};

void addTraders(Department *, int );
void removeTraders(Department *, int);

int main(){

char addOrRemove;
Department departments[10] = {
    {"Bank Loan", 10},
    {"Conservative Allocation", 9},
    {"Europe Stock", 10},
    {"Domestic", 21},
    {"Asia", 10},
    {"Large Growth", 5},
    {"Long-term Bond", 5},
    {"Money Market", 25},
    {"Emerging Market", 18},
    {"Large Blend", 12}
};

int choice, numberToAdd, numberToRemove;

Department* p_departments = departments;

for(int i = 0; i < 10; i++){
    cout << "Department # " << (i + 1) << ", Name: " << p_departments[i].deptName <<
        ", Traders: " << p_departments[i].numTraders << endl;
}
cout << endl;

do{

cout << "Enter 0 to quit, or choose a department number: ";
cin >> choice;

cout << "Add or remove traders (A or R) ? ";
cin >> addOrRemove;

if(addOrRemove == 'A' || 'a'){
    cout << "how many traders to add" << endl;
    cin >> numberToAdd;
    addTraders(&departments[choice-1] ,numberToAdd);
}
else if(addOrRemove == 'R' || 'r'){
    cout << "how many traders to remove" << endl;
    cin >> numberToRemove;
    removeTraders(&departments[choice-1],numberToRemove);
}
else{
    cout << addOrRemove << " is not a valid selection. \n";
}

for(int i = 0; i < 10; i++){
    cout << "Department # " << (i + 1) << ", Name: " << p_departments[i].deptName <<
        ", Traders: " << p_departments[i].numTraders << endl;
}
cout << endl;

}while(count != 0);

system("pause");
return 0;
}

void addTraders(Department *dept, int numAdd){

dept->numTraders += numAdd;
}

void removeTraders(Department *dept, int numRemove){

dept->numTraders += numRemove;
} 

Answer 1

以下條件始終被評估為true ； 即使是false || 'a' false || 'a' ， 'a' a'〜> true ：

if(addOrRemove == 'A' || 'a'){ ...

這是命中注定的：

if(addOrRemove == 'A' || addOrRemove == 'a'){ ...

但是，當addOrRemove聲明為char ，則：

cin >> addOrRemove;

可能只是讀取換行符或空白。 將addOrRemove聲明為std::string並將條件更改為：

if(addOrRemove == "A" || addOrRemove == "a"){ ...

在您讀取了choice並且它是0 ，您應該break循環，以便它不會嘗試訪問索引0 - 1元素：

cin >> choice;
if (choice == 0) break;   // <-- THIS

Answer 2

首先代替

if(addOrRemove == 'A' || 'a'){

你應該寫

if(addOrRemove == 'A' || addOrRemove == 'a'){

其次，您應該定義變量count，因為似乎編譯器認為count是標准算法std :: count的名稱。