我试图将用户输入的笔记存储在一个数组中。验证工作正常,但是当我在循环中输入最后一个值时,我得到:
Debug Assertion Failed!
Expression: "(_Ptr_user & (_BIG_ALLOCATION_ALIGNMENT - 1))==0"&&0
An invalid parameter was passed to a function that considers invalid parameters fatal.
我正在努力了解问题所在以及如何解决问题。
#include "stdafx.h"
#include <iostream>
#include <string>
using namespace std;
typedef string noteName;
noteName getNoteName(int i)
{
bool flag = true;
noteName Namein;
do
{
cout << "Please enter note name no. " << i + 1 << ": ";
cin >> Namein;
cout << "------------------------------------\n";
if (Namein.length() > 3 || Namein.length() < 2)
{
cout << "Sorry, a note name must be 2 or 3 characters long. Please try again.\n";
flag = false;
}
else if (Namein.length() == 3 && Namein[1] != '#')
{
cout << "Sorry, the second character of a sharp note name must be #. Please try again.\n";
flag = false;
}
else if ((Namein[0] < 'a' || Namein[0] > 'g') && (Namein[0] < 'A' || Namein[0] > 'G'))
{
cout << "Sorry, the first character of a note name must be a letter between A and G. Please try again.\n";
flag = false;
}
else if (isdigit(Namein.back()) == false)
{
cout << "Sorry, the last character of a note name must be a number. Please try again.\n";
flag = false;
}
else
{
flag = true;
}
} while (flag == false);
return Namein;
}
int main()
{
const int numNotes = 4;
noteName NoteNames[numNotes];
cout << "Hello\n";
for (int i = 0; i <= numNotes; i++)
{
NoteNames[i] = getNoteName(i);
}
cout << "Thank you, the note names and lengths you entered were: \n\n";
for (int i = 0; i <= numNotes; i++)
{
cout << i << ". " << NoteNames[i] << "\n";
}
cout << "Done!";
return 0;
}
我想说这与getNoteName()
string
返回类型有关,因为我还没有回复{{{{{{{ 1}}。
答案 0 :(得分:1)
noteName NoteNames[numNotes];
定义了一个数组,其中NoteNames[numNotes - 1]
是您可以访问的最大元素。
你走得更远。这样做的行为是 undefined ,它表现为你观察到的崩溃。
使用for (int i = 0; i < numNotes; i++)
或类似内容替换您的循环限制。
(您还有类别名称和变量名称的CamelCase约定从正常情况切换,这会使您的代码难以阅读。)
(我还希望看到constexpr int numNotes = 4;
:Google了解详情。)