我最近一直致力于一个以名字作为输入的程序,最终会对其进行排序。二进制搜索它们。但是,在尝试使数组成为动态大小(每次循环迭代时会增加一个)时,它会遇到各种问题。
我可以使字符串数组由20个元素组成,并且程序可以正常工作,但我的任务的额外功劳是使其成为动态大小。目前程序在达到" getline(cin,Names [x]);"时没有任何错误代码崩溃。 我一直在四处搜寻,我知道在这种情况下我更容易做一个向量而不是一个数组,但是我不相信我允许使用向量关于这项任务。
由于
原始代码
using namespace std;
#include <iostream>
#include <string>
void main()
{
int x = 0;
string * Names = new string[x];
bool NameInputEnd(0);
cout << " Enter your names to be sorted\n";
cout << "To exit just press [Enter] at any time\n";
do
{
cout << x << endl;
cout << "\n< Name " << (x + 1) << " > = ";
!!**CRASHES HERE**!!
getline(cin, Names[x]);
if (Names[x].empty() || x == 19)
{
cout << "\nFinal Name Amount = " << (x + 1) << endl << endl;
NameInputEnd = 1;
continue;
}
x++;
} while (NameInputEnd == 0);
delete [] Names;
}
更改
int tempsize(1), x(0);
string * Names = new string[tempsize];
...
do
{
...
x++;
tempsize++;
}while (NameInputEnd == 0);
答案 0 :(得分:1)
创建数组后,无法调整其大小。您必须销毁它并使用现有数据的副本创建一个新数组。例如:
Accept
你真的应该使用#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
void main()
{
int x = 0;
int capacity = 20;
string * Names = new string[capacity];
string Name;
cout << " Enter your names to be sorted\n";
cout << "To exit just press [Enter] at any time\n";
do
{
cout << x << endl;
cout << "\n< Name " << (x + 1) << " > = ";
if ((!getline(cin, Name)) || Name.empty())
break;
if (x == capacity)
{
int newCapacity = capacity + 20;
string *newNames = new string[newCapacity];
copy(Names, Names + x, newNames);
delete [] Names;
Names = newNames;
capacity = newCapacity;
}
Names[x] = Name;
++x;
}
while (true);
cout << "\nFinal Name Amount = " << x << endl << endl;
delete [] Names;
}
,但是:
std::vector