Question

我试图从用户读取一个字符串，将其更改为一个int数组，然后通过int显示它。这样我可以获得非常大的int数，但是当我运行它时，它允许我输入一个数字，但是它会给我一个错误，调试断言失败并且我应该中止，重试或忽略。有一个无效的空指针。我不知道是什么意思。这就是我所拥有的：

  //LargeInt.h File
  #pragma once
  #include <iostream>
  #include <string>

  using namespace std;

  class LargeInt
  {
  private:
  string number1;
  string number2;
  string sum;

  public:
  LargeInt();
  ~LargeInt();


  //This function will take the 
  //the number entered by user 
  //as int digits and print out
  //each digit. 
 void ReadNumber(int[]);
 };


 //LargeInt.cpp file
 #include "stdafx.h"
 #include "LargeInt.h"
#include <iostream>

 LargeInt::LargeInt() {     
 number1 = " ";
  }


 LargeInt::~LargeInt() { 
 }


  //This function will take the 
  //the number entered by user 
  //as int digits and print out 
  //each digit.  
 void LargeInt::ReadNumber(int
     number[]) {

    for (int i = 0; i < 75; i++) {      
       cout << number[i];   }

     }

 //Main File
 #include "stdafx.h"
 #include <string>
 #include "LargeInt.h"
 #include <iostream>

 using namespace std;

 int main()
{
string number1 = " ";
string number2 = " ";
string sum = " ";
int summ[75];

//Get number 1 from user and output it
int numberOne[76] = {};
cout << "Enter first number: ";
getline(cin, number1);
cout << endl;

//Check to see if it is more than 75 digits
if (number1.length() > 75) {
    cout << "Invalid number!" << endl;
}
else {
    int length1 = number1.length();
    cout << "First Number: " << endl;
    int k = 75;
    for (int i = length1; i >= 0; i--) {
        numberOne[k] = number1[i] - 48;
        k--;
    }
}
LargeInt num1;
num1.ReadNumber(numberOne);
cout << endl;

system("pause");
return 0;
}

Answer 1

int length1 = number1.length();

是否将length1设置为字符串的大小。然后在

的第一次迭代中

for (int i = length1; i >= 0; i--) {
    numberOne[k] = number1[i] - 48;
    k--;
}

您正在使用number1[i]访问i = length1，因为字符串位置为0，您将在字符串末尾访问。这是未定义的行为，在这种情况下会抛出一个断言。要解决此问题，您需要将i设置为length1 - 1。

我收到调试断言失败的错误？

1 个答案: