Chi*_*ffa 1 c++ string assertion
我的代码在MS VS 2012中编译后,接受输入,然后崩溃并使用以下报告:
Debug Assertion Failed!
...\include\xstring
Line:1662
Expression:string subscript out of range
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代码如下:
#include "stdafx.h"
#include <iostream>
#include <string>
#include <vector>
#include <cassert>
#include <time.h>
using namespace std;
const unsigned short MAX_STRINGS = 10;
const unsigned int MAX_SIZE=10000;
vector<string> strings;
unsigned int len;
string GetLongestCommonSubstring( string string1, string string2 );
inline void readNumberSubstrings();
inline const string getMaxSubstring();
void readNumberSubstrings()
{
cin >> len;
assert(len > 1 && len <=MAX_STRINGS);
strings.resize(len);
for(unsigned int i=0; i<len;i++)
strings[i]=string(MAX_SIZE,0);
for(unsigned int i=0; i<len; i++)
cin>>strings[i];
}
const string getMaxSubstring()
{
string maxSubstring=strings[0];
long T=clock();
for(unsigned int i=1; i < len; i++)
maxSubstring=GetLongestCommonSubstring(maxSubstring, strings[i]);
cout << clock()-T << endl;
return maxSubstring;
}
string GetLongestCommonSubstring( string string1, string string2 )
{
const int solution_size = string2.length()+ 1;
int *x=new int[solution_size]();
int *y= new int[solution_size]();
int **previous = &x;
int **current = &y;
unsigned int max_length = 0;
unsigned int result_index = 0;
unsigned int j;
unsigned int length;
int J=string2.length() - 1;
for(unsigned int i = string1.length() - 1; i >= 0; i--)
{
for(j = J; j >= 0; j--)
{
if(string1[i] != string2[j])
(*current)[j] = 0;
else
{
length = 1 + (*previous)[j + 1];
if (length > max_length)
{
max_length = length;
result_index = i;
}
(*current)[j] = length;
}
}
swap(previous, current);
}
string1[max_length+result_index]='\0';
return &(string1[result_index]);
}
int main()
{
readNumberSubstrings();
cout << getMaxSubstring() << endl;
return 0;
}
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我想它必须是非常简单的东西(比如一些循环条件或类似的东西)导致它崩溃,但我没有看到它.
你的问题在于使用unsigned int.看看这一行:
for(j = J; j >= 0; j--)
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这里,在j变为0之后,它会减少.但是你的j是unsigned,如此而不是成为-1(并退出对下一次迭代周期),j成为4158584613和循环下去,试图在索引访问明显超出债券的元素4158584613在该行:
(*current)[j] = 0;
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制作i,j以及length签署将解决你的问题.