【发布时间】:2016-11-05 12:19:44
【问题描述】:
(首先sorry英语不好,因为我不擅长)
我已经为幻方写了一个代码,它工作得很好并且给出了真正的平方。
当它编译并运行程序时,没有任何问题。
但是当它到达main() 中的return 0 时,我得到了这个错误
在正常阻塞后检测到堆损坏 (#180)
我搜索了一下,发现我的square 类没有遵循三规则(What is The Rule of Three?)
所以我试了一下,又报错了。
这是我的代码。
#include<iostream>
using namespace std;
class square {
public:
//a n*n square which n=size
int size;
//a set of blocks in square which is declared
//by pointers cuz i couldn't simply write block[size][size]
int** block = new int*[size];
square(const int _size) {
size = _size;
for (int c = 0;c < size;c++) {
block[c] = new int[size];
}
//putting 0 value in all blocks
for (int i = 0; i < size;i++)
for (int j = 0;j < size;j++) {
block[i][j] = 0;
}
}
//distructor
~square() {
for (int c = 0; c < size; c++)
delete block[c];
delete block;
}
//copy constructor
square(const square& that) {
for (int c = 0; c < size; c++)
delete block[c];
delete block;
size = that.size;
block = new int*[size];
for (int c = 0;c < size;c++) {
block[c] = new int[size];
}
for (int i = 0; i < size;i++)
for (int j = 0;j < size;j++) {
block[i][j] = that.block[i][j];
}
}
//add or change the value of a block
void add_block(int &value, int i, int j) {
block[i][j] = value;
}
//get the value stored in a block
int get_block(int i, int j) {
return block[i][j];
}
//get the size of the square
int get_size() {
return size;
}
//display the square
void disp() {
for (int i = 0; i < size;i++) {
for (int j = 0;j < size;j++) {
cout << block[i][j]<<' ';
}
cout << endl;
}
}
};
//just to write clean code i made this function
//to get the size of the square from input
int intro() {
cout << "magic square" << endl;
cout << "what's the size of rows and columns?(just enter an integer and must be one of the following numbers)" << endl;
cout << "3 , 5 , 7 , ... , (2n+1)" << endl;
int s_size;
cin >> s_size;
while (s_size % 2 == 0) {
cout << "!!! must be one of the following numbers : 3 , 5 , 7 , ... , (2n+1) !!!" << endl;
cout << "enter again" << endl;
cin >> s_size;
}
return s_size;
}
//look for next valid block to add the next number
//and adding the next number
//the algorithm is go up.go left.if not filled before
//put new number. else go back to last position.go down.put the number.
void add_next_magic_block(square &s,int &v,int &i , int &j) {
int s_size = s.get_size();
int ii = i;
int jj = j;
if (i == 0)
i = s_size - 1;
else
i--;
if (j == s_size - 1)
j = 0;
else
j++;
if (s.get_block(i, j) == 0)
s.add_block(v, i, j);
else {
s.add_block(v, ++ii, jj);
i = ii;
j = jj;
}
}
//magic square which starts from [0][j/2]
//and ends when it reaches to the
//size*size
//which is the last number
void do_magic_square_on(square &s) {
int end = s.get_size()*s.get_size();
int count=1;
int i = 0;
int j = s.get_size() / 2;
s.add_block(count, 0, j);
count++;
while (count <= end) {
add_next_magic_block(s, count, i, j);
count++;
}
}
int main() {
int s_size = intro();
square my_s(s_size);
do_magic_square_on(my_s);
my_s.disp();
system("pause");
return 0;
}
不是关于问答 这是编辑后的代码 这是运行新代码的链接 http://cpp.sh/72mmg
#include<iostream>
using namespace std;
class magic_square {
public:
int size;
int** block;
magic_square(int _size) :size(_size) , block(new int*[size]) {
size = _size;
for (int c = 0;c < size;c++) {
block[c] = new int[size];
}
for (int i = 0; i < size;i++)
for (int j = 0;j < size;j++) {
block[i][j] = 0;
}
}
~magic_square() {
for (int c = 0; c < size; c++)
delete block[c];
delete block;
}
magic_square(const magic_square& that) {
for (int c = 0; c < size; c++)
delete block[c];
delete block;
size = that.size;
block = new int*[size];
for (int c = 0;c < size;c++) {
block[c] = new int[size];
}
for (int i = 0; i < size;i++)
for (int j = 0;j < size;j++) {
block[i][j] = that.block[i][j];
}
}
void add_block(int &value, int i, int j) {
block[i][j] = value;
}
int get_block(int i, int j) {
return block[i][j];
}
int get_size() {
return size;
}
int sum_of_row(int row) {
int sum = 0;
int c = 0;
for (int c = 0; c < this->get_size(); c++)
sum += this->get_block(row, c);
return sum;
}
int sum_of_col(int col) {
int sum = 0;
int c = 0;
for (int c = 0; c < this->get_size(); c++)
sum += this->get_block(c, col);
return sum;
}
void disp() {
for (int i = 0; i < size;i++)
cout << '\t' << "["<<i+1<<"]";
cout <<'\t'<<"row sum"<< endl << endl;
for (int i = 0; i < size;i++) {
cout << "[" << i + 1 << "]";
for (int j = 0;j < size;j++) {
cout<<'\t'<< block[i][j];
}
cout <<'\t'<<sum_of_row(i) <<endl;
}
cout <<endl <<"col sum";
for (int i = 0; i < size;i++)
cout << '\t'<< sum_of_col(i);
cout << endl;
}
};
int intro() {
cout << "magic square" << endl;
cout << "what's the size of rows and columns?(just enter an integer and must be one of the following numbers)" << endl;
cout << "3 , 5 , 7 , ... , (2n+1)" << endl;
int s_size;
cin >> s_size;
while (s_size % 2 == 0) {
cout << "!!! must be one of the following numbers : 3 , 5 , 7 , ... , (2n+1) !!!" << endl;
cout << "enter again" << endl;
cin >> s_size;
}
return s_size;
}
void add_next_magic_block(magic_square &s,int &v,int &i , int &j) {
int s_size = s.get_size();
int ii = i;
int jj = j;
if (i == 0)
i = s_size - 1;
else
i--;
if (j == s_size - 1)
j = 0;
else
j++;
if (s.get_block(i, j) == 0)
s.add_block(v, i, j);
else {
s.add_block(v, ++ii, jj);
i = ii;
j = jj;
}
}
void normal_do_magic_square_on(magic_square &s) {
int end = s.get_size()*s.get_size();
int count = 1;
int i = 0;
int j = s.get_size() / 2;
s.add_block(count, 0, j);
//cout <<endl<< " <------STEP------> "<< "[" << count << "]" << endl<<endl;
//s.disp();
//cout << endl;
count++;
while (count <= end) {
add_next_magic_block(s, count, i, j);
//cout << endl << " <------STEP------> " << "["<<count<<"]" << endl << endl;
//s.disp();
//cout << endl;
count++;
}
}
void step_do_magic_square_on(magic_square &s) {
int end = s.get_size()*s.get_size();
int count=1;
int i = 0;
int j = s.get_size() / 2;
s.add_block(count, 0, j);
cout <<endl<< " <------STEP------> "<< "[" << count << "]" << endl<<endl;
s.disp();
cout << endl;
count++;
while (count <= end) {
add_next_magic_block(s, count, i, j);
cout << endl << " <------STEP------> " << "["<<count<<"]" << endl << endl;
s.disp();
cout << endl;
count++;
}
}
//to decide if show the steps or no
void out_put(magic_square &s) {
cout << "want to see the steps?(y,n)";
char in;
cin >> in;
if (in == 'n' || in == 'N') {
normal_do_magic_square_on(s);
s.disp();
}
else if(in=='y' || in=='Y')
step_do_magic_square_on(s);
}
int main() {
int s_size = intro();
magic_square my_s(s_size);
out_put(my_s);
system("pause");
return 0;
}
【问题讨论】:
-
请edit您的问题提供minimal reproducible example。
-
@BaummitAugen 我不知道您所说的完整是什么意思。它已经完成了。
-
虽然不是很小。
-
请评论您的代码。什么是next(/previous),为什么是
s.add_block(v, ++ii, jj);? -
@greybeard 我使用了一种以这种方式工作的算法。 first :将第一个数字放在 [0][j/2] 的位置。然后是 [i-1][j+1] 位置的下一个数字。因此,如果在将数字放在 [i+1][j] 的位置之前填充下一个块并继续直到正方形已满
标签: c++ heap-memory heap-corruption rule-of-three