为什么我会收到此段错误(核心转储)?我最初被告知这是因为我的指针 p_p_tictactoe = new char*[cols],但我被告知的是不对的。每组代码的用途在整个代码中都有注释。代码正在运行,但我得到了这个结果。我知道 for 循环必须是这个的主要问题。
Please enter a number of rows: 4
Please enter number of columns: 3
Enter a single character for position( << i << ): a
Enter a single character for position( << j << ): b
Segmentation fault (core dumped)
#include <iostream>
using namespace std;
int main()
{
// TODO: Below each Bp_tictactoe` of type pointer-to-a-pointer-to-a-char
char **p_p_tictactoe;
// 2. Prompt your user to enter a number of rows, then a number of columns.
// store their answers in the variables `rows` and `cols`.
char rows;
char cols;
cout << "Please enter a number of rows: ";
cin >> rows;
cout << "Please enter number of columns: ";
cin >> cols;
// 3. Allocate a 1-dimensional array of pointers-to-chars (length == `rows`)
// and store its address in `p_p_tictactoe`
p_p_tictactoe = new char*[rows];
// 4. Use a for-loop to allocate a dynamic array of chars (length == `cols`)
// for i from 0 to rows - 1 and store its address in `p_p_tictactoe[i]`.
for (int i = 0; i < rows - 1; i++)
{
p_p_tictactoe = new char*[cols];
}
// 5. Use a for-loop to prompt the user to enter a char for each position in
// (" << i << ", " << j << "): "
// As you read each char, store it in the array.
// 6. Use a nested for-loop to print the array, one row per line. The chars
// for each row should be space-separated. For example, if the array is
// 2 x 3 and stores the values A, B, C, X, !, &, the output should look
// like:
// A B C
// X ! &\
char new_input1;
char new_input2;
for (int i = 0; i < rows; i++)
{
for (int j = 0; j < cols; j++)
{
cout << "Enter a single character for position( << i << ): ";
cin >> new_input1;
cout << "Enter a single character for position( << j << ): ";
cin >> new_input2;
*p_p_tictactoe[i] = new_input1;
*p_p_tictactoe[j] = new_input2;
cout << *p_p_tictactoe[i] <<endl;
}
}
// *** Prevent memory leaks by deallocating dynamic memory when you are done
// using it. ***
// 7. Use a for-loop to delete each row of the dynamic array.
// 8. Delete the pointer-to-a-pointer to release the array of row pointers,
// and set its value to NULL to avoid accessing invalid memory.
for (int i = 0; i < 3; i++)
{
delete[] p_p_tictactoe[i];
delete[] p_p_tictactoe;
}
cout << "Bye!" << endl;
return 0;
}
最佳答案
除其他外,您对 p_p_tictactoe
的分配不正确。这是一个双指针,这仅仅意味着它是一个指向指针数组的指针。你的两步分配是正确的想法,但你在 for 循环中的内容是不正确的。在 p_p_tictactoe = new char*[rows]
行之后,您现在有一个指向 char*
类型数组的指针。因此,如果 rows
为 4,则内存中的内容现在如下所示:
p_p_tictactoe[0] == char* --> junk
[1] == char* --> junk
[2] == char* --> junk
[3] == char* --> junk
您现在必须遍历这 4 个 char*
中的每一个并为它们分配空间。每个 char*
必须指向一个 chars
数组。这是注释为您提供有关循环和使用 p_p_tictactoe[i]
索引的提示的地方:
for (int i = 0; i < rows - 1; i++)
{
p_p_tictactoe[i] = new char[cols];
}
现在,对于 cols == 3
,在内存中你有:
p_p_tictactoe[0] == char* --> 3 consecutive bytes
[1] == char* --> 3 consecutive bytes
[2] == char* --> 3 consecutive bytes
[3] == char* --> 3 consecutive bytes
您发布的代码是内存泄漏。每次执行 p_p_tictactoe = new char*[#]
时,操作系统都会进入堆以获得足够的内存来分配。您没有跟踪先前的指针,也没有先释放它,因此分配的内存现在没有指向它的东西。
同样的理论也适用于释放内存。你最后得到的不太正确。解除分配始终是分配的镜像。这显然是一项家庭作业,所以我不会发布该代码,但它与分配相同,只是相反。
我强烈推荐使用 gdb
,这是一个用于 linux 的文本调试器(或任何等效的调试器)。如果你希望在 C/C++ 中成功编码,你必须了解内存在堆栈和堆上的工作方式,你必须学习如何正确管理它,否则你将陷入痛苦的境地。 gdb
起初有点令人生畏,但它会让您打印出内存地址并检查内存,这对学习和强化您认为自己知道的内容非常有帮助。
关于c++ - 分段故障。是因为坏指针吗?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35423603/