typedef struct Tuple
{
int row;
int col;
int data;
} Tuple;
int** createMatrix(int r, int c)
{
int** mat = (int**)malloc(sizeof(int*) * r);
for (int i = 0; i < r; i++)
mat[i] = malloc(sizeof(int) * c);
return mat;
}
Tuple* createTuple(int** mat, int r, int c)
{
int count = 0;
for (int i = 0; i < r; i++)
for (int j = 0; j < c; j++)
if (mat[i][j])
count++;
Tuple* tuple = (Tuple*)malloc(sizeof(Tuple) * count);
tuple[0].col = c;
tuple[0].row = r;
tuple[0].data = count;
int k = 1;
for (int i = 0; i < r; i++)
for (int j = 0; j < c; j++)
if (mat[i][j])
{
tuple[k].col = j;
tuple[k].row = i;
tuple[k++].data = mat[i][j];
}
return tuple;
}
Tuple* fastTranspose(Tuple* tuple)
{
Tuple* trans = (Tuple*)calloc((tuple[0].data + 1),sizeof(Tuple) );
trans[0].col = tuple[0].row;
trans[0].row = tuple[0].col;
trans[0].data = tuple[0].data;
int* rowTerms = (int*)malloc(sizeof(int) * tuple[0].col);
int* startingPos = (int*)malloc(sizeof(int) * tuple[0].col);
for (int i = 0; i < tuple[0].col; i++)
rowTerms[i] = 0;
for (int i = 1; i <= tuple[0].data; i++)
rowTerms[tuple[i].col]++;
startingPos[0] = 1;
for (int i = 1; i < tuple[0].col; i++)
startingPos[i] = startingPos[i - 1] + rowTerms[i - 1];
for (int i = 1; i <= tuple[0].data; i++)
{
int j = startingPos[tuple[i].col]++;
trans[j].col = tuple[i].row;
trans[j].row = tuple[i].col;
trans[j].data = tuple[i].data;
}
free(rowTerms);
free(startingPos);
return trans;
}
void printTuple(Tuple* tuple)
{
printf("nRow Col Val:n");
for (int i = 0; i <= tuple[0].data; i++)
{
printf("%3d%4d%4dn", tuple[i].row, tuple[i].col, tuple[i].data);
}
}
int main()
{
int** mat = createMatrix(3, 4);
printf("Enter data:n");
for (int i = 0; i < 3; i++)
for (int j = 0; j < 4; j++)
scanf("%d", &mat[i][j]);
Tuple* tuple = createTuple(mat, 3, 4);
printf("Original tuple:");
printTuple(tuple);
Tuple* trans = fastTranspose(tuple);
printf("Tuple transpose:");
printTuple(trans);
free(trans);
free(tuple);
for (int i = 0; i < 3; i++)
free(mat[i]);
free(mat);
}
为什么断点是在函数fastTranspose的第一行触发的，尽管我没有在那里放一个？
我研究过类似的问题，但没能找到解决办法。这是一个求矩阵转置的程序。我们首先动态地分配一个二维矩阵。然后将其转换为稀疏矩阵表示形式，该形式使用结构来存储行、列和数据的值。我尝试打印矩阵和元组的值。它奏效了。但是当进入快速转置函数时，会突然触发一个断点。
我们通常自己在visual studio中设置一个断点。为什么IDE会放一个？
Visual Studio在那里放置了一个断点以提供帮助，因为您在那个地方遇到了错误。如果我在最近的Linux(Mint 20(上使用最近的clang(12.0.0(，我会得到以下输出：
$ echo -e '1n2n3n4n5n6n7n8n9n10n11n12' | ./so_tmat
Enter data:
Original tuple:
Row Col Val:
3 4 12
0 0 1
0 1 2
0 2 3
0 3 4
1 0 5
1 1 6
1 2 7
1 3 8
2 0 9
2 1 10
2 2 11
2 3 12
tuple[0].data + 1 = 13, sizet(Tuple) = 12
so_tmat: malloc.c:2379: sysmalloc: Assertion `(old_top == initial_top (av) && old_size == 0) || ((unsigned long) (old_size) >= MINSIZE && prev_inuse (old_top) && ((unsigned long) old_end & (pagesize - 1)) == 0)' failed.
Aborted (core dumped)
GDB(GNU调试器(的回溯指向与VS相同的行。
如果我(你的里程数可能会有所不同，因为：谷歌(将其原样复制并粘贴到谷歌中，我会得到为什么我会得到C malloc断言失败？作为第一支安打。公认的答案指向正确的方向，我们将向Valgrind寻求一些提示。
$ echo -e '1n2n3n4n5n6n7n8n9n10n11n12' | valgrind --track-origins=yes --leak-check=full --show-leak-kinds=all ./so_tmat
==16047== Memcheck, a memory error detector
==16047== Copyright (C) 2002-2017, and GNU GPL'd, by Julian Seward et al.
==16047== Using Valgrind-3.15.0 and LibVEX; rerun with -h for copyright info
==16047== Command: ./so_tmat
==16047==
Enter data:
==16047== Invalid write of size 8
==16047== at 0x40144D: createTuple (so_tmat.c:38)
==16047== by 0x40144D: main (so_tmat.c:89)
==16047== Address 0x4a796a0 is 0 bytes after a block of size 144 alloc'd
==16047== at 0x483B7F3: malloc (in /usr/lib/x86_64-linux-gnu/valgrind/vgpreload_memcheck-amd64-linux.so)
==16047== by 0x401377: createTuple (so_tmat.c:28)
==16047== by 0x401377: main (so_tmat.c:89)
==16047==
==16047== Invalid write of size 4
==16047== at 0x401452: createTuple (so_tmat.c:39)
==16047== by 0x401452: main (so_tmat.c:89)
==16047== Address 0x4a796a8 is 8 bytes after a block of size 144 alloc'd
==16047== at 0x483B7F3: malloc (in /usr/lib/x86_64-linux-gnu/valgrind/vgpreload_memcheck-amd64-linux.so)
==16047== by 0x401377: createTuple (so_tmat.c:28)
==16047== by 0x401377: main (so_tmat.c:89)
==16047==
...
我们可以在这里停止，因为我们找到了原因：您有缓冲区溢出，因为您在createMatrix(int r, int c)中从count = 0开始，这导致缺少最后一个Tuple所需的内存。将其更改为count = 1，它应该可以工作。
让我们检查一下：
$ echo -e '1n2n3n4n5n6n7n8n9n10n11n12' | valgrind --track-origins=yes --leak-check=full --show-leak-kinds=all ./so_tmat
==16215== Memcheck, a memory error detector
==16215== Copyright (C) 2002-2017, and GNU GPL'd, by Julian Seward et al.
==16215== Using Valgrind-3.15.0 and LibVEX; rerun with -h for copyright info
==16215== Command: ./so_tmat
==16215==
Enter data:
Original tuple:
Row Col Val:
3 4 13
0 0 1
0 1 2
0 2 3
0 3 4
1 0 5
1 1 6
1 2 7
1 3 8
2 0 9
2 1 10
2 2 11
2 3 12
==16215== Invalid read of size 4
==16215== at 0x4014A0: printTuple (so_tmat.c:78)
==16215== by 0x4014A0: main (so_tmat.c:91)
==16215== Address 0x4a796b4 is 8 bytes after a block of size 156 alloc'd
==16215== at 0x483B7F3: malloc (in /usr/lib/x86_64-linux-gnu/valgrind/vgpreload_memcheck-amd64-linux.so)
==16215== by 0x40137A: createTuple (so_tmat.c:28)
==16215== by 0x40137A: main (so_tmat.c:89)
...
仍然存在一个问题，现在它在printTuple(Tuple* tuple)中，因为我们更改了count的初始化，但没有调整tuple[0].data中的值，所以for (int i = 0; i <= tuple[0].data; i++)行中的上限太大。在这里或那里纠正它，没有太大关系，我会在循环中纠正上限，因为tuple[0].data包含正确数量的非零条目。因此，将循环更改为for (int i = 0; i < tuple[0].data; i++)
再次检查：
$ echo -e '1n2n3n4n5n6n7n8n9n10n11n12' | valgrind --track-origins=yes --leak-check=full --show-leak-kinds=all ./so_tmat
==16328== Memcheck, a memory error detector
==16328== Copyright (C) 2002-2017, and GNU GPL'd, by Julian Seward et al.
==16328== Using Valgrind-3.15.0 and LibVEX; rerun with -h for copyright info
==16328== Command: ./so_tmat
==16328==
Enter data:
Original tuple:
Row Col Val:
3 4 13
0 0 1
0 1 2
0 2 3
0 3 4
1 0 5
1 1 6
1 2 7
1 3 8
2 0 9
2 1 10
2 2 11
2 3 12
tuple[0].data + 1 = 14, sizet(Tuple) = 12
==16328== Invalid read of size 4
==16328== at 0x4015B1: fastTranspose (so_tmat.c:56)
==16328== by 0x4015B1: main (so_tmat.c:92)
==16328== Address 0x4a796b0 is 4 bytes after a block of size 156 alloc'd
==16328== at 0x483B7F3: malloc (in /usr/lib/x86_64-linux-gnu/valgrind/vgpreload_memcheck-amd64-linux.so)
==16328== by 0x40137A: createTuple (so_tmat.c:28)
==16328== by 0x40137A: main (so_tmat.c:89)
==16328==
==16328== Invalid read of size 4
==16328== at 0x4017D0: fastTranspose (so_tmat.c:63)
==16328== by 0x4017D0: main (so_tmat.c:92)
==16328== Address 0x4a796b0 is 4 bytes after a block of size 156 alloc'd
==16328== at 0x483B7F3: malloc (in /usr/lib/x86_64-linux-gnu/valgrind/vgpreload_memcheck-amd64-linux.so)
==16328== by 0x40137A: createTuple (so_tmat.c:28)
==16328== by 0x40137A: main (so_tmat.c:89)
==16328==
==16328== Invalid read of size 4
==16328== at 0x4017DF: fastTranspose (so_tmat.c:64)
==16328== by 0x4017DF: main (so_tmat.c:92)
==16328== Address 0x4a796ac is 0 bytes after a block of size 156 alloc'd
==16328== at 0x483B7F3: malloc (in /usr/lib/x86_64-linux-gnu/valgrind/vgpreload_memcheck-amd64-linux.so)
==16328== by 0x40137A: createTuple (so_tmat.c:28)
==16328== by 0x40137A: main (so_tmat.c:89)
==16328==
==16328== Invalid read of size 4
==16328== at 0x4017ED: fastTranspose (so_tmat.c:66)
==16328== by 0x4017ED: main (so_tmat.c:92)
==16328== Address 0x4a796b4 is 8 bytes after a block of size 156 alloc'd
==16328== at 0x483B7F3: malloc (in /usr/lib/x86_64-linux-gnu/valgrind/vgpreload_memcheck-amd64-linux.so)
==16328== by 0x40137A: createTuple (so_tmat.c:28)
==16328== by 0x40137A: main (so_tmat.c:89)
...
看看有问题的行(你的行号可能不同(：这是和以前一样的上限问题。更改并再次检查。
$ echo -e '1n2n3n4n5n6n7n8n9n10n11n12' | valgrind --track-origins=yes --leak-check=full --show-leak-kinds=all ./so_tmat
==16420== Memcheck, a memory error detector
==16420== Copyright (C) 2002-2017, and GNU GPL'd, by Julian Seward et al.
==16420== Using Valgrind-3.15.0 and LibVEX; rerun with -h for copyright info
==16420== Command: ./so_tmat
==16420==
Enter data:
Original tuple:
Row Col Val:
3 4 13
0 0 1
0 1 2
0 2 3
0 3 4
1 0 5
1 1 6
1 2 7
1 3 8
2 0 9
2 1 10
2 2 11
2 3 12
tuple[0].data + 1 = 14, sizet(Tuple) = 12
Tuple transpose:
Row Col Val:
4 3 13
0 0 1
0 1 5
0 2 9
1 0 2
1 1 6
1 2 10
2 0 3
2 1 7
2 2 11
3 0 4
3 1 8
3 2 12
==16420==
==16420== HEAP SUMMARY:
==16420== in use at exit: 0 bytes in 0 blocks
==16420== total heap usage: 10 allocs, 10 frees, 5,548 bytes allocated
==16420==
==16420== All heap blocks were freed -- no leaks are possible
==16420==
==16420== For lists of detected and suppressed errors, rerun with: -s
==16420== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 0 from 0)
很好，没有问题了。至少对于那个精确的输入。但我在calloc()之前挤了一点printf("tuple[0].data + 1 = %d, sizet(Tuple) = %zun",tuple[0].data + 1, sizeof(Tuple) );，上面写着
tuple[0].data + 1 = 14, sizet(Tuple) = 12
这是一个Tuple太多；保存一些内存并通过删除CCD_ 15中的CCD_。就是这样。
我在这里使用了Unix工具，其中大多数都适用于Windows，但VS也有所有必要的工具，它们只是有不同的名称，使用方式也不同。试着找到它们，并与它们一起检查您的原始代码(您现在知道其中的错误(，以熟悉该工具链。你真的需要这些知识！