c语言 - 操作 2D 数组时出现"Invisible"错误。(0 错误， 0 警告)

Question

应用程序崩溃，但我看不出原因。我已经检查了代码很多次，但仍然有一个问题。

我有一个二维数组'a'(NxM)，我必须对它进行转置。新数组为'p'(MxN)。

下面是程序的一部分:

/// 7: Transposing array. ( NxM ---> MxN ).   
int **p = NULL;
p = (int **)malloc(M*sizeof(int *));
if ( NULL == p)
{
    printf("Failed to allocate memory.");
    return 1;
}
for ( i = 0; i < M; i++ )
    p[i] = (int *)malloc(N*sizeof(int ));
// Filling in the new array called "p" with '0's.
for ( i = 0; i < M; i++ )
    for ( j = 0; j < N; j++ )
        p[i][j] = 0;
// Transposing array.
for ( i = 0; i < N; i++ )
{
    for ( j = 0; j < M; j++ )
    {
        a[i][j] = p[j][i];
    }
}
// Displaying ARRAY
printf(">>>nn");
for ( i = 0; i < M; i++ )
{
    for ( j = 0; j < N; j++ )
        printf("%4d ", p[i][j]);
    printf("n");
}

请告诉我出了什么事。

更新:我为我犯的错误道歉…

// Filling in the new array called "p" with '0's.
for ( i = 0; i < M; i++ )
    for ( j = 0; j < N; j++ )
        p[i][j] = 0;   // there will be 'p', not 'a'.

Answer 1

由于a的大小为N x M，因此问题来自此循环，您正在访问M x N元素而不是N x M

// Filling in the new array called "p" with '0's.
for ( i = 0; i < M; i++ )
    for ( j = 0; j < N; j++ )
        a[i][j] = 0;

这个循环应该是

for ( i = 0; i < N; i++ ) // M is swapped with N
    for ( j = 0; j < M; j++ ) // N is swapped with M
        a[i][j] = 0;

并且，从您的问题来看，a是需要转置到p的输入数组。因此，处理步骤应该是

// Transposing array.
for ( i = 0; i < N; i++ )
{ 
    for ( j = 0; j < M; j++ )
    {
        p[j][i] = a[i][j];
    }
}

通过这些更改，代码可以按预期工作。在不同阶段打印数据的示例代码上传到http://cfiddle.net/zoZazB

Answer 2

问题在这里:因为a是N *M的，但你不小心把它变成了M*N。

 // Filling in the new array called "p" with '0's.
 //^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^You are doing with a not p below
 for ( i = 0; i < M; i++ )
    for ( j = 0; j < N; j++ )
         a[i][j] = 0;
 //here it seems that a is of M*N not N*M
 //your comment and code does not match

大小不匹配将导致您访问不属于a的内存，因此，崩溃。