gdb-如何查看指针数组的内容? [英] gdb - how to view contents of an array of pointer?

问题描述

在获取变量'arr'的内容时遇到了问题，该变量是一个指针数组. 我尝试了p *arr@n，但是它给出了以下输出:$1 = {0x603010, 0x603030}. 我该怎么办?

I am having a problem in getting the contents of the variable 'arr' which is an array of pointers. I tried,p *arr@n, but it gives the following output: $1 = {0x603010, 0x603030}. What should I do?

int n, q;
scanf("%d %d", &n, &q);

int lastAnswer=0, index_size[n], *arr[n];  // <-- here
for(int i=0; i<n; i++)
    index_size[i] = 0;

for(int i=0; i<n; i++) {

    int *temp = malloc(sizeof(int)*n);
    arr[i] = temp;
}

while(q--) {
    int w, x, y, seq;
    scanf("%d %d %d", &w, &x, &y);

    if(w == 1) {

        seq = ((x ^ lastAnswer) % n);
        arr[seq][index_size[seq]++] = y;
    }
    else {

        seq = ((x ^ lastAnswer) % n);
        lastAnswer = y%n;
        printf("%d\n", lastAnswer);
    }
}

return 0;

推荐答案

如果您自己打印出一个指针，它只会在您的内存块中提供一个地址.

If you print out a pointer itself, it would just give you an address in your memory block.

所以print *arr@n只会为您提供第一维的内容(输出中的地址数组)

So print *arr@n would simply give you the content of the first dimension (an array of address in your output)

如果要打印出更深的内容.您可能想要做这样的事情:

If you want to print out the deeper content. You might want to do something like this:

print **arr@n;

或

print *arr[0]@n

另一种方法是在程序内部定义一个漂亮的打印函数，然后在gdb中调用它.

Another method would be define a pretty print function inside your program and call it in gdb.

void print(int arr[][], n, m)
{
    int i, j;
    for (i = 0; i < n; ++i)
    {
        for (j = 0; j < m; ++j)
            printf("%d ", arr[i][j]);
        printf("\n");
    }
}

然后在gdb中调用它

call print(arr, n, m)

我不认为gdb本身支持打印2D数组，为什么? 因为print *array@3的定义不是打印array中的前三个元素，所以它是定价*array(或array[0])以及array[0]之后的三个元素.

I don't think gdb support printing 2D array itself, why? Because the definition of print *array@3 isn't printing the first three elements in array, instead, it is "priting *array (or array[0]) and the three elements following array[0].

print **arr@n@n

在这种情况下将不起作用(尽管它会打印出不错的格式)

would not work in this case, (although it print out an nice format)

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