LeetCode 报错AddressSanitizer: SEGV on unknown address 0x000000619d20 (pc 0x000000401a6a bp 0x7ffd35b3

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#LeetCode

LeetCode90子集 II

题目描述：

给定一个可能包含重复元素的整数数组 nums，返回该数组所有可能的子集（幂集）。

说明：解集不能包含重复的子集。

示例:

输入: [1,2,2]

输出:

[

[2],

[1],

[1,2,2],

[2,2],

[1,2],

[]

]

2. 代码实现

int comp(const void *a, const void *b)
{
    return *(int*)a - *(int*)b;
}

int count(int num, int k) {
    int count = 0;
    for (int i = 0; i < k; i++) {
        if(num & 1)
            count++;
        num >>= 1;
    }
    return count;
}

int** subsetsWithDup(int* nums, int numsSize, int* returnSize, int** returnColumnSizes){
    qsort(nums, numsSize, sizeof(int), comp);//从小到大排序

    int t = (int)pow(2, numsSize);
    *returnSize = t;
    int **result = (int**)malloc(t * sizeof(int*));
    returnSize[0] = (int*)malloc(t * sizeof(int));

    int k, size;
    for (int i = 0; i < t; i++) {
        returnColumnSizes[0][i] = count(i, numsSize);
        result[i] = (int*)malloc(returnColumnSizes[0][i] * sizeof(int));
        k = i;
        size = 0;
        for (int j = 0; j < numsSize; j++) {
            if (k & 1)
                result[i][size++] = nums[j];
            k >>= 1;
        }
    }
    return result;
}

3.报错信息

AddressSanitizer:DEADLYSIGNAL

=================================================================

==29==ERROR: AddressSanitizer: SEGV on unknown address 0x000000619d20 (pc 0x000000401a6a bp 0x7ffd35b3f980 sp 0x7ffd35b3f930 T0)

==29==The signal is caused by a WRITE memory access.

#2 0x7ff33e0882e0 in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x202e0)

AddressSanitizer can not provide additional info.

==29==ABORTING

从报错信息看，是内存泄漏问题，和数组指针操作有关，通过排查发现是在代码行

returnSize[0] = (int*)malloc(t * sizeof(int));

实际改为

returnColumnSizes[0] = (int*)malloc(t * sizeof(int));

4.修改后正确代码

int comp(const void *a, const void *b)
{
    return *(int*)a - *(int*)b;
}

int count(int num, int k) {
    int count = 0;
    for (int i = 0; i < k; i++) {
        if(num & 1)
            count++;
        num >>= 1;
    }
    return count;
}

int** subsetsWithDup(int* nums, int numsSize, int* returnSize, int** returnColumnSizes){
    qsort(nums, numsSize, sizeof(int), comp);//从小到大排序

    int t = (int)pow(2, numsSize);
    *returnSize = t;
    int **result = (int**)malloc(t * sizeof(int*));
    returnColumnSizes[0] = (int*)malloc(t * sizeof(int));

    int k, size;
    for (int i = 0; i < t; i++) {
        returnColumnSizes[0][i] = count(i, numsSize);
        result[i] = (int*)malloc(returnColumnSizes[0][i] * sizeof(int));
        k = i;
        size = 0;
        for (int j = 0; j < numsSize; j++) {
            if (k & 1)
                result[i][size++] = nums[j];
            k >>= 1;
        }
    }
    return result;
}