c++ - 为返回的引用分配空间

Question

我创建了一个函数来计算 2 个时间函数的差值。这是该函数的基本版本。

struct timeval* getDifference(struct timeval* startTime, struct timeval* finishTime)
{
    struct timeval* difference = new struct timeval;
    difference->tv_sec = finishTime->tv_sec - startTime->tv_sec;
    difference->tv_usec = finishTime->tv_usec - startTime->tv_usec;
    return difference;
}

int main()
{
    struct timeval a,b;
    struct timeval* c;
    gettimeofday(&a, NULL);
    usleep(100000);
    gettimeofday(&b, NULL);

    c = getDifference(&a, &b);
 }

我想知道使用 struct timeval* c; 是否安全？还是应该像 struct timeval* c = new struct timeval;？

如果它是安全的，那么释放 c 会释放函数内部分配的 difference = new struct timeval; 吗？

Answer 1

我建议您完全不要使用指针。让参数成为引用并复制返回值:

struct timeval getDifference(const struct timeval& startTime, const struct timeval& finishTime)
{
    struct timeval difference;
    difference.tv_sec = finishTime.tv_sec - startTime.tv_sec;
    difference.tv_usec = finishTime.tv_usec - startTime.tv_usec;
    return difference;
}