下面是二维数组的示例.
int s[5][2] = {
{0, 1},
{2, 3},
{4, 5},
{6, 7},
{8, 9}
};
int (*p)[2];
如果我写,p = &s[0];那就没有错误.但是如果我写的p = s[0];是有错误,即使&s[0]并且s[0]会给出相同的地址.
请告诉我为什么会有不同之处,即使两者都给出相同的地址.
I have few queries with respect to below code snapshot.
1) With respect to pthread_create(), assume Thread_1 creates Thread_2. To my understanding Thread_1 can exit without join, but still Thread_2 will keep running. Where as in below example without join() I am not able to run thread and I am seeing exceptions.
2) In few examples I am seeing thread creation without thread object as below. But when I do the same, code is terminated.
std::thread(&Task::executeThread, this);
I am compiling …作为向量的一部分，使用 [] 我无法初始化。
std::vector<int> vectorVar;
vectorVar[0] = 0;
看到未定义的行为。
而只有在已经创建的情况下才能使用。
vectorVar.push_back(0);
vectorVar[0] = 5;
但如果有地图，即使没有条目也可以使用。
std::map<int, std::vector<int>> mapVar;
mapVar[0] = vectorVar;
请澄清。
感谢和问候 毗湿奴比玛
如果“vectorVar[0] = 0;”，则会出现未定义的行为