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C语言面试题大汇总之华为面试题
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华为公司 java 面试题
2. What are three ways in which a thread can enter the waiting state?
解析: 操作系统问题
4. In C++, there're four type of Casting Operators, please enumerate and explain them especially the difference.
解析: C++类型转换问题
答案:
7 以下代码有什么问题,如何修改?
#include <iostream>
#include <vector>
using namespace std;
void print(vector<int>);
int main()
{
vector<int> array;
array.push_back(1);
array.push_back(6);
array.push_back(6);
array.push_back(3);
//删除array数组中所有的6
vector<int>::iterator itor;
vector<int>::iterator itor2;
itor=array.begin();
for(itor=array.begin(); itor!=array.end(); )
{
if(6==*itor)
{
itor2=itor;
array.erase(itor2);
}
itor++;
}
print(array);
return 0;
}
void print(vector<int> v)
{
cout << "n vector size is: " << v.size() << endl;
vector<int>::iterator p = v.begin();
}
我的答案是,迭代器问题,只能删除第一个6,以后迭代器就失效了,不能删除之后的元素。
但我不知道怎么改
void print(const vector<int>&);
int main()
{
vector<int> array;
array.push_back(1);
array.push_back(6);
array.push_back(6);
array.push_back(3);
//删除array数组中所有的6
array.erase( remove( array.begin(), array.end(), 6 ) , array.end() );
print(array);
return 0;
}
void print(const vector<int>& v)
{
cout << "n vector size is: " << v.size() << endl;
copy(v.begin(), v.end(), ostream_iterator<int>(cout, " ") );
}
#include <iostream>
#include <vector>
using namespace std;
int main()
{
vector<int> array;
array.push_back(1);
array.push_back(6);
array.push_back(6);
array.push_back(6);
array.push_back(6);
array.push_back(6);
array.push_back(3);
array.push_back(9);
array.push_back(8);
array.push_back(5);
//ɾ³ýarrayÊý×éÖÐËùÓеÄ6
vector<int>::iterator itor;
itor=array.begin();
for(itor=array.begin(); itor!=array.end();++itor )
{
if(6==*itor)
{
itor=array.erase(itor);
--itor;
}
}
cout << "vector size is: " << array.size() << endl;
for(itor=array.begin(); itor!=array.end();++itor )
{
cout<<*itor<<" ";
}
system("pause");
return 0;
}
2. What are three ways in which a thread can enter the waiting state?
尝试答一下
CPU调度给优先级更高的thread,原先thread进入waiting
阻塞的thread获得资源或者信号,进入waiting
还有什么
static_cast 数制转换
dynamic_cast 用于执行向下转换和在继承之间的转换
const_cast 去掉const
reinterpret_cast 用于执行并不安全的orimplmentation_dependent类型转换
我给你一个完整的解释:
执行itor=array.erase(itor);这句话后,itor不会移动,而只是把删除的数后面的数都往前移一位,所以删除了第一个6后,指针指向第2个6,然后在来个itor++,指针就指向array.end()了,给你画个草图:
1 6 6 3 array.end() //最开始指针itor指向第一个6;
1 6 3 array.end() //删除第一个6后,指向第二个6
1 6 3 array.end() //itor++后,就指向3了,所以不能删除