访问未初始化的元素时，std向量不会抛出out_of_range异常[重复]
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这个问题在这里已有答案：
我读了这个教程
std::vector beginners tutorial
并且还看到了这个问题：
然而，当我运行我的简单示例时，我没有看到预期的结果，这是 - >没有抛出qazxsw poi异常。
我在这里误解了什么吗？
我运行的示例代码如下（代码运行并成功终止，即 - 没有抛出异常）：
std::out_of_range笔记：
示例代码使用-std = c ++ 11选项进行编译。
编译器版本是g ++ 5.4（在我的Ubuntu 16.04机器上）。
谢谢，
盖伊。
向量#include <iostream>
#include <vector>
using namespace std;
class MyObjNoDefualtCtor
{
public:
MyObjNoDefualtCtor(int a) : m_a(a)
{
cout << "MyObjNoDefualtCtor::MyObjNoDefualtCtor - setting m_a to:" << m_a << endl;
}
MyObjNoDefualtCtor(const MyObjNoDefualtCtor& other) : m_a(other.m_a)
{
cout << "MyObjNoDefualtCtor::copy_ctor - setting m_a to:" << m_a << endl;
}
~MyObjNoDefualtCtor()
{
cout << "MyObjNoDefualtCtor::~MyObjNoDefualtCtor - address is" << this << endl;
}
// just to be sure - explicitly disable the defualt ctor
MyObjNoDefualtCtor() = delete;
int m_a;
};
int main(int argc, char** argv)
{
// create a vector and reserve 10 int's for it
// NOTE: no insertion (of any type) has been made into the vector.
vector<int> vec1;
vec1.reserve(10);
// try to access the first element - due to the fact that I did not inserted NOT even a single
// element to the vector, I would except here an exception to be thrown.
size_t index = 0;
cout << "vec1[" << index << "]:" << vec1[index] << endl;
// now try to access the last element - here as well: due to the fact that I did not inserted NOT even a single
// element to the vector, I would excpet here an excpetion to be thrown.
index = 9;
cout << "vec1[" << index << "]:" << vec1[index] << endl;
// same thing goes for user defined type (MyObjNoDefualtCtor) as well
vector<MyObjNoDefualtCtor> vec2;
vec2.reserve(10);
// try to access the first element - due to the fact that I did not inserted NOT even a single
// element to the vector, I would except here an exception to be thrown.
index = 0;
cout << "vec2[" << index << "]:" << vec2[index].m_a << endl;
// now try to access the last element - here as well: due to the fact that I did not inserted NOT even a single
// element to the vector, I would except here an exception to be thrown.
index = 9;
cout << "vec2[" << index << "]:" << vec2[index].m_a << endl;
return 0;
}
函数可能会也可能不会进行边界检查。具有边界检查的实现通常仅用于调试构建。 GCC及其标准库没有。
另一方面，operator[]函数确实有强制边界检查，并将保证抛出at异常。
这里发生的事情只是你走出界限并拥有out_of_range。
是undefined behavior执行范围检查，而不是（必然）at()。
您的代码具有未定义的行为。
如果您想确保获得例外，请使用
operator[]代替
vec1.at(index)
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