索引具有负索引的std :: vector [英] Indexing an std::vector with a negative index

问题描述

我尝试使用负索引来索引向量。 vector :: at（）成员函数检查指定的索引是否在向量的边界内，如果没有发生，则 out_of_range 抛出异常。

I tried to index a vector using a negative index. The vector::at() member function checks whether the specified index is within the bounds of the vector, and if this does not occur, an out_of_range exception is thrown.

vector<float> array;   // sample vector
array.push_back(0.123);
array.push_back(1.234);
array.push_back(2.345);
array.push_back(3.456);
array.push_back(4.567);

int index = -1;
float f = array.at(index);
cout << f << endl;

的签名vector :: at（）成员函数要求指定的参数为向量< T> :: size_type 类型，并且此类型为 unsigned int 用于向量，因此编译器应该执行从 int （ index 变量的类型）到 unsigned int 。由于在上面的例子中 index 的值是 -1 ，隐式转换的 code>是 4294967295 （即 unsigned int 类型的最大值）：此值传递到 vector :: at（）成员函数，它引发一个 out_of_range 异常。

The signature of vector::at() member function requires that the specified parameter is of vector<T>::size_type type, and this type is unsigned int for the vector, so the compiler should perform an implicit conversion from int (the type of the index variable) to unsigned int. Since the index value is -1 in the above example, the implicitly converted index is 4294967295 (that is the max value of the unsigned int type): this value is passed to vector::at() member function, which throws an out_of_range exception.

换句话说，这个异常不会抛出，因为 vector :: at（）成员函数看到 / code>小于零，而是因为隐式转换的 index 大于向量的当前大小。这是正确的解释吗？

In other words, this exception is not thrown because the vector::at() member function sees that the index is less than zero, but rather because the implicitly converted index is greater than the current size of the vector. Is this a correct explanation?

推荐答案

是的，这是一个正确的解释。 （除了在处接受一个 std :: size_t ，这是一些未指定的无符号整数类型（通常是位宽您的系统上的指针），它允许您的系统上 unsigned int ，但不是标准强制的。）

Yes, that is a correct explation. (Except at takes a std::size_t, which is some unspecified unsigned integer type (usually the bit-width of pointers on your system). It being unsigned int on your system is allowed, but not mandated by the standard.)

另外，请注意无符号到有符号的转换：标准不要求它是一个往返签名 - >无符号 - >负值，一些编译器积极地优化意想不到的方式。

As an aside, be careful with unsigned to signed conversion: the standard does not require it be a round-trip going signed->unsigned->signed for negative values, and some compilers aggressively optimize in unexpected ways.

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