为什么随机访问迭代器的算术运算符接受/返回int而不是size_t? [英] Why do random access iterator's arithmetic operators accept / return int and not size_t?
问题描述
因为 std :: vector
上的大多数操作都需要/ return size_t
- 这是我用于索引的类型。但现在我已经启用所有的编译器警告,修复一些签名/未签名的转换问题,我知道我有,这个消息让我惊讶:
Since most operations on std::vector
require / return size_t
- that's the type I use for indexing. But now I've enabled all compiler warnings to fix some signed / unsigned conversion issues that I know I have, and this message surprised me:
warning C4365:'argument':从'size_t'转换为'__w64 int',签名/未签名不匹配
warning C4365: 'argument' : conversion from 'size_t' to '__w64 int', signed/unsigned mismatch
此代码:
std::vector<int> v;
size_t idx = 0;
v.insert(v.begin() + idx + 1, 0);
我有很多其他类似的消息,建议迭代器的算术运算符接受并返回 int
。为什么 size_t
?修复所有这些消息是一种痛苦,并且不会使我的代码更漂亮!
I've got a lot of other similar messages suggesting that iterator's arithmetic operators accept and return int
. Why not size_t
? Fixing all these messages is a pain, and doesn't make my code prettier!
推荐答案
有很多其他类似的消息,表明迭代器的算术运算符接受并返回
int
。
不一定 int
。它是由迭代器类型的 iterator_traits
定义的(signed) difference_type
。对于大多数迭代器类型,默认为 ptrdiff_t
。
Not necessarily int
. It's the (signed) difference_type
defined by the iterator type's iterator_traits
. For most iterator types, this defaults to ptrdiff_t
.
$ c> size_t ?
因为算术需要使用带符号值正确工作;人们会期望 it +(-1)
等同于 it-1
。
Because arithmetic needs to work correctly with signed values; one would expect it + (-1)
to be equivalent to it - 1
.
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