c++ - s.compare 成员函数在下面的 C++ 代码中的行为如何？

Question

我有下面这个 C++ 代码来解决作业，在我用 Code::Blocks 运行它之后，它告诉我 i=0，这意味着表达式 s.compare(t)<0是假的。但是，在我看来，情况恰恰相反:( s<t ，因为 AbcA < AAbcA )。有人可以给我解释一下吗？

#include <iostream>
#include <string>

using namespace std;

int main(void) {
        string s = "Abc", t = "A";
        s=s+t;
        t=t+s;
        int i = s.compare(t)<0;
        int j = s.length()<t.length();
        cout<<i+j<<endl;
        return 0;
}

Answer 1

根据引用std::string::compare返回:

negative value if *this appears before the character sequence specified by the arguments, in lexicographical order

zero if both character sequences compare equivalent

positive value if *this appears after the character sequence specified by the arguments, in lexicographical order

Lexicographical comparison被定义为:

Lexicographical comparison is a operation with the following properties:

• Two ranges are compared element by element. The first mismatching element defines which range is lexicographically less or greater than the other.
• If one range is a prefix of another, the shorter range is lexicographically less than the other.
• If two ranges have equivalent elements and are of the same length, then the ranges are lexicographically equal.
• An empty range is lexicographically less than any non-empty range.
• Two empty ranges are lexicographically equal.

"AbcA" 按字典顺序出现在 "AAbc" 之后，因为第一个非等号字符 'b' ( ASCII 0x62) 在 'A' (ASCII 0x41)