string.find() 在使用 ==-1 时返回 true，但在使用 <0 时返回 false

我试图在字符串中查找一个字符，但我得到了意想不到的结果.我的理解是 string::find(char c) 在找不到时返回 -1 .但是，我得到了一些意想不到的结果.

即使字符串不包含 '8'，它仍然返回 true.

std::string s = "123456799";if(s.find('8')<0)cout <<未找到"<<结束；别的cout <<找到"<<结束；//输出:找到

但是，当使用 == 时，代码会按预期工作.

std::string s = "123456799";if(s.find('8')==-1)cout <<未找到"<<结束；别的cout <<找到"<<结束；//输出:未找到

解决方案

我的理解是string::find(char c)在找不到的时候返回-1.

这不准确.根据文档:

<块引用>

返回值
找到的子字符串的第一个字符的位置，如果没有，则为 npos找到了这样的子字符串.

准确地说，当找不到 std::string::find 时将返回 std::string::npos.重点是std::string::npos的类型是std::string::size_type，是无符号整数类型.即使它是从 -1 的值初始化的，它也不是 -1;它仍然没有签名.所以 s.find('8')<0 将永远是 false 因为不可能是负数.

std::string::npos 的文档:

<块引用>

static const size_type npos = -1;

这是一个特殊值，等于 size_type 类型可表示的最大值.

所以你应该使用 std::string::npos用于检查结果，以避免这种混乱.

if (s.find('8') == std::string::npos)cout <<未找到"<<结束；别的cout <<找到"<<结束；

<小时>

if(s.find('8')==-1) 工作正常，因为 operator== 这里是无符号的，右边的是有符号的.根据算术运算符的规则，

<块引用>

• 否则，如果无符号操作数的转换等级大于或等于有符号操作数的转换等级，则将有符号操作数转换为无符号操作数的类型.

所以-1会被转换成unsigned，也就是std::string::npos的值，然后一切都按预期工作.

I am trying to find a character within a string but I am getting unexpected results. My understanding is that string::find(char c) returns -1 when it is not found. However, I am getting some unexpected results.

Even though the string does not include an '8', it is still returning true.

std::string s = "123456799";
if(s.find('8')<0)
    cout << "Not Found" << endl;
else
    cout <<  "Found" << endl;

//Output: Found

However, when using == instead the code works as expected.

std::string s = "123456799";
if(s.find('8')==-1)
    cout << "Not Found" << endl;
else
    cout <<  "Found" << endl;

//Output: Not Found

解决方案

My understanding is that string::find(char c) returns -1 when it is not found.

It's not accurate. According to the documentation:

Return value
Position of the first character of the found substring or npos if no such substring is found.

So to be precise, when not found std::string::find will return std::string::npos. The point is that the type of std::string::npos is std::string::size_type, which is an unsigned integer type. Even it's initialized from value of -1, it's not -1; it's still unsigned. So s.find('8')<0 will always be false because it's not possible to be negative.

Documentation of std::string::npos:

static const size_type npos = -1;

This is a special value equal to the maximum value representable by the type size_type.

So you should use std::string::npos for checking the result, to avoid such kind of confusing.

if (s.find('8') == std::string::npos)
    cout << "Not Found" << endl;
else
    cout <<  "Found" << endl;

---

if(s.find('8')==-1) works fine, because the left-hand operand of operator== here is unsigned, the right-hand one is signed. According to the rules for arithmetic operators,

• Otherwise, if the unsigned operand's conversion rank is greater or equal to the conversion rank of the signed operand, the signed operand is converted to the unsigned operand's type.

So -1 will be converted to unsigned, which is the value of std::string::npos and then all work as expected.