c++ - 当没有字符串匹配时，在 String 中查找函数会输出垃圾

Question

我有以下代码:-

    string name = "hello world";
    size_t find = name.find("world");

    if (find != string::npos) {
        cout<<"match found at "<<find;
    }
    else{
        cout<<find;
    }

该程序运行良好，并按预期打印 6 个输出。

但是如果我将其更改为 size_t find = name.find('\n'); 它将垃圾值打印为 18446744073709551615。 find函数在没有找到匹配的字符串时会打印垃圾值吗？？

Answer 1

引自http://www.cplusplus.com/reference/string/string/find/

Return Value

The position of the first character of the first match. If no matches were found, the function returns string::npos.

size_t is an unsigned integral type (the same as member type string::size_type).

因此，您的函数会打印 std::string::npos，例如在 MSVS std 库实现中是

basic_string<_Elem, _Traits, _Alloc>::npos =
        (typename basic_string<_Elem, _Traits, _Alloc>::size_type)(-1);

64 位系统上的最大 unsigned int 值是:18,446,744,073,709,551,615，请参阅 https://msdn.microsoft.com/en-us/library/s3f49ktz.aspx

您可以将代码更改为

string name = "hello world";
size_t find = name.find("world");

if (find != string::npos) {
    cout<<"match found at "<<find;
}
else {
    cout<<"match not found";
}