c ++获取字符串中的最后一个（n）字符
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c++ get the last (n) char in a string
提问by Castro Roy
i have a string and i want to get, for example, the position of the last (.) in the string, or whatever char i want to check, but untill now i just get a headeach.
我有一个字符串，我想获得，例如，字符串中最后一个 (.) 的位置，或者我想检查的任何字符，但直到现在我才得到一个 headeach。
thanks
谢谢
回答by Kos
Is find_last_ofwhat you need?
是find_last_of你需要什么？
size_type find_last_of( const basic_string& str, size_type pos = npos ) const;
Finds the last character equal to one of characters in the given character sequence. Search finishes at pos, i.e. only the substring [0, pos] is considered in the search. If npos is passed as pos whole string will be searched.
查找与给定字符序列中的一个字符相等的最后一个字符。搜索在 pos 结束，即在搜索中只考虑子串 [0, pos]。如果 npos 作为 pos 传递，则将搜索整个字符串。
回答by take-it
string lastN(string input)
{
return input.substr(input.size() - n);
}
回答by thbusch
If your string is a char array:
如果您的字符串是字符数组：
#include <cstdio>
#include <cstring>
int main(int argc, char** argv)
{
char buf[32] = "my.little.example.string";
char* lastDot = strrchr(buf, '.');
printf("Position of last dot in string: %i", lastDot - buf);
return 0;
}
..or a std::string:
.. 或 std::string:
#include <cstdio>
#include <string>
int main(int argc, char** argv)
{
std::string str = "my.little.example.string";
printf("Position of last dot in string: %i", str.find_last_of('.'));
return 0;
}
回答by user744951
#include <string>
/**
* return the last n characters of a string,
* unless n >= length of the input or n <= 0, in which case return ""
*/
string lastN(string input, int n)
{
int inputSize = input.size();
return (n > 0 && inputSize > n) ? input.substr(inputSize - n) : "";
}
