从 C++ 字符串中删除最后一个字符
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/2310939/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Remove last character from C++ string
提问by skazhy
How can I remove last character from a C++ string?
如何从 C++ 字符串中删除最后一个字符？
I tried st = substr(st.length()-1);But it didn't work.
我试过了，st = substr(st.length()-1);但没有用。
采纳答案by Matthieu M.
For a non-mutating version:
对于非变异版本：
st = myString.substr(0, myString.size()-1);
回答by mpgaillard
Simple solution if you are using C++11. Probably O(1) time as well:
如果您使用的是 C++11，那么简单的解决方案。也可能是 O(1) 时间：
st.pop_back();
回答by Steve314
if (str.size () > 0) str.resize (str.size () - 1);
An std::erase alternative is good, but I like the "- 1" (whether based on a size or end-iterator) - to me, it helps expresses the intent.
std::erase 替代方案很好，但我喜欢“- 1”（无论是基于大小还是最终迭代器）-对我来说，它有助于表达意图。
BTW - Is there really no std::string::pop_back ? - seems strange.
顺便说一句 - 真的没有 std::string::pop_back 吗？——似乎很奇怪。
回答by RC.
buf.erase(buf.size() - 1);
This assumes you know that the string is not empty. If so, you'll get an out_of_rangeexception.
这假设您知道该字符串不为空。如果是这样，你会得到一个out_of_range例外。
回答by ribamar
str.erase( str.end()-1 )
str.erase( str.end()-1 )
Reference: std::string::erase() prototype 2
no c++11 or c++0x needed.
不需要 c++11 或 c++0x。
回答by jcrv
That's all you need:
这就是你所需要的：
#include <string> //string::pop_back & string::empty
if (!st.empty())
st.pop_back();
回答by codaddict
int main () {
string str1="123";
string str2 = str1.substr (0,str1.length()-1);
cout<<str2; // output: 12
return 0;
}
回答by Zebra
str.erase(str.begin() + str.size() - 1)
str.erase(str.begin() + str.size() - 1)
str.erase(str.rbegin())does not compile unfortunately, since reverse_iteratorcannot be converted to a normal_iterator.
str.erase(str.rbegin())不幸的是无法编译，因为reverse_iterator无法转换为 normal_iterator。
C++11 is your friend in this case.
在这种情况下，C++11 是你的朋友。
回答by Michael Goldshteyn
With C++11, you don't even need the length/size. As long as the string is not empty, you can do the following:
使用 C++11，您甚至不需要长度/大小。只要字符串不为空，您就可以执行以下操作：
if (!st.empty())
st.erase(std::prev(st.end())); // Erase element referred to by iterator one
// before the end
回答by Jan Glaser
If the length is non zero, you can also
如果长度不为零，您也可以
str[str.length() - 1] = '##代码##';
