Remove last character from C++ string

Question

How can I remove last character from a C++ string?

I tried st = substr(st.length()-1); But it didn't work.

Do you want a new string with the last character removed or the same string without the last character ? — Matthieu M., Feb 22 '10 at 12:58
For a MFC Visual C++ CString: `CString str=CString("Hello world"); str.Delete(str.GetLength()-1);` — sergiol, Oct 11 '17 at 18:20

score 565 · Answer 1 · edited Oct 28 '16 at 17:35

565

Simple solution if you are using C++11. Probably O(1) time as well:

st.pop_back();

edited Oct 28 '16 at 17:35

user2891462

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answered May 19 '13 at 00:29

mpgaillard

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As a FYI - its being supported only from GCC 4.7 (off course along w/ the -std=c++11 compilation switch) – Shmil The Cat Feb 07 '14 at 08:15
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Don't forget to check `length()`. – Jul 25 '14 at 16:01
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Would be good to add the most convenient way to remove the last X characters as well. – FvD Sep 06 '18 at 10:29
@user1382306 most implementation already check for empty-ness – fiorentinoing Feb 22 '19 at 13:00
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`The behavior is undefined if the string is empty. ` from [here](https://en.cppreference.com/w/cpp/string/basic_string/pop_back) – Raffi Jun 28 '19 at 12:36

score 260 · Accepted Answer · answered Feb 22 '10 at 12:59

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For a non-mutating version:

st = myString.substr(0, myString.size()-1);

answered Feb 22 '10 at 12:59

Matthieu M.

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@MatthieuM. Your example is confusing, I think that the essence of the question is to modify the original string, in your example you're not modifying the original string, because in your example the original string is called "myString" which gives to the confusion, in the question it is "st". Your code should be: `st = st.substr(0, st.size()-1)`. But it still doesn't look the right way, I think that the proper way is to use the function that is intended for this task, it's called erase() and the code is: `st.erase(st.size()-1)`. This would be called a "mutating version". – Czarek Tomczak Jan 19 '13 at 14:46
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@CzarekTomczak: I understand this is not exactly what was asked for, thus the disclaimer before the actual gist. – Matthieu M. Jan 19 '13 at 15:05
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@MattPhillips: his solution is C++11 specific though (`pop_back` did not exist in C++03) and it is also an in-place modification (and the OP never clarified whether he wanted in-place or not)... as such, he has **a** correct answer, but not the only possible one. – Matthieu M. May 21 '13 at 06:24

jcrv · Answer 3 · 2017-12-07T22:05:00.520

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That's all you need:

#include <string>  //string::pop_back & string::empty

if (!st.empty())
    st.pop_back();

edited Dec 07 '17 at 22:05

answered Dec 07 '17 at 19:24

jcrv

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score 28 · Answer 4 · edited Jan 24 '23 at 05:56

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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.

BTW - Is there really no std::string::pop_back? - seems strange.

edited Jan 24 '23 at 05:56

ggorlen

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answered Feb 22 '10 at 13:13

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There is no `std::string::pop_back` in C++03; it's been added in C++0x, though. – James McNellis Feb 22 '10 at 14:20
OK - thanks. It caused be a bit of confusion - I could swear I've used it, yet it's not there. Maybe I have a non-standard library in some compiler somewhere (between VC++2003, VC++2008, MinGW GCC3 MinGW GCC 4 and Linux GCC 4, you do get a few differences). More likely, I'm just getting confused with other types. – Feb 22 '10 at 14:31
resize() is probably not intended for such use, it's a memory related function, erase() is for deleting characters. – Czarek Tomczak Jan 19 '13 at 14:36
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@Czarek Tomczak - sorry for the absurdly late reply, but `resize` is a resizing function, and no more a memory function that anything else that might increase the memory needed. For example, if you `resize` to a smaller size, it will *not* reduce the memory reserved. I think you're thinking of `reserve`, which at least *might* reduce memory allocated if asked to - see [resize here](http://www.cplusplus.com/reference/string/string/resize/) and [reserve here](http://www.cplusplus.com/reference/string/string/reserve/). – Mar 05 '14 at 06:53
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if (!str.empty()) is preferred over size – ericcurtin Jul 09 '18 at 13:09

RC. · Answer 5 · 2012-08-01T11:52:14.137

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buf.erase(buf.size() - 1);

This assumes you know that the string is not empty. If so, you'll get an out_of_range exception.

edited Aug 01 '12 at 11:52

answered Feb 22 '10 at 13:09

RC.

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buf[buf.size()-1] = '\0'; doesn't remove anything - it just changes the character that was there to have the value zero. std:;strings can happily contain such characters. – Feb 22 '10 at 13:18
Neil is correct. I probably should have clarified this in my answer. The second option will effectively change the value of the last character so it won't print, but the string length will stay the same. Using erase actually "removes" the last character and will change the size of the string. – RC. Feb 22 '10 at 13:43
@RC It will print, assuming you use something like cout << buf. How it appears will depend on your platform. And you can always clarify by editing your answer. – Feb 22 '10 at 20:04
What's much better about `size()` instead of `end()` of another answer? – ribamar Sep 08 '17 at 12:36

score 19 · Answer 6 · edited Mar 17 '14 at 11:11

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str.erase( str.end()-1 )

Reference: std::string::erase() prototype 2

no c++11 or c++0x needed.

edited Mar 17 '14 at 11:11

MKroehnert

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answered Mar 12 '14 at 20:21

ribamar

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This can lead to weird situation: the size of the string had been reduced but the last character is not set to '\0'. – Deqing Mar 18 '14 at 07:33
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@Deqing can you give more details of what happens in such case? – ribamar Mar 26 '14 at 19:05
For example if you have `string s("abc");`, after erase it seems working: `cout< – Deqing Mar 27 '14 at 10:13
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easily fixed: just `str[str.length()-1] = 0; str.erase(str.end()-1);` – taxilian Mar 19 '15 at 21:25
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@Dequing: your example is invalid. The erase reduces the size of the string, so the access to `s[2]` is illegal. – EML Mar 29 '15 at 13:11
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EML is correct. That's why it is important to use c_str() any time you're going to access character by character and expect a \0 terminal. The only correct way to access a std:string otherwise is with an iterator. – srm Mar 02 '16 at 17:53

score 13 · Answer 7 · answered Feb 22 '10 at 13:01

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int main () {

  string str1="123";
  string str2 = str1.substr (0,str1.length()-1);

  cout<<str2; // output: 12

  return 0;
}

answered Feb 22 '10 at 13:01

codaddict

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score 3 · Answer 8 · answered Apr 24 '13 at 20:27

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With C++11, you don't even need the length/size. As long as the string is not empty, you can do the following:

if (!st.empty())
  st.erase(std::prev(st.end())); // Erase element referred to by iterator one
                                 // before the end

answered Apr 24 '13 at 20:27

Michael Goldshteyn

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score 2 · Answer 9 · edited Mar 17 '14 at 11:09

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str.erase(str.begin() + str.size() - 1)

str.erase(str.rbegin()) does not compile unfortunately, since reverse_iterator cannot be converted to a normal_iterator.

C++11 is your friend in this case.

edited Mar 17 '14 at 11:09

MKroehnert

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answered Mar 02 '13 at 05:30

Zebra

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This method has problem as well, the last character is not set to '\0'. – Deqing Mar 18 '14 at 07:34
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Any particular reason for not doing `str.erase(str.end() - 1)` ? – vallentin Dec 23 '16 at 21:26

Pneumonoultramicrosco Picsilic · Answer 10 · 2022-11-07T05:55:22.677

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Don't worry about boundary check or empty string with ternary operator:

str.erase(str.end() - ((str.length() > 0) ? 1 : 0), str.end());

edited Nov 07 '22 at 05:55

answered Nov 07 '22 at 05:53

Pneumonoultramicrosco Picsilic

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score 0 · Answer 11 · answered Mar 26 '21 at 03:52

#include<iostream>
using namespace std;
int main(){
  string s = "Hello";// Here string length is 5 initially
  s[s.length()-1] = '\0'; //  marking the last char to be null character
  s = &s[0]; // using ampersand infront of the string with index will render a string from the index until null character discovered
  cout<<"the new length of the string "<<s + " is " <<s.length();
  return 0;
}

score -8 · Answer 12 · answered Mar 25 '18 at 21:25

-8

If the length is non zero, you can also

str[str.length() - 1] = '\0';

answered Mar 25 '18 at 21:25

Jan Glaser

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Setting the last character to `\0` does not change the length of the string. `str.length()` will be inaccurate. – jww Nov 05 '18 at 19:42
Yes, I see it now. We are considering C++. You are correct, it does not change the length of the string – Jan Glaser Apr 18 '19 at 22:47

Remove last character from C++ string

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