Anonymous
Chapter 6 "Circulation" took 2 days and I couldn't understand
Leo
In my opinion, that book isn't good for beginers.
Anonymous
I think so too. It's too comprehensive. I just want to lay some foundation of C language for postgraduate computer
Nils
Hi, can I split a string by fast space?
So "Hello nice world!" turns into: {"Hello", "nice world!"}
Anonymous
Nils
Hi, can I split a string by fast space?
So "Hello nice world!" turns into: {"Hello", "nice world!"}
And no, I do not want to use boost
Dima
Nils
std::list<std::string> split(std::string s, char delimiter, int times) {
std::string currpart;
std::list<std::string> splitted;
int timesdone = 0;
for (std::string::iterator strit = s.begin(); strit != s.end(); strit++) {
currpart.push_back(*strit);
if (*strit == delimiter and timesdone < times) {
splitted.push_back(currpart);
currpart = "";
times++;
}
}
splitted.push_back(currpart);
return splitted;
}
(Or: https://hastebin.com/kulujukula.cpp)
Do you think this is okay?
Francisco
Nils
What does the "times" parameter do?
It says how many times the string shoulds be split by delimiter
Francisco
Well, you don't use the parameter, so...
Nils
sure I do
Francisco
Oh okay, you edited it
Nils
Yep, sorry
Anonymous
@leoleoecdsa_bot Thanks, I am already watching mooc
Nils
auto cmdsplit = strsplit(commandstr, ' ', 1);
std::string command = cmdsplit[0];
std::string argsstr = cmdsplit[1];
Is there any better way to solve this?
Francisco
klimi
Nils
What do you want to achieve?
First element of list in variable command, second one in variable argsstr
Nils
strsplit() returns std::list<std::string>
Francisco
strsplit() returns std::list<std::string>
Keep in mind that std::list is not random access, so you can't do stuff like my_list[i];
Nils
oh.
Nils
but nvm I solved it using iterator
Francisco
Also, why are you using lists instead of vectors?
Francisco
There're very few situations where lists actually beat vectors performance-wise
Anonymous
std::list<std::string> split(std::string s, char delimiter, int times) {
std::string currpart;
std::list<std::string> splitted;
int timesdone = 0;
for (std::string::iterator strit = s.begin(); strit != s.end(); strit++) {
currpart.push_back(*strit);
if (*strit == delimiter and timesdone < times) {
splitted.push_back(currpart);
currpart = "";
times++;
}
}
splitted.push_back(currpart);
return splitted;
}
(Or: https://hastebin.com/kulujukula.cpp)
Do you think this is okay?
the parameter s is never modified, change to std::string_view instead. the for loop can be replaced with calls to the various find algorithms (<algorithm> or string_view members or string members) and substr (string_view or string member)
Nils
Are vectors used exactly the same as lists?
Nils
Anonymous
it is not intended to be modified
ye so don't copy it. just use string_view as the parameter
Anonymous
clangd should warn about these things
Nils
strsplit(std::basic_string_view s, char delimiter, int times)
Like this?
Anonymous
std::list<std::string> split(std::string s, char delimiter, int times) {
std::string currpart;
std::list<std::string> splitted;
int timesdone = 0;
for (std::string::iterator strit = s.begin(); strit != s.end(); strit++) {
currpart.push_back(*strit);
if (*strit == delimiter and timesdone < times) {
splitted.push_back(currpart);
currpart = "";
times++;
}
}
splitted.push_back(currpart);
return splitted;
}
(Or: https://hastebin.com/kulujukula.cpp)
Do you think this is okay?
the times parameter is unnecessary since 1) it is declared as a local variable and never returned, 2) the size of the returned container would contain the number of times you split the string
i did not see the timesdone < times part
Anonymous
strsplit(std::basic_string_view s, char delimiter, int times)
Like this?
just std::string_view or std::basic_string_view<char>
Nils
Nils
just std::string_view or std::basic_string_view<char>
error: ‘string_view’ is not a member of ‘std’
Nils
std::string stringlisttostring(std::list<std::string> strlist) {
std::string liststr;
liststr.append("[");
for (std::list<std::string>::iterator listit = strlist.begin(); listit-- != strlist.end(); listit++) {
liststr.append(*listit);
if (listit != strlist.end())
liststr.append(", ");
}
liststr.append("]");
return liststr;
}
Also why does this sigfault?
Nils
What compiler are you using? And what version?
g++ (Ubuntu 9.2.1-9ubuntu2) 9.2.1 20191008
Copyright (C) 2019 Free Software Foundation, Inc.
This is free software; see the source for copying conditions. There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
Francisco
Are you sure you included the correct header?
Nils
yes
Nils
#include <string_view>
Nils
std::string stringlisttostring(std::list<std::string> strlist) {
std::string liststr;
liststr.append("[");
for (std::list<std::string>::iterator listit = strlist.begin(); listit-- != strlist.end(); listit++) {
liststr.append(*listit);
if (listit != strlist.end())
liststr.append(", ");
}
liststr.append("]");
return liststr;
}
Also why does this sigfault?
terminate called after throwing an instance of 'std::bad_alloc'
what(): std::bad_alloc
Aborted (core dumped)
Nils
It does not
Anonymous
std::list<std::string> split(std::string s, char delimiter, int times) {
std::string currpart;
std::list<std::string> splitted;
int timesdone = 0;
for (std::string::iterator strit = s.begin(); strit != s.end(); strit++) {
currpart.push_back(*strit);
if (*strit == delimiter and timesdone < times) {
splitted.push_back(currpart);
currpart = "";
times++;
}
}
splitted.push_back(currpart);
return splitted;
}
(Or: https://hastebin.com/kulujukula.cpp)
Do you think this is okay?
#include <vector>
#include <string>
#include <string_view>
std::vector<std::string> split(std::string_view s, char delimiter, std::vector<std::string>::size_type times = 0) {
std::vector<std::string> to_return;
decltype(s.size()) start = 0, finish = 0;
while ((finish = s.find_first_of(delimiter, start)) != std::string_view::npos) {
to_return.emplace_back(s.substr(start, finish - start));
start = finish + 1;
if (to_return.size() == times) { break; }
}
to_return.emplace_back(s.substr(start));
return to_return;
}
here's how i would do it
Nils
Anonymous
Anonymous
Nils
ah
Nils
I changed my code to vecotr now, getting logic error... :-/
Nils
I changed my code to vecotr now, getting logic error... :-/
https://hastebin.com/guyuronava.cpp
Any idea?
Nils
I solved my string_view problem now btw
Francisco
https://hastebin.com/guyuronava.cpp
Any idea?
Nobody likes a 170 loc snippet. Send a minimal example that reproduces your problem
Nils
Uhm… 🤔
Nils
Very hard to do
Nils
https://hastebin.com/ufufihaheq.cpp
This is just execute and see command, I can't get it any more minimal
Nils
the problem function is execline()
Nils
mainlyx the for loop starting at line 132 but I am not sure
Francisco
Very hard to do
I've reduced a couple thousands of locs to less than 50, you should be able to reduce your 170 to at most 50
Nils
That would take me more time than finding the error myself… Okay I'll try
Nils
Okay I simply make a paste only containing the problematic funcion
Nils
https://hastebin.com/iyewipepez.cpp
Anonymous
https://hastebin.com/ufufihaheq.cpp
This is just execute and see command, I can't get it any more minimal
std::vector<std::string>::iterator cmdsplit_it = strsplit(commandstr, ' ', 1).begin();
std::string command = *cmdsplit_it;
cmdsplit_it++;
line 119: idk if this is something you should do, isn't the vector a temporary that goes out of scope as soon as this line is done?
Francisco
That would take me more time than finding the error myself… Okay I'll try
You're starting to understand the thing. By reducing the code to the bare minimum to reproduce your problem, you begin to find some hints of where the bug might be