我有以下方法：
string Company::cheap(list<Candidate*>& candidates) {
candidates.sort(candidateSalaryCompare);
for (std::list<Candidate*>::iterator iter = candidates.begin(); iter
!= candidates.end(); ++iter) {
}
int m(candidates.front()->getExpectedSalary());
list<Candidate*> potentialList;
for (std::list<Candidate*>::iterator iter = candidates.begin(); (*iter)->getExpectedSalary()
== m && iter != candidates.end(); ++iter)
potentialList.push_back(*iter);
if (potentialList.size() > 0)
potentialList.sort(candidateIdCompare);
return potentialList.front()->getId();
}
按原样运行并且我的程序正常工作，但是如果我在开头删除空FOR循环（没有做任何事情），我会遇到分段错误。 任何线索？
修改
候选课，实际上我不知道我在哪个方面得到了段错误， 我正在使用eclipse，调试器似乎无法正常工作
#include "../include/Candidate.h"
#include <iostream>
#include "../include/AppLogger.h"
#include <sstream>
Candidate::Candidate(string id, list<Skill> skills, list<
string> desiredJobs, double expectedSalary) :
id_(id), dateJoined_(), skills_(skills),
desiredJobs_(desiredJobs), expectedSalary_(expectedSalary),
originalSalary_(expectedSalary), gotJob_(0) {
}
void Candidate::compromise(const DateTime& currentDate) {
double salaryAfter30(0.9*this->originalSalary_);
double salaryAfter60(0.8*this->originalSalary_);
double salaryAfter90(0.7*this->originalSalary_);
Timespan duration = currentDate - this->dateJoined_;
if (duration.days() == 30 || duration.days() == 60 || duration.days() == 90) {
if (duration.days() == 30 && (this->expectedSalary_
== this->originalSalary_)) {
this->expectedSalary_ = salaryAfter30;
std::stringstream sstm;
sstm << "Candidate "<< this->getId() <<" is willing to compromise, and his expected salary is " <<this->expectedSalary_ << ".";
CAppLogger::Instance().Log(sstm.str(),
Poco::Message::PRIO_WARNING);
return;
}
else if (duration.days()==30)
poco_bugcheck_msg("Error, 30 days passed, worker already compromised");
if (duration.days() == 60 && (this->expectedSalary_ == salaryAfter30)) {
this->expectedSalary_ = salaryAfter60;
std::stringstream sstm;
sstm << "Candidate "<< this->getId() <<" is willing to compromise, and his expected salary is " <<this->expectedSalary_ << ".";
CAppLogger::Instance().Log(sstm.str(),
Poco::Message::PRIO_WARNING);
return;
}
else if (duration.days()==60)
poco_bugcheck_msg("Error, 60 days passed, worker already compromised");
if ((duration.days() == 90) && (this->expectedSalary_ == salaryAfter60)) {
this->expectedSalary_ = salaryAfter90;
std::stringstream sstm;
sstm << "Candidate "<< this->getId() <<" is willing to compromise, and his expected salary is " <<this->expectedSalary_ << ".";
CAppLogger::Instance().Log(sstm.str(),
Poco::Message::PRIO_WARNING);
return;
}
else if (duration.days()==90)
poco_bugcheck_msg("Error, 90 days passed, worker already compromised");
}
else poco_bugcheck_msg("Error, worker told to compromise when not needed");
}
list<Skill> Candidate::getSkills() const {
return this->skills_;
}
list<string> Candidate::getDesiredJobs() const {
return this->desiredJobs_;
}
double Candidate::getExpectedSalary() const {
return this->expectedSalary_;
}
DateTime Candidate::getDateJoined() const {
return this->dateJoined_;
}
DateTime Candidate::getDateLeft() const {
return this->dateLeft_;
}
void Candidate::setDateLeft(const DateTime& date) {
this->dateLeft_ = date;
}
string Candidate::getId() const {
return this->id_;
}
void Candidate::setDateJoined(const DateTime& date) {
this->dateJoined_=date;
this->setGotJob();
}
void Candidate::setGotJob() {
if (this->gotJob_==1)
std::cerr<<"error, setting gotJob while already has job"<<std::endl;
this->gotJob_=1;
}
bool Candidate::gotJob() const {
return this->gotJob_;
}
void Candidate::setQl(double ql){
jobQl_=ql;
}
int Candidate::getQl() const{
return this->jobQl_;
}
应用提供的解决方案后，我收到以下错误：
assignment2(48823) malloc: *** mmap(size=140734799806464) failed (error code=12)
*** error: can't allocate region
*** set a breakpoint in malloc_error_break to debug
terminate called after throwing an instance of 'std::bad_alloc'
what(): std::bad_alloc
修改
将int m更改为double m，现在似乎可以正常工作 因为得到的预期薪水会增加一倍 但为什么会导致这个错误？
答案 0 :(得分：9)
请看看Cris Hopman回答here，因为他的回答比我的回答更加完整和详细，指出了原始代码中可能/将导致未定义行为（分段错误）的其他问题。
这种情况是错误的：
for (std::list<Candidate*>::iterator iter = candidates.begin(); (*iter)->getExpectedSalary()
== m && iter != candidates.end(); ++iter)
您应重新排序，以便end()的测试在取消引用之前进行并保持短路评估。实际上，您将取消引用指向容器末尾的指针：
for (std::list<Candidate*>::iterator iter = candidates.begin();
iter != candidates.end() && (*iter)->getExpectedSalary() == m; ++iter)
答案 1 :(得分：7)
让我稍微分解一下......
...
int m(candidates.front()->getExpectedSalary());
你认为候选人是非空的，如果记录在案，但我会添加
，这很好assert(!candidates.empty())
在函数的开头。这两者都强制执行约束，并使读取代码的任何人都明白约束。
其次，getExpectedSalary（）返回一个你要转换为int的double ...稍后详细介绍。
list<Candidate*> potentialList;
for (std::list<Candidate*>::iterator iter = candidates.begin();
(*iter)->getExpectedSalary() == m && iter != candidates.end(); ++iter)
potentialList.push_back(*iter);
这里有两个问题：iter != candidates.end()应该是条件的第一部分，否则你可能会在第一部分中取消引用无效的迭代器。其次，平等检查很糟糕。即使列表中的第一项也可能失败，因为如果d是双精度则不一定是(int)d == d的情况。关键是m应该是双倍的。通过该更改，测试可能仍会意外失败，因为它是浮点比较，请参阅here。
if (potentialList.size() > 0)
potentialList.sort(candidateIdCompare);
return potentialList.front()->getId();
这是不一致的。在if中，您假设potentialList可能为空，但在返回时您假设它不能为空。实际上，potentialList不应该为空（因为它至少应该包含candidate.front（）），因此应该更改为：
assert(!potentialList.empty());
potentialList.sort(candidateIdCompare);
return potentialList.front()->getId();
如果您只进行了更改，您会发现断言将失败。然后，您将意识到potentialList是空的，因为上面讨论过从double到int的转换。
最后，您似乎将货币值存储为浮点数。这可能会引起问题，要小心。为此只使用定点通常更安全，但不是必需的。
作为最后一点，这个功能本身过于复杂。以下应该有效（无保证）......
string Company::cheap(list<Candidate*>& candidates) {
typedef list<Candidate*> list_t;
assert(!candidates.empty());
candidates.sort(candidateSalaryCompare);
pair<list_t, list_t> rng = equal_range(candidates.begin(), candidates.end(),
candidates.front(), candidateSalaryCompare);
assert(rng.first != rng.second);
return (*min_element(rng.first, rng.second, candidateIdCompare))->getId();
}
这也方便地将所有工资比较卸载到另一个功能，因此只需要正确完成一次。
答案 2 :(得分：2)
你的第二个循环，特别是条件，可能是危险的：
(*iter)->getExpectedSalary() == m && iter != candidates.end()
这是因为在实际检查迭代器是否有效之前，它会将iter与m 进行比较。您需要反转检查的顺序，以便条件变为：
iter != candidates.end() && (*iter)->getExpectedSalary() == m