基于C++ list中erase与remove函数的使用详解
erase的作用是,使作为参数的迭代器失效,并返回指向该迭代器下一参数的迭代器。
如下:
代码如下:
list ParticleSystem;
list::iterator pointer;
if(pointer->dead == true)
{
pointer = ParticleSystem.erase(pointer);
}
using namespace std;
int main()
{
std::listtest_list;
std::list::iterator test_list_it;
test_list.push_back(1);
test_list_it = test_list.begin();
for(;test_list_it != test_list.end();test_list_it++)
{
test_list.erase(test_list_it);
}
}
问题:该程序不能跳出循环
原因:test_list.erase(test_list_it);每次做erase时都有可能使迭代器失效,test_list_it++就发生错误了。可以参见effective stl一书。所有容器做erase操作时都有可能使迭代器失效。
改为:
代码如下:
for(;test_list_it != test_list.end();)
{
test_list.erase(test_list_it++);
}
for(;test_list_it != test_list.end();)
{
std::list::iterator iter_e=test_list_it++;
test_list.erase(iter_e);
}
for(;test_list_it != test_list.end();test_list_it++;) {
std::list::iterator iter_e=test_list_it;
test_list.erase(iter_e);
}
这样仍然是错误的,原因是:iter_e=test_list_it 是指针值的复制,它俩其实指向同一个位置,所以iter_e失效那么test_list_it也会失效,所以test_list_it++就会有问题
如果是
代码如下:
for(;test_list_it != test_list.end();)
{
std::list::iterator iter_e=test_list_it++;
test_list.erase(iter_e);
}
则没有问题。
remove函数也存在erase函数同样的问题,但remove函数返回值是空,erase返回指向下一个元素的迭代器。
#include "stdafx.h"
#include <stdio.h>
#include <string.h>
#include <malloc.h>
#include <list>
using namespace std;
int _tmain(int argc, _TCHAR* argv[])
{
printf("------------------------------ Start\n");
list<int> ls;
printf("ls.empty() = %d \n", ls.empty());
printf("ls.max_size() = %d \n", ls.max_size());
printf("ls.size() = %d \n", ls.size());
ls.push_back(1);
ls.push_back(2);
ls.push_back(3);
printf("\n--------- after push 1, 2, 3 ---------\n");
printf("ls.empty() = %d \n", ls.empty());
printf("ls.max_size() = %d \n", ls.max_size());
printf("ls.size() = %d \n", ls.size());
for (list<int>::iterator i = ls.begin(); i != ls.end(); i++) {
printf("%d, ", *i);
}
printf("\n------------------------------\n");
for (list<int>::iterator i = ls.begin(); i != ls.end(); ) {
printf("erase %d \n", *i);
ls.erase(i++);
}
printf("\n--------- after erase ---------\n");
printf("ls.empty() = %d \n", ls.empty());
printf("ls.max_size() = %d \n", ls.max_size());
printf("ls.size() = %d \n", ls.size());
printf("\n------------------------------\n");
ls.push_back(1);
ls.push_back(2);
ls.push_back(3);
for (list<int>::iterator i = ls.begin(); i != ls.end(); ) {
printf("remove %d \n", *i);
ls.remove(*i++);
}
printf("\n--------- after remove ---------\n");
printf("ls.empty() = %d \n", ls.empty());
printf("ls.max_size() = %d \n", ls.max_size());
printf("ls.size() = %d \n", ls.size());
printf("\n------------------------------ End\n");
getchar();
return 0;
}
for (list<int>::iterator i = ls.begin(); i != ls.end(); ) {
printf("erase %d \n", *i);
ls.erase(i++);
}
也可以写成下面的形式,因为erase函数的返回值就是指向下一个元素的迭代器。
代码如下:
for (list<int>::iterator i = ls.begin(); i != ls.end(); ) {
printf("erase %d \n", *i);
i = ls.erase(i);
}
输出结果如下:
------------------------------ Start
ls.empty() = 1
ls.max_size() = 1073741823
ls.size() = 0
--------- after push 1, 2, 3 ---------
ls.empty() = 0
ls.max_size() = 1073741823
ls.size() = 3
1, 2, 3,
------------------------------
erase 1
erase 2
erase 3
--------- after erase ---------
ls.empty() = 1
ls.max_size() = 1073741823
ls.size() = 0
------------------------------
remove 1
remove 2
remove 3
--------- after remove ---------
ls.empty() = 1
ls.max_size() = 1073741823
ls.size() = 0
------------------------------ End