C++ list中erase和remove函式的用法
erase的作用是,使作為引數的迭代器失效,並返回指向該迭代器下一引數的迭代器。
如下:
list ParticleSystem;
list::iterator pointer;
if(pointer->dead == true)
{
pointer = ParticleSystem.erase(pointer);
}
有一段關於錯誤使用erase的程式
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); } }
問題:該程式不能跳出迴圈
原因:test_list.erase(test_list_it);每次做erase時都有可能使迭代器失效,test_list_it++就發生錯誤了。可以參見effective stl一書。所有容器做erase操作時都有可能使迭代器失效。
改為:
for(;test_list_it != test_list.end();)
{
test_list.erase(test_list_it++);
}
or
for(;test_list_it != test_list.end();) { std::list::iterator iter_e=test_list_it++; test_list.erase(iter_e); }
注意:
for(;test_list_it != test_list.end();test_list_it++;) {
std::list::iterator iter_e=test_list_it;
test_list.erase(iter_e);
}
這樣仍然是錯誤的,原因是:iter_e=test_list_it 是指標值的複製,它倆其實指向同一個位置,所以iter_e失效那麼test_list_it也會失效,所以test_list_it++就會有問題
如果是
for(;test_list_it != test_list.end();) { std::list::iterator iter_e=test_list_it++; test_list.erase(iter_e); }
則沒有問題。
remove函式也存在erase函式同樣的問題,但remove函式返回值是空,erase返回指向下一個元素的迭代器。
下面是一個簡單的例子。
#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;
}
其中:
for (list<int>::iterator i = ls.begin(); i != ls.end(); ) {
printf("erase %d \n", *i);
ls.erase(i++);
}
也可以寫成下面的形式,因為erase函式的返回值就是指向下一個元素的迭代器。
for (list<int>::iterator i = ls.begin(); i != ls.end(); ) {
printf("erase %d \n", *i);
i = ls.erase(i);
}
輸出結果如下:
------------------------------ 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