我有一个基于范围的 for 循环来迭代 foobar 中的元素,如下所示:
#include 此代码产生以下输出:
{1, 2} {2, 3} {3, 4}{1, 1} {2, 2} {3, 3}第一行被修改并打印在 for 循环中,第二行应该打印相同的修改值.为什么输出不匹配?对 std::map 的更改是否仅在循环范围内有效?有没有办法不仅可以访问而且可以修改这些值?
可以在 cpp.sh 上找到 此代码的运行版本.
为了清楚起见,此处给出的示例经过修改以匹配接受的答案.
你可以把 auto 变成 auto& 如果你想改变/修改容器,例如:
#include 编译和输出
<前>{1, 2} {2, 3} {3, 4}现场示例
I have a range based for loop to iterate over elements in foobar as follows:
#include <map>
#include <iostream>
int main()
{
std::map<int, int> foobar({{1,1}, {2,2}, {3,3}});
for(auto p : foobar)
{
++p.second;
std::cout << "{" << p.first << ", " << p.second << "} ";
}
std::cout << std::endl;
for(auto q : foobar)
{
std::cout << "{" << q.first << ", " << q.second << "} ";
}
std::cout << std::endl;
}
This code produces the following output:
{1, 2} {2, 3} {3, 4}
{1, 1} {2, 2} {3, 3}
The first line is modified and printed inside a for loop and the second line supposedly prints the same modified values. Why don't the outputs match? Are changes to std::map only effective in the scope of the loop? Is there a way I can not only access but modify these values?
A running version of this code can be found on cpp.sh.
EDIT: The example given here was modified to match the accepted answer for clarity.
You can turn auto into auto& if you want to mutate/modify the container, for instance:
#include <map>
#include <iostream>
int main()
{
std::map<int, int> foobar({{1,1}, {2,2}, {3,3}});
for(auto& p : foobar) {
++p.second;
std::cout << '{' << p.first << ", " << p.second << "} ";
}
std::cout << std::endl;
}
compiles ands outputs
{1, 2} {2, 3} {3, 4}
live example
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