I want to iterate through each element in the map<string, int> without knowing any of its string-int values or keys.
What I have so far:
void output(map<string, int> table)
{
map<string, int>::iterator it;
for (it = table.begin(); it != table.end(); it++)
{
//How do I access each element?
}
}
You can achieve this like following :
map<string, int>::iterator it;
for (it = table.begin(); it != table.end(); it++)
{
std::cout << it->first // string (key)
<< ':'
<< it->second // string's value
<< std::endl;
}
With C++11 ( and onwards ),
for (auto const& x : table)
{
std::cout << x.first // string (key)
<< ':'
<< x.second // string's value
<< std::endl;
}
With C++17 ( and onwards ),
for (auto const& [key, val] : table)
{
std::cout << key // string (key)
<< ':'
<< val // string's value
<< std::endl;
}
auto const& [key, val] : symbolTable format!
Try the following
for ( const auto &p : table )
{
std::cout << p.first << '\t' << p.second << std::endl;
}
The same can be written using an ordinary for loop
for ( auto it = table.begin(); it != table.end(); ++it )
{
std::cout << it->first << '\t' << it->second << std::endl;
}
Take into account that value_type for std::map is defined the following way
typedef pair<const Key, T> value_type
Thus in my example p is a const reference to the value_type where Key is std::string and T is int
Also it would be better if the function would be declared as
void output( const map<string, int> &table );
The value_type of a map is a pair containing the key and value as it's first and second member, respectively.
map<string, int>::iterator it;
for (it = symbolTable.begin(); it != symbolTable.end(); it++)
{
std::cout << it->first << ' ' << it->second << '\n';
}
Or with C++11, using range-based for:
for (auto const& p : symbolTable)
{
std::cout << p.first << ' ' << p.second << '\n';
}
As @Vlad from Moscow says,
Take into account that value_type for std::map is defined the following way:
typedef pair<const Key, T> value_type
This then means that if you wish to replace the keyword auto with a more explicit type specifier, then you could this;
for ( const pair<const string, int> &p : table ) {
std::cout << p.first << '\t' << p.second << std::endl;
}
Just for understanding what auto will translate to in this case.
As P0W has provided complete syntax for each C++ version, I would like to add couple of more points by looking at your code
const & as argument as to avoid extra copies of the same object.unordered_map as its always faster to use. See this discussionhere is a sample code:
#include <iostream>
#include <unordered_map>
using namespace std;
void output(const auto& table)
{
for (auto const & [k, v] : table)
{
std::cout << "Key: " << k << " Value: " << v << std::endl;
}
}
int main() {
std::unordered_map<string, int> mydata = {
{"one", 1},
{"two", 2},
{"three", 3}
};
output(mydata);
return 0;
}
& which avoids creating the multiple copies and when we take the reference, we do not want it to be modified hence we use constif you just want to iterate over the content without changing values do:
for(const auto & variable_name : container_name(//here it is map name)){
cout << variable_name.first << " : " << variable_name.second << endl;
}
If you want to modify the contents of the map, remove the const and keep & (if you want to modify directly the contents inside container). If you want to work with a copy of the container values, remove the & sign too; after that, you can access them by using .first and .second on "variable_name".
it can even be done with a classic for loop.
advancing the iterator manually.
typedef std::map<int, int> Map;
Map mymap;
mymap['a']=50;
mymap['b']=100;
mymap['c']=150;
mymap['d']=200;
bool itexist = false;
int sizeMap = static_cast<int>(mymap.size());
auto it = mymap.begin();
for(int i = 0; i < sizeMap; i++){
std::cout << "Key: " << it->first << " Value: " << it->second << std::endl;
it++;
}
Other way :
map <int, string> myMap = {
{ 1,"Hello" },
{ 2,"stackOverflow" }
};
for (auto iter = cbegin(myMap); iter != cend(myMap); ++iter) {
cout << iter->second << endl;
}
const map<string, int>& table