Why is the output different even though the expressions are the same in lambda expression?-CodePudding

I have problem about lambda expression like this:

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std; 

int sum(vector<int>& v){
    int total = 0;
    auto lambda = for_each(v.begin(), v.end(), [&total](int n){total  = n;});
    lambda; // lambda expression doesn't work.
    for_each(v.begin(), v.end(), [&total](int n){total  = n;}); // work same as I intended.
    return total;
}

int main(void){
    vector<int> v = {1, 2, 3, 4, 5};
    cout << sum(v) << endl; // 30 (I think this should be 45.)
}

I thought that lambda; could do same things like for_each algorithm. Why lambda; doesn't works?

CodePudding user response：

according to cppreference, the for_each(...) call returns the UnaryFunction which was passed to the function.

In this case, the for_each returns the UnaryFunction

[&total](int n){total  = n;}

a lambda(5) would increase the total value by 5.

A solution would be to put the for_each call in a separate function - which is actually a "sum" function. This is already done by the std:accumulate function

std::accumulate(v.begin(), v.end(), 0);

CodePudding user response：

If do not take into account the side effect of accumulating the variable total this declaration

auto lambda = for_each(v.begin(), v.end(), [&total](int n){total  = n;});

in fact is equivalent to

auto lambda = [&total](int n){total  = n;};

So this statement

lambda; // lambda expression doesn't work.

does not make a sense.

So the declaration of the lambda expression is just redundant.

In fact you called the algorithm std::for_each two times in this declaration

auto lambda = for_each(v.begin(), v.end(), [&total](int n){total  = n;});

and in this statement

for_each(v.begin(), v.end(), [&total](int n){total  = n;});