How to pass a local variable to lambda function?

Question

I have an integer 153, and I need to determine whether or not he is equal to the sum of pow(digit, digits). so 153 = 1^3 + 5^3 + 3^3 = 153 returns true.

I chose to find the number of digits using a basic loop, then pushing all the integers into a vector so I could use the std::for_each function to iterate and use pow() on each of the elements.

The problem I am facing is the changing of each element. I don't want to resort to a for-loop yet, but I can't seem to figure out how to do it via the for_each.

Errors are commented in the code.

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <numeric>

bool narcissistic(int value)
{
    int k{};
    std::vector <int> sum{};
    while (value > 0) {
        k++;
        sum.push_back(value % 10);
        value /= 10;
    }
    std::for_each(sum.begin(), sum.end(), [](int& n) {pow(n, k); }); // error: 'k' is not captured
                                                              // (warning) note: the lambd has no capture-default
    if (std::accumulate(sum.begin(), sum.end(), 0) == value) return true;
    return false;
}

int main() 
{
    std::cout << narcissistic(153) << std::endl;
    return 0;
}

Possible duplicate of [What is a lambda expression in C++11?](https://stackoverflow.com/questions/7627098/what-is-a-lambda-expression-in-c11) The error message gave you the key word "capture" -- look for that word in the answers. — JaMiT, Aug 27 '19 at 03:41

Dinesh · Accepted Answer · 2019-08-27T03:25:04.310

You can let the lambda function to capture k

by value (the lambda function gets a copy) by using [k](int &n) { pow(n, k); }
by reference by using [&k](int &n) { pow(n, k); }.

Read more about lambda captures at https://isocpp.org/wiki/faq/cpp14-language#lambda-captures

That being said, I wouldn't use a std::vector to solve your problem. You can directly use unsigned int k = std::floor(std::log10(n)) + 1; to get the number of digits in a number, and just accumulate the sum in your while loop. Make sure to check for n > 0 at the beginning though.

JeJo · Answer 2 · 2019-08-27T03:51:48.273

but I can't seem to figure out how to do it via the for_each.

You have a couple of problems there

In the while-loop you are changing the original passed value here
```
value /= 10;
```
Therefore, latter you can not compare with it at the line.
```
if(std::accumulate(sum.begin(),sum.end(),0) == value)
```
You should preserve the original value for the latter comparison.

Secondly, you need to sum the pow(n, k) up not the sum vector as it only contains the individual digits of the passed value.

double sum{};   // since, std::pow returns the floating points 
std::for_each(powVec.begin(), powVec.end(), [k, &sum](int n) { sum += std::pow(n, k); });
//                                                            ^^^^^^^^^^^^^^^^^^^^^ -> sum the powers of digits
//                                                    ^^^^^^ -> no need of taking reference here
//                                           ^^^^^^^^^ -> capture `k` by value, `sum` by reference

Thirdly, the std::accumulate(sum.begin(),sum.end(),0) == value is wrong as value would have been already 0 at this time and the std::accumulate returns the sum of the individual digits(which is not what you wanted)

That means, change the program to: (See online live)

bool narcissistic(const int value)
//                ^^^^ --> cons because, passed value can only read
{
    int k{};
    std::vector<int> powVec{}; // named as a vector of power of digits in the `value`
    int value_copy{ value };   // copy of original value
    while (value_copy > 0) {
        k++;
        powVec.push_back(value_copy % 10);
        value_copy /= 10;      // work with the copy of value
    }
    double sum{};  // since, std::pow returns the floating points
    std::for_each(powVec.begin(), powVec.end(),
        [k, &sum](int n) { sum += std::pow(n, k); });
    //   ^^^^^^^^-->capture `k` by value `sum` by reference
    // if (std::accumulate(sum.begin(), sum.end(), 0) == value) return true; // not required
    return static_cast<int>(sum) == value;
}

You can read more about the lambdas here:

How to pass a local variable to lambda function?

2 Answers2