merging two arrays

[ejkang62]

I have written a function to merge two string arrays as below
I want the result array to to have arr1 elements first followed by arr2 elements

void mergeArr(const string arr1[], int n1, const string arr2[], int n2, string result[]) {
	result[10000] = {};

	for (size_t i = 0; i < n1; i++) {
		result[i] = arr1[i];
	}
	for (size_t i = n1; i < n1 + n2 -1; i++) {
		for (size_t k = 0; k < n2; k++)
		{
			result[i] = arr2[k];
		}
	}

	cout << result << endl;
}

this doesn't allow me to merge two arrays and gives an error. what would i need to correct?

[Troaat]

Hi @ejkang62,
The problem in your code is that you implemented a loop within a loop, when it is not necesseary.
When the program enters "for (i=n1; ...)", it starts with the value n1 for i.
Then, for the same i value, k will get all values of the arr2 string.
To fix that I would do something like this:

size_t k=0;
for (size_t i=n1; i<n1+n2-1; i++)
{
    result[i]=arr2[k];
    k++;
}

That way you'll parse all n2 elements of arr2 and add them to result.
Also I don't know about the first instruction, "result[10000]={};". This sounds like you want to access the 10000th element of result.
Just to make sure, I'd try doing this:

result.resize(10000);

And even better, if you want to have the minimum number of elements in result, do this:

result.resize(arr1.length()+arr2.length());
//or
int
length1=arr1->length(),
length2=arr2->length();
result->resize(length1+length2);

Also, I don't know what error it is that you're getting.
If this didn't solve it, could you say a little more about the problem?
Hope this helps.

EDIT:
Also, since you're using std::string object, it's for the best to use its overloaded operator.
Instead of:

cout<<result<<endl;

You might want to do this:

cout<<result+'\n';

This is concatenation, which makes the output WAY faster.

[gunnerfunner]

you've declared result[] with a seemingly large size, 10000, but what if n1 == 10050? in that case all of arr1[] could not be read into result despite it's 'large' size let along arr2[]. so the fool-proof method would be to declare the new, merged array dynamically of such size that can hold the contents of both arr1[] and arr2[]:

#include <iostream>
#include <string>

void mergeArray(const std::string* arr1, const size_t size_1, const std::string* arr2, const size_t size_2)
{
   std::string* p = new std::string [size_1 + size_2];
    for (size_t i = 0; i < size_1; ++i)
    {
        *(p + i) = *(arr1 + i );
    }
    for (size_t i = 0; i < size_2; ++i)
    {
        *(p + size_1 + i) = *(arr2 + i);
    }
    for (size_t i = 0; i < size_1 + size_2; ++i)
    {
        std::cout << *(p + i) << " ";
    }
    delete []p;
}

int main()
{
    const std::string  arr1[] = {"today", "is", "friday"};
    const std::string arr2[] {"and", "yesterday", "was", "thursday"};

    mergeArray(arr1, 3, arr2, 4);
}

needless to say std::vector is usually the container of choice over C-style arrays but perhaps this is a class exercise for you