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Anuj Saharan
Anuj Saharan
  • 43
  • 7
#include <iostream>
#include <vector>

bool is_lucky(int check_num)
{
while(check_num!=0)
    {
    if((check_num%10!=4)&&(check_num%10!=7))
        {
        return false;
        }
    check_num/=10;
    }
return true;
}

int main()
{
std::vector <long long> lucky;
for(int in_num=1;in_num<1000;in_num++)
    {
    if(is_lucky(in_num))
        {
        lucky.push_back(in_num);
        }
    }
}

Even though I am still using the for-if anti-pattern as Deduplicator pointed out, I was finally able to make the is_lucky function which maybe less efficient and a lesser intelligent way to do it than the other ones posted, seems to be the shortest way to do the task.

#include <iostream>
#include <vector>

bool is_lucky(int check_num)
{
while(check_num!=0)
    {
    if((check_num%10!=4)&&(check_num%10!=7))
        {
        return false;
        }
    check_num/=10;
    }
return true;
}

int main()
{
std::vector <long long> lucky;
for(int in_num=1;in_num<1000;in_num++)
    {
    if(is_lucky(in_num))
        {
        lucky.push_back(in_num);
        }
    }
}

Even though I am still using the for-if anti-pattern as Deduplicator pointed out, I was finally able to make the is_lucky function which maybe less efficient than the other ones posted, seems to be the shortest way to do the task.

#include <iostream>
#include <vector>

bool is_lucky(int check_num)
{
while(check_num!=0)
    {
    if((check_num%10!=4)&&(check_num%10!=7))
        {
        return false;
        }
    check_num/=10;
    }
return true;
}

int main()
{
std::vector <long long> lucky;
for(int in_num=1;in_num<1000;in_num++)
    {
    if(is_lucky(in_num))
        {
        lucky.push_back(in_num);
        }
    }
}

Even though I am still using the for-if anti-pattern as Deduplicator pointed out, I was finally able to make the is_lucky function which maybe less efficient and a lesser intelligent way to do it than the other ones posted, seems to be the shortest way to do the task.

Source Link
Anuj Saharan
Anuj Saharan
  • 43
  • 7

#include <iostream>
#include <vector>

bool is_lucky(int check_num)
{
while(check_num!=0)
    {
    if((check_num%10!=4)&&(check_num%10!=7))
        {
        return false;
        }
    check_num/=10;
    }
return true;
}

int main()
{
std::vector <long long> lucky;
for(int in_num=1;in_num<1000;in_num++)
    {
    if(is_lucky(in_num))
        {
        lucky.push_back(in_num);
        }
    }
}

Even though I am still using the for-if anti-pattern as Deduplicator pointed out, I was finally able to make the is_lucky function which maybe less efficient than the other ones posted, seems to be the shortest way to do the task.