Question

I don’t get why I get 0 when I use printf and %d to get the size of my vector:

vector<long long> sieve; int size; ... //add stuff to vector ... size = sieve.size(); printf('printf sieve size: %d \n', size); //prints 'printf sieve size: 0' std::cout << 'cout sieve size: '; std::cout << size; std::cout << ' \n '; //prints 'cout sieve size: 5 (or whatever the correct sieve size is)' 

If I iterate through the vector via

if(i=0;i<sieve.size();i++)  

I get the correct number of iterations.

What am I doing wrong or what is up with printf? size() returns an int right??

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Here’s my entire little script:

#include <iostream> #include <vector> #include <stack> #include <math.h>  int main (int argc, char * const argv[]) {     unsigned long long answer = 0;     unsigned long long cur = 2;     std::vector<long long> sieve;     unsigned long long limit;     unsigned long long value;     unsigned int i;     int size;     bool isPrime;     std::cout << 'Provide a value to find its largest prime factor: ';     std::cin >> value;     limit = ceil(sqrt(value));     sieve.push_back(2);     while(cur++ < limit){       isPrime = true;       sieve.begin();       for(i=0; i<sieve.size();i++){         if(!(cur % sieve[i])){           isPrime = false;           break;         }       }       if(isPrime){           if(!(value % cur)){           std::printf('Is prime factor: %d\n', cur);           sieve.push_back(cur);           answer = sieve[sieve.size() - 1];           size = sieve.size();           std::printf('current last: %d sieve size: %ld\n', answer, size);           for(i=0; i<sieve.size();i++){             std::printf('sieve iter: %d sieve val: %d\n', i, sieve[i]);             std::cout << size;             std::cout << ' wtf\n';           }         }       }     }     answer = sieve[sieve.size() - 1];     size = sieve.size();     std::printf('Limit: %d Answer: %d sieve size: %ld\n', limit, answer, size);     return 0; } 

Answer 1

Now, with the complete source, it is clear.

You declared:

int size; 

Then you used:

std::printf('current last: %d sieve size: %ld\n', answer, size); std::printf('Limit: %d Answer: %d sieve size: %ld\n', limit, answer, size); 

If size is int, you should use ‘%d’, not ‘%ld’. A good compiler would have warned you about this. GCC gives these warnings for your original version:

test.cpp: In function ‘int main(int, char* const*)’: test.cpp:17: warning: converting to ‘long long unsigned int’ from ‘double’ test.cpp:30: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘long long unsigned int’ test.cpp:34: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘long long unsigned int’ test.cpp:34: warning: format ‘%ld’ expects type ‘long int’, but argument 3 has type ‘int’ test.cpp:36: warning: format ‘%d’ expects type ‘int’, but argument 3 has type ‘long long int’ test.cpp:45: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘long long unsigned int’ test.cpp:45: warning: format ‘%d’ expects type ‘int’, but argument 3 has type ‘long long unsigned int’ test.cpp:45: warning: format ‘%ld’ expects type ‘long int’, but argument 4 has type ‘int’ 

This tells a lot.

You should declare size as:

std::vector<long long>::size_type size; 

Then you should use it as:

std::printf('current last: %llu sieve size: %llu\n', (unsigned long long) answer, (unsigned long long) size); std::printf('Limit: %llu Answer: %llu sieve size: %llu\n', (unsigned long long) limit, (unsigned long long) answer, (unsigned long long) size); 

Of course, using iostream avoids you these problems, specially the ugly casting in printf() to transform size to a type known to printf.