C程序崩溃

Question

I have found a challenge on Reddit to make a program which given a number A , will find the smallest possible value of B+C , if B*C = A . Here A , B , and C must all be positive integers. For some reason, my program crashes and I can't find the reason why. I'm still a beginner in C and I'm still trying to learn the language. Any help and tips would be helpful.

#include <stdio.h>
#define MAXDIM 128

int factor1[MAXDIM];
int factor2[MAXDIM];
int sum[MAXDIM];
int counter;

void factors(int a);
void sumf(const int *f1,const int *f2);
int smallestSum(const int *s);

int main(void){
    int a;
    scanf("%d",&a);
    factors(a);
    sumf(&factor1[0],&factor2[0]);
    printf("%d => %d",a,smallestSum(&sum[0]));
    return 0;
}

void factors(int a){
    int i=1;
    int j=0;
    while(i<a/2){
        if(a%i==0){
            /*two arrays which hold all factors 
            e.g. 
            a = 10 
            factor1 = 1 2
            factor2 = 10 5
            */
            factor1[j]=i;
            factor2[j]=a/i;
            j++;
        }
        i++;
    }
    counter=i; // counter which holds length of factor arrays, used in other functions
}

void sumf(const int *f1,const int *f2){
    int i=0;
    while(i<counter+1){
        // array which holds sums of factor1 and factor2
        sum[i]=*(f1+i)+*(f2+i);
        i++;
    }
}

int smallestSum(const int *s){
    int min;
    int i=0;
    min=*s;
    while(i<counter+1){
        if(*(s+i)<min){
            min=*(s+i);
        }
    }
    return min; 
}

Answer 1

To start off with, in smallestSum i will be forever 0, so it will run forever. I think you need to do i++. But I would think you have more bugs in there. What input are you providing when running it ?

Answer 2

one possible crash source is here:

counter=i;

it should be:

counter=j;

j is keeping factors array index, not i . Another crash source is here:

void sumf(const int *f1,const int *f2){
    int i=0;
    while(i<counter+1){ // <-- here
         ...
    }
}

Because you have counter element, it should be like this:

while(i<counter)

smallestSum() has same problem too. In addition yo have forgotten to increase i in the loop in smallestSum() .

I should say these are bugs that I found and there may be more that I'm missing. But generally this code can be much better when you get more experience.

Answer 3

When you debug your program use debugger and breakpoints. It also helps to print out intermediate results to see if the algorithm works properly. Doing so you can notice that counter gets initialized to improper value all the time. You would notice that smallestSum() function never ends because the loop never breaks. You need to increase i at the proper place inside that function. Check if your condition while(i<counter+1) is correct, since loop starts from 0 comparing to counter is enough.

The program works after all those corrections:

#include <stdio.h>
#define MAXDIM 128

int factor1[MAXDIM];
int factor2[MAXDIM];
int sum[MAXDIM];

int counter;

void factors(int a);
void sumf(const int *f1, const int *f2);
int smallestSum(const int *s);

void factors(int a){
    int i = 1;
    int j = 0;     
    while( i < a/2 ){

        if( a%i == 0){
            /*two arrays which hold all factors 
            e.g. 
            a = 10 
            factor1 = 1 2
            factor2 = 10 5
            */
            factor1[j] = i;
            factor2[j] = a/i;

            printf("i=%d j=%d factor1[j]=%d factor12[j]=%d\n",i,j,factor1[j],factor2[j]  );

            j++;
        }           
        i++;
    }

    counter = j; // counter which holds length of factor arrays
}

void sumf(const int *f1, const int *f2){

    int i = 0; 
    while(i<counter){
        // array which holds sums of factor1 and factor2
        sum[i] = *(f1+i) + *(f2+i);

         printf("i=%d sum %d %d %d\n", i,  sum[i], *(f1+i), *(f2+i) );
        i++;
    }
}

int smallestSum(const int *s){

    int min;
    int i = 0;
    min =*s;

    while(i<counter){

        if( *(s+i) < min){
             min= *(s+i);
        }
        i++;
    }

    return min;
}

int main(void){

    int a;
    scanf("%d",&a);

    factors(a);

    sumf( &factor1[0], &factor2[0] );

    printf("For a=%d the smallest sum is => %d", a, smallestSum(&sum[0]));    
    return 0;
}

Test:

90                                                                                                                                             
i=1 j=0 factor1[j]=1 factor12[j]=90                                                                                                            
i=2 j=1 factor1[j]=2 factor12[j]=45                                                                                                            
i=3 j=2 factor1[j]=3 factor12[j]=30                                                                                                            
i=5 j=3 factor1[j]=5 factor12[j]=18                                                                                                            
i=6 j=4 factor1[j]=6 factor12[j]=15                                                                                                            
i=9 j=5 factor1[j]=9 factor12[j]=10                                                                                                            
i=10 j=6 factor1[j]=10 factor12[j]=9                                                                                                           
i=15 j=7 factor1[j]=15 factor12[j]=6                                                                                                           
i=18 j=8 factor1[j]=18 factor12[j]=5                                                                                                           
i=30 j=9 factor1[j]=30 factor12[j]=3                                                                                                           
i=0 sum 91 1 90                                                                                                                                
i=1 sum 47 2 45                                                                                                                                
i=2 sum 33 3 30                                                                                                                                
i=3 sum 23 5 18                                                                                                                                
i=4 sum 21 6 15                                                                                                                                
i=5 sum 19 9 10                                                                                                                                
i=6 sum 19 10 9                                                                                                                                
i=7 sum 21 15 6                                                                                                                                
i=8 sum 23 18 5                                                                                                                                
i=9 sum 33 30 3                                                                                                                                
For a=90 the smallest sum is => 19