在C中，我的程序输出的数字不在用户输入数组内或数组中任何数字之和不输出任何内容

Question

I am doing an assignment for class. The prompt is:

Write a program to find the smallest positive integer that does not appear in the array. The number can't be formed by sum of 2 different or sum of same numbers from the array. for e1,2,3 and 4 appear in this array. 4 can be formed as 3+1, 5 as 3+2, 6 as 3+3 , 7 as 3+4 and 8 as 4+4 (it is ok to use a number twice). 9 does not appear in the array and it can not be formed as sum of 2 numbers in the array. So, 9 is the solution for this array. Use functions

 int issumof2(int data[], int size, int number) int inarray(int data[], int size, int number) 

• issumof2 returns 1 if number is sum of 2 elements in the data and returns 0 otherwise.
• inarray returns 1 if the number is in the data and returns 0 otherwise.

#include <stdio.h>
//function declaration
int issumof2(int data[], int size, int number);
int inarray(int data[], int size, int number);
int size = 7, sum2, sum, number, data[7], i, j;

int main()
{
    //get user input
    printf("Enter 7 numbers: ");
    for (i = 0; i < size; i++)
        scanf("%d", &data[i]);  
    //function call
    issumof2(data, size, number);
    inarray(data, size, number);
    number = 0;
    //loop while condition is true and stop when condition becomes false
    while (inarray(data, size, number) == 1 || issumof2(data, size, number) == 1)
    {
        number = number + 1;//increment number till loop stop
         //print smallest number
        printf("Smallest positive integer: %d\n", number);
    }
    return 0;
}
Expected output:
Enter 7 numbers
1 2 2 3 4 3 1
Smallest positive Integer = 9
//1,2,3,4 are in array. 3+1=4,3+2=5,3+3=6,3+4=7,4+4=8(number can be used twice).
//9 is not inarray or not a  sum of 2 numbers in the array

int issumof2(int data[], int size, int number)
{
    //add data numbers in array
    for (i = 0; i < size; i++)
    {
        for (j = 0 + i; j < size; j++)
        {
            sum = data[i] + data[i];//add same numbers        
            sum2 = data[i] + data[j];//add all 2 combinations of different number             
            if (sum == data[i] || sum2 == data[i])//comparing with data value          
                return 1;//if sum  or sum2 exists in data
        }
    }
    return 0;//if sum or sum2 don't exist
}

//goes to infinite loop
int inarray(int data[], int size, int number)
{
    for (number = 1; number <=size; number++)
    {
        printf("inarray number=%d\n", number);
        //loop through data array
        for (i = 0; i < size; i++)
        {
            printf("data[i]=%d\n", data[i]);
            if (number == data[i])//if this is true it goes to infinite loop
                                  //if false it just stops comparing
            {
                return 1;
                printf("\n");
            }
        }
    }
    return 0;
}

Answer 1

A quick run of what you've posted actually results in an infinite loop, unless i've missed something. A couple of errors I see:

• You have set size equal to three, but then use it to attempt to scan seven numbers.
• In both of the functions inarray and issumof2 , there are conditionals that return either one or zero depending on the result of the conditional, so they only ever run one iteration of the innermost loops. You need to move the else statements outside of the loops.
• The smallest possible integer that meets the criteria given the sample input is in fact eight.
• It seems the logic in both functions are off as they never return 1.
  • In issumof2 there is no use of number , yet that is what you must compare any sums to if I am not mistaken. Additionally, i think you may have inadvertently used i in place of j or vice versa.

Here is some of your code commented with remarks that hopefully lead you to fix the mistakes:

int main() {
    int data[7];
    int i, number = 0;

    // size is three so we won't scan 7 integers if we enter them line by line.
    printf("Enter 7 numbers: ");
    for (i = 0; i < size; i++) 
        scanf("%d", &data[i]);

    // the problem states that both of these functions must return zero for the
    // expected answer, yet this loop breaks whenever number does not meet both
    // criteria, which isn't what we want.
    while(inarray(data, size, number) == 0 && issumof2(data, size, number) == 0) {
        number = number + 1;
        // if the conditional matched the problem statement, then we would print
        // this line for every integer that fails the criteria, but that isn't
        // what we want.
        printf("Smallest positive integer: %d\n", number);
    }
    return 0;
}

int issumof2(int data[], int size, int number) {
    int i, j, sum = 0;
    for (i = 0; i < size; i++) {
        // here we've doubled the value at i and saved within a temporary
        // variable -- but why? 
        sum2 =data[i]+data[i];
        for (j = 0; j < size; j++) {
            // here we're combining different values to test their sum against
            // number -- good job 
            sum = data[i] + data[j];
            // why are we comparing the data at i with twice itself (sum2), and
            // its addition with the data at j (sum)? we must compare sum2
            // against number.
            if (sum==data[i] || sum2==data[i])
                // here we return 1 if the condition is met. this would be good
                // if our conditional was representative of the problem.
                return 1;
            else
                // otherwise we return 0 -- why? i guess we don't need to check
                // any other sums...
                return 0;
        }
    }
    // alright we're done here. or are we? the prototype says we'll return an int
}

int inarray(int data[], int size, int number)
{
    int i;
    // so we're looping over all input -- good job
    for (i = 0; i < size; i++) {
        // here we test for equality of index i with the data at index i -- why?
        // we must compare the data at index i with number.
        if (i == data[i])
            return 1;
        else // again we can't just call it a day after checking just one number
            return 0;
    }
    // and as before we need to return some value if we complete the loop
}

And here is a corrected solution:

#include<stdio.h>

int issumof2(int[], int, int);
int inarray(int[], int, int);

int size = 7;

int issumof2(int data[], int size, int number) {
    int i, j;
    for (i = 0; i < size; i++) {
        for (j = 0; j < size; j++) {
            if (i == j)
               continue;
           if (data[i] + data[j] == number)
               return 1;
        }
    }
    return 0;
}

int inarray(int data[], int size, int number) {
    int i;
    for (i = 0; i < size; i++) {
        if (data[i] == number)
            return 1;
    }
    return 0;
}

int main() {
    int data[7];
    int i, number = 1;
    printf("Enter 7 numbers: ");
    for (i = 0; i < size; i++)
        scanf("%d", &data[i]);

    while(inarray(data, size, number) == 1 || issumof2(data, size, number) == 1)
        number++;

    printf("Smallest positive integer: %d\n", number);

    return 0;
}

Answer 2

First Here is the corrected answer for your logic

#include<stdio.h>

int issumof2(int data[], int size, int number);
int inarray(int data[], int size, int number);
int size=7,sum2;
int main()
{
   int data[7];
   int i, number=1;
   printf("Enter 7 numbers: ");
   for (i = 0; i < size; i++)
       scanf("%d", &data[i]);
   //I am a little confused about this part. I wasn't sure how to write it.
   while(inarray(data, size, number)==1 || issumof2(data, size, number)==1)//till functions are false
   {
        number=number+1;
   }

    printf("Smallest positive integer: %d\n", number);                      
    return 0;
}

int issumof2(int data[], int size, int number)
{
   int i, j, sum = 0;
   for (i = 0; i < size; i++)
   {
        sum2 =data[i]+data[i];
       for (j = 0; j < size; j++)
       {
           sum = data[i] + data[j];           
           if (sum==number || sum2==number) {
           return 1;

       }
       }

   }

   return 0;
}

int inarray(int data[], int size, int number)
{

   int i;
   for (i = 0; i < size; i++)
   {
       if (number == data[i]) {
           return 1;
       }
    }

   return 0;
}

Also I have prepared one more set of code explaining your mistakes. Hope it helps for you to learn

#include<stdio.h>

int issumof2(int data[], int size, int number);
int inarray(int data[], int size, int number);
int size=7,sum2; // Size has be changed from 3 to 7 , since you want to check the for 7 numbers

int main()
{
   int data[7];
   int i, number=0;
   printf("Enter 7 numbers: ");
   for (i = 0; i < size; i++)
       scanf("%d", &data[i]);
   //I am a little confused about this part. I wasn't sure how to write it.
//   while(inarray(data, size, number)==0 && issumof2(data, size, number)==0)//till functions are false 

//   I feel its best to check the followin way
//   If any one of the condition is true, then that is not the required number, so increment it
//   else leave the loop which is the required number
   while(inarray(data, size, number)==1 || issumof2(data, size, number)==1)//till functions are false 
    {
        number=number+1;
        //printf("Smallest positive integer: %d\n", number);   
        // You will print the required number after the loop, not inside the loop

    }
        printf("Smallest positive integer: %d\n", number);                      
    return 0;
}

int issumof2(int data[], int size, int number)
{
   int i, j, sum = 0;
   for (i = 0; i < size; i++)
   {
        sum2 =data[i]+data[i];
       for (j = 0; j < size; j++)
       {
           sum = data[i] + data[j];           
//           if (sum==data[i] || sum2==data[i]) 
           if (sum==number || sum2==number)  // You are supposed to check the Sum with Number. Not with data irself
               return 1;
//            else
//                return 0; // Returning 0 here is not correct, It should be returned if all the condition are failed. 
       }
   }

   return 0; // Correct place to return 0;
}

int inarray(int data[], int size, int number)
{

   int i;
   for (i = 0; i < size; i++)
   {
//       if (i == data[i])
       if (i == number) // You are supposed to check with numbner not with data itself
           return 1;
//       else 
 //           return 0; // You are not supposed to return 0 here. Return is after all the condition is failed
    }

   return 0; // Corect place to return 0
}

Answer 3

I compiled and ran your code. It will print, but runs off into the hundreds of thousands.

$ gcc code.c
$ ./a.out
>Enter 7 numbers: 1 2 3 4 5 6 7 --press enter here--
...

I could somewhat recreate your issue if I pressed enter after a single number. Make sure you're typing your input as a single line.