合并排序C中的逻辑错误是什么

Question

编辑：这绝不是我之前的问题的重复。我的代码已根据我之前的帖子收到的答案进行了相应更改，但我仍然收到这些错误，这与我上次询问的错误不同。

我正在编写代码来排序用户使用合并排序输入的间隔。它从用户输入，它存储在结构interval中，用于按升序排序。它使用正值和负值来检查间隔侧是打开还是关闭，但这与正在执行的代码无关。下面是我的代码和我得到的输出类型的示例。非常感谢大家提前寻求帮助。

#include <stdio.h>

typedef struct interval{
    int num_left, denum_left;
    int num_right, denum_right;
    int left_state, right_state;
}interval;

interval combine(interval x, interval y);
int combineCheck(interval x, interval y);
int valueCheck(interval x, interval y);
void mergeSort(interval x[], int l, int r);
void merge(interval x[], int l, int m, int r);
interval takeInput();

int main(){
    int response, i;
    char d;
    printf("Enter the number of intervals to input: ");
    scanf("%d", &response);
    interval data[response];

    for(i = 0; i < response; i++){
        data[i] = takeInput();
    }

    mergeSort(data, 0, response-1);
    printf("%d %d %d\n", data[0].num_left, data[1].num_left, data[2].num_left);
    for(i = 0; i < response; i++){
        printf("%d/%d   %d/%d\n", data[i].num_left, data[i].denum_left, data[i].num_right, data[i].denum_right);
    }

}

interval takeInput(){
    interval temp;
    printf("Enter left numerator: ");
    scanf("%d", &temp.num_left);
    printf("Enter left denominator: ");
    scanf("%d", &temp.denum_left);
    printf("Enter right numerator: ");
    scanf("%d", &temp.num_right);
    printf("Enter right denominator: ");
    scanf("%d", &temp.denum_right);
    printf("\n");

    if(temp.num_left < 0){
        temp.num_left = temp.num_left*-1;
        temp.left_state = -1;}
    else{
        temp.left_state = 0;}

    if(temp.num_right < 0){
        temp.num_right = temp.num_right*-1;
        temp.right_state = -1;}
    else{
        temp.right_state = 0;}
    return temp;
}

int combineCheck(interval x, interval y){
    int left, right;
    left = x.num_right * y.denum_left;          //used to find relationship between 2 fractions
    right = y.num_left * x.denum_right;

    if(left == right && (x.right_state + x.left_state) == 0){
        return 1;
    }
    else if(left > right){
        return 1;
    }
    return 0;
}

interval combine(interval x, interval y){
    int left, right;                        //used to check if one interval is all encompassing
    left = x.num_right * y.denum_right;
    right = x.denum_right * y.num_right;

    interval temp;
    temp.num_left = x.num_left;
    temp.denum_left = x.denum_left;
    temp.left_state; 
    if(left > right){
        temp.num_right = x.num_right;
        temp.denum_right = x.denum_right;
        temp.right_state = x.right_state;
        return temp;
    }
    temp.num_right = y.num_right;
    temp.denum_right = y.denum_right;
    temp.right_state = y.right_state;
    return temp;
}

int valueCheck(interval x, interval y){
    int first, second;                  //used to check values
    first = x.num_left * y.denum_left;
    second = y.num_left * x.denum_left;
    if(first > second){
        return 1;
    }
    return -1;
}

void mergeSort(interval x[], int l, int r){
    if(l < r){
        int m = l + (r-l)/2;

        mergeSort(x, l, m);
        mergeSort(x, m+1, r);
        merge(x, l, m, r);
    }
}

void merge(interval arr[], int l, int m, int r){
    int i, j, k;
    int n1 = m-l +1;
    int n2 = r-m;

    interval L[n1], R[n2];

    for(i = 0; i < n1; i++)
        L[i] = arr[l + i];
    for(j = 0; j < n2; j++)
        R[j] = arr[m + 1 + j];
    j = 0;
    i = 0;
    k = l;
    while(i < n1 && j < n2){
        if(valueCheck(L[i], R[j]) == -1){
            arr[k] = L[i];
            i++;
        }
        else{
            arr[k] = R[j];
            j++;
        }
    }

    while(i < n1){
        arr[k] = L[i];
        i++;
        k++;
    }

    while(j < n2){
        arr[k] = R[j];
        j++;
        k++;
    }
}

输出为文字：

Enter the number of intervals to input: 5
Enter Left numerator: 1
Enter left denominator: 1
Enter right numerator: 1
Enter right denominator: 1

Enter Left numerator: 2
Enter left denominator: 1
Enter right numerator: 2
Enter right denominator: 1

Enter Left numerator: 3
Enter left denominator: 1
Enter right numerator: 3
Enter right denominator: 1

Enter Left numerator: 4
Enter left denominator: 1
Enter right numerator: 4
Enter right denominator: 1


Enter Left numerator: 5
Enter left denominator: 1
Enter right numerator: 5
Enter right denominator: 1

5 5 3
5/1 5/1
5/1 5/1
3/1 3/1
5/1 5/1
5/1 5/1

Answer 1

在条件

之后错过了K的迭代