我在这里检查了几乎所有类似的帖子，但是我不知道该怎么做。我正在尝试在C程序中输入数字4，该程序在数组中返回以下数字：

1
2
3
4
12
13
14
23
24
34
123
134
124
1234

更清楚地说：如果输入数字为4，那么我想使用数字1-4，并生成所有可能的数字组合（从1位数字组合到4位数字组合）而无数字重复。

我尝试了以下代码：

#include <stdio.h>

/* Prints out a combination like {1, 2} */
void printc(int comb[], int k) {
    printf("{");
    int i;
    for (i = 0; i < k; ++i)
        printf("%d, ", comb[i] + 1);
    printf("\\b\\b}\\n");
}


    int next_comb(int comb[], int k, int n) {
    int i = k - 1;
    ++comb[i];
    while ((i >= 0) && (comb[i] >= n - k + 1 + i)) {
        --i;
        ++comb[i];
    }

    if (comb[0] > n - k) /* Combination (n-k, n-k+1, ..., n) reached */
        return 0; /* No more combinations can be generated */

    /* comb now looks like (..., x, n, n, n, ..., n).
    Turn it into (..., x, x + 1, x + 2, ...) */
    for (i = i + 1; i < k; ++i)
        comb[i] = comb[i - 1] + 1;

    return 1;
}

int main(int argc, char *argv[]) {
    int n = 5; /* The size of the set; for {1, 2, 3, 4} it's 4 */
    int k = 3; /* The size of the subsets; for {1, 2}, {1, 3}, ... it's 2 */
    int comb[16]; /* comb[i] is the index of the i-th element in the
            combination */

    /* Setup comb for the initial combination */
    int i;
    for (i = 0; i < k; ++i)
        comb[i] = i;

    /* Print the first combination */
    printc(comb, k);

    /* Generate and print all the other combinations */
    while (next_comb(comb, k, n))
        printc(comb, k);

    return 0;
}

上面的程序打印结果。我想以某种方式获得结果。但是我不能，因为上面的代码以奇怪的方式打印结果。