C语言实现bitmap,取两个数组的交集等操作

bitmap的基本操作：

#include
#include
#define WORD 32
#define SHIFT 5 ////移动5个位,左移则相当于乘以32,右移相当于除以32取整
#define MASK 0x1F //16进制下的31
#define N 10000000
/*
 * 置位函数——用"|"操作符,i&MASK相当于mod操作
 * m mod n 运算，当n = 2的X次幂的时候,m mod n = m&(n-1)
 */
void set(int *bitmap, int i) {
    bitmap[i >> SHIFT] |= (1 << (i & MASK));
}
/* 清除位操作，用&~操作符 */
void clear(int *bitmap, int i) {
    bitmap[i >> SHIFT] &= ~(1 << (i & MASK));
}
/* 测试位操作用&操作符 */
int test(int *bitmap, int i) {
    return bitmap[i >> SHIFT] & (1 << (i & MASK));
}

排序：

void sort() {
    int bitmap[1 + N / WORD];
    FILE *in = fopen("in.txt", "r");
    FILE *out = fopen("out.txt", "w");
    if (in == NULL || out == NULL) {
        exit(-1);
    }
    int i = 0;
    int m;
    for (i = 0; i < N; i++) {
        clear(bitmap, i);
    }
    while (!feof(in)) {
        fscanf(in, "%d", &m);
        printf("%d/n", m);
        set(bitmap, m);
    }
    printf("abnother");
    for (i = 0; i < N; i++) {
        if (test(bitmap, i)) {
            printf("%d/n", i);
            fprintf(out, "%d/n", i);
        }
    }
    fclose(in);
    fclose(out);
}

求交集：

//合并两个数组，去交集
void merge() {
    int A[] = { 1, 3, 5, 7, 9, 435, 3454, 345543, 3453455 };
    int B[] = { 4, 5, 6, 8, 9, 435, 3454, 345543 };
    int bitmapA[1 + N / WORD];
    int bitmapB[1 + N / WORD];
    int i = 0;
    int j = 0;
    for (i = 0; i < N; i++) {
        clear(bitmapA, i);
        clear(bitmapB, i);
    }
    for (i = sizeof(A) / sizeof(*A) - 1; i >= 0; i--) {
        set(bitmapA, *(A + i));
    }
    for (j = sizeof(B) / sizeof(*B) - 1; j >= 0; j--) {
        set(bitmapB, *(B + j));
    }
    for (i = 0; i < N; i++) {
        if (test(bitmapA, i) & test(bitmapB, i)) {//交集
            printf("%d/n", i);
        }
    }
}
int main(void) {
    //sort();
    merge();
    return EXIT_SUCCESS;
}