c - 在C中使用printf格式化二进制搜索的输出？

这个程序的输出有点问题。我需要在一行上打印动词，如果没有动词，我需要单独打印一个语句。例如。
"talk and walk"应该打印"The verbs are: talk walk"
而"hello there"应该打印"There are no verbs"

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int binary_search(char *list_of_words[], int size, char *target){
    int bottom= 0;
    int mid;
    int top = size - 1;
    int found = 0;

    while(bottom <= top && !found){
        mid = (bottom + top)/2;
        if (strcmp(list_of_words[mid], target) == 0){
            //printf("%s found at location %d.\n", target, mid+1);
            found = 1;
        } else if (strcmp(list_of_words[mid], target) > 0){
            top    = mid - 1;
        } else if (strcmp(list_of_words[mid], target) < 0){
            bottom = mid + 1;
        }
    }
    if (found == 1)
        return mid;
    else
        return -1;
}

int main(int argc, char* argv[]){
    char *input = strtok(argv[1], " \"\n");
    char *verbs[5] = { "do", "make", "take", "talk", "walk" };
    int position;
    int check = 0;
    while (input != NULL) {
        //printf("%s\n", input);
        position = binary_search(verbs, 5, input);
        if (position != -1)
            printf("The verbs are: %s\n", verbs[position]);
            check = 1;
        input = strtok(NULL, " ");
    }
    if (check == 0){
        printf("There are no verbs\n");
    }
    return 0;
}

有什么想法吗？

最佳答案

它看起来运行良好，但您需要在周围添加括号

    if (position != -1)
        printf("The verbs are: %s\n", verbs[position]);
        check = 1;

像在

    if (position != -1) {
        printf("The verbs are: %s\n", verbs[position]);
        check = 1;
    }

否则，check总是在循环中设置为1。
如果您不想重复“动词是：”，请添加一个检查

    if (position != -1) {
        if (first) {
            printf("The verbs are:");
            first = 0;
            check = 1;
        }
        printf(" %s", verbs[position]);

    }

关于c - 在C中使用printf格式化二进制搜索的输出？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/16789585/