Question

我正在编写一个有趣的程序（不是为了学校），并且我很难弄清楚为什么scanf函数在我的循环的每次迭代中都没有执行 - 我已经同时玩弄了两个＆＃39;循环和＆＃39;而＃39;循环。

我知道，根据我写scanf函数的方式（即scanf("%s", &variablename); VS scanf("%99[^\n]s", &variablename);）会有所不同，但我已经尝试了所有内容，而且我绝望了！ / p>

当我对来自printf的输入进行scanf检查时，在每次迭代时，每次迭代只会输入一个字符串，所以如果我在第一次输入中输入两个单词，则需要要处理的两次迭代 - 每个一个单词。以下是我所描述的代码段：

int main(void){
    int tries = 0;
    int score = 0;
    char question[100];
    char useranswer[100];
    const char *phrase = {"our favorite saying\0"};

    printf("\nQuestion #3 (10 points): What is our secret saying?\n");
        sleep(1);
        tries = 1;

    while (tries<=3){
        printf("YOUR ANSWER:");
        scanf("%s[^\n]", useranswer); 

        if(strncmp(useranswer, phrase, 15) != 0){
            printf ("Nope, try again!\n");
            printf("You have used %d out of 3 tries!\n", tries);
            if (tries == 2){
                printf("Here's your final hint:xxx...\n");
            }
            if (tries == 3){
            printf("You didn't get it. The answer is: our favorite saying!\n");
            }
            tries++;
        }   
        if (strncmp(useranswer, phrase, 15) == 0){
            printf("Damn, you're good.  Well done.\n");
            score += 10;
            break;
        }
    }

此代码的输出为：

Question #3 (10 points): What is our secret saying?
YOUR ANSWER:our favorite saying
Nope, try again!
You have used 1 out of 3 tries!
YOUR ANSWER:Nope, try again!
You have used 2 out of 3 tries!
Here's your final hint:xxx...
YOUR ANSWER:Nope, try again!
You have used 3 out of 3 tries!
You didn't get it. The answer is: our favorite saying!

（它只允许我输入一次，然后输入＆＃34;我们最喜欢的一句＆＃34;。）

Answer 1

在评论中，您可以找到scanf中的格式说明符不起作用的原因。

另一种方法是使用fgets代替，也许在辅助函数中，它可以处理读取用户输入时可能出现的一些极端情况：

#include <ctype.h>

char *read_line( char *buf, size_t n, FILE *pfin )
{
    ssize_t length = 0;
    int ch;

    if ( !buf || n == 0 )
        return NULL;
    /*  Consume trailing control characters, like '\0','\n', '\r', '\f'...
        but also '\t'. Note that ' ' is not skipped. */
    while ( (ch = fgetc(pfin)) != EOF  &&  iscntrl(ch) ) { }
    if ( ch == EOF )
        return NULL;
    /*  At least one char is printable  */
    *buf = ch;
    ++length;

    /*  Read from file till a newline or up to n-2 chars. The remaining chars
        are left in the stream buffer. Return NULL if no char is read.      */
    if ( fgets(buf + 1, n - 1, pfin) )
    {
        /*  Trim the string to the first control character                  */
        while ( !iscntrl(buf[length]) ) 
        {
            ++length;
        }
        buf[length] = '\0';       
    }
    return buf;
}

我也改变了以下逻辑。 OP多次使用strncmp(useranswer, phrase, 15)，但该幻数15低于phrase的大小，因此最终仅比较子字符串。

while ( tries <= 3 ) {
    printf("YOUR ANSWER:");
    if ( !read_line(useranswer, sizeof useranswer, stdin) ) {
        printf("Error: Unexpected end of input.\n");
        exit(EXIT_FAILURE);
    }
    if( strcmp(useranswer, phrase) == 0 ) {
        printf("Damn, you're good.  Well done.\n");
        score += 10;
        break;
    } else {
        printf ("Nope, try again!\n");
        printf("You have used %d out of 3 tries!\n", tries);
        if (tries == 2) {
            printf("Here's your final hint:xxx...\n");
        }
        if (tries == 3) {
            printf("You didn't get it. The answer is: our favorite saying!\n");
        }
        tries++;
    }
}

<小时/> 作为最后一点，我发现OP phrase的声明有点奇怪（可能是一个错字）：

const char *phrase = {"our favorite saying\0"};
//            string literals are already ^^ null terminated...

虽然我们可以使用简单的数组声明，例如：

const char phrase[] = "our favorite saying";

还要考虑在这两种不同情况下sizeof phrase返回的值。

<小时/> 感谢 @chux 提供了所有有价值的提示和提供的有趣链接：
https://stackoverflow.com/a/27729970/4944425
https://stackoverflow.com/a/28462221/4944425
并且 @Dmitri 在他的评论中指出，一旦我们确定两个字符串都以空值终止，我们就可以使用strcmp代替strncmp。

scanf不会在每次循环迭代时执行

1 个答案: