C 程序数组超过 1 个字

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这个问题来自 HackerRank，我尝试使用 %[^\n]s 来表示长单词。但是，输出继续产生 .0

如何将 %[^\n]s 替换为其他内容以使字符串接收输入？

这是输入:

12
4.0
is the best place to learn and practice coding!

这是我的输出:

16
8.0
HackerRank  .0

这是预期的输出:

16
8.0
HackerRank is the best place to learn and practice coding!

这是我的完整代码，如您所见，它无法识别 %[^\n]s。如何解决这个问题呢？谢谢。

完整代码:

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

int main() {
    int i = 4;
    double d = 4.0;
    char s[] = "HackerRank ";

    // Declare second niteger, double, and String variables.
    int value1, sum1, value2;
    double e = 2.0, sum2;
    char t[30];

    // Read and save an integer, double, and String to your variables.
    scanf(" %d", &value1);
    scanf("%d", &value2);
    scanf("%[^\n]s", t); //** POINT OF INTEREST **

    // Print the sum of both integer variables on a new line.
    sum1 = value1 + i;
    printf("%d\n", sum1);

    // Print the sum of the double variables on a new line.
    sum2 = d * e;
    printf("%.1lf\n", sum2);

    // Concatenate and print the String variables on a new line
    // The 's' variable above should be printed first.
    printf("%s %s", s, t);

    return 0;
}

Answer 1

考虑到您的输入输出示例，我修改了您的代码，如下所示:

char t[256]; // the string "is the best place to learn and practice coding!" MUST FIT!!!
...
scanf("%d", &value1);
scanf("%lf", &d); // NOT %d, %lf !!! &d or &e - I don't know - depends on you
scanf("\n%[^\n]", &t);
...
printf("%s%s", s, t); // you don't need a space, since your "s" already contains it.

对我来说效果很好。

UPD:现在它实际上工作得很好。