为什么代码在 c 中以如此奇怪的顺序执行？

Question

int is_valid(char *input)
{
    int i;

    for (i = 0;; i++)
    {
        // Check for null terminator
        if ( ! input[i])
        {
            printf("stop executing is_valid()");
            return 0;
        }

        // Will be executed for null terminator... o_O
        printf("%c isdigit: %d\n", input[i], isdigit(input[i]));
    }

    return 1;
}

int main()
{
    char input[80];
    fgets(input, sizeof(input), stdin);
    is_valid(input);
    return 0;
}

输出：

1 isdigit: 1
0 isdigit: 1
1 isdigit: 1

 isdigit: 0 // Why is that here? null terminator after return?!
stop executing is_valid()invalid input!

为什么在 return 之前通过 isdigit 处理空终止符？而且..好的，为什么在它之后执行条件？

Answer 1

Why null terminator is processed via isdigit before return?

不是。它对 fgets 读取的 '\n' 个字符执行。

C11-§7.21.7.2

The fgets function reads at most one less than the number of characters specified by n from the stream pointed to by stream into the array pointed to by s. No additional characters are read after a new-line character (which is retained) or after end-of-file. A null character is written immediately after the last character read into the array.

为输入

里面有 5 个字符

101\n[=11=]

\n 当然不是数字。

您也需要为 \n 添加条件

if (input[i] == '[=12=]' || input[i] == '\n')
{
    printf("stop executing is_valid()");
    return 0;
}

Answer 2

使用scanf("%d",&char[i]) 获取输入。由于 fgets 也接受换行符，这就是您获得的额外输出，fgets.

对 "\n" 的检查

为什么代码在 c 中以如此奇怪的顺序执行？

Why code is being executed in such strange sequence in c?

c

fgets