为什么将4096个元素用于char数组缓冲区？ [英] Why use 4096 elements for a char array buffer?

问题描述

我发现一个程序接受标准输入

I found a program that takes in standard input

int main(int argc, char **argv) {
    if (argc != 2) {
        fprintf(stderr, "Usage: %s <PATTERN>\n", argv[0]);
        return 2;
    }

    /* we're not going to worry about long lines */
    char buf[4096]; // 4kibi

    while (!feof(stdin) && !ferror(stdin)) { // when given a file through input redirection, file becomes stdin
        if (!fgets(buf, sizeof(buf), stdin)) { // puts reads sizeof(buf) characters from stdin and puts it into buf; fgets() stops reading when the newline is read
            break;
        }
        if (rgrep_matches(buf, argv[1])) {
            fputs(buf, stdout); // writes the string into stdout
            fflush(stdout);
        }
    }

    if (ferror(stdin)) {
        perror(argv[0]); // interprets error
        return 1;
    }

    return 0;
}

为什么buf设置为4096个元素？是不是因为每行上的最大字符数只能是4096？

Why is the buf set to 4096 elements? Is it because the maximum number of characters on each line can only be 4096?

推荐答案

答案是在您粘贴的代码中：

The answer is in the code you pasted:

/* we're not going to worry about long lines */
char buf[4096]; // 4kibi

行数可能超过4096个字符，但作者并不认为它们值得关注关于。

Lines longer than 4096 characters can occur, but the author didn't deem them worth caring about.

还要注意 fgets ：

fgets（）从流中读取最多小于大小的字符，并将其存储到s指向的缓冲区中。在EOF或换行符之后停止读取。如果读取换行符，则将其存储到缓冲区中。终止的空字节（ \0 ）存储在缓冲区的最后一个字符之后。

fgets() reads in at most one less than size characters from stream and stores them into the buffer pointed to by s. Reading stops after an EOF or a newline. If a newline is read, it is stored into the buffer. A terminating null byte (\0) is stored after the last character in the buffer.

因此，如果一行的长度超过4095个字符（因为第4096个字节保留给空字节），它将是拆分了 while 循环的多次迭代。

So if there is a line longer than 4095 characters (since the 4096'th is reserved for the null byte), it will be split across multiple iterations of the while loop.

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