I want to know why when we're reading a file do we have set the buf equal to '\0' before we can close fd? I'll demonstrate in an example.

ssize_t read_textfile(const char *filename, size_t letters)
{
    int fd;
    int i, y;
    char *buf;
    if (!filename)
        return (0);
    fd = open(filename, O_RDONLY);
    if (fd < 0)
        return (0);
    buf = malloc(sizeof(char) * letters);
    if (!buf)
        return (0);
    i = read(fd, buf, letters);
    if (i < 0)
    {
        free(buf);
        return (0);
    }
    buf[i] = '\0';        // This is what I'm talking about
    close(fd);
    y = write(STDOUT_FILENO, buf, i);
    if (y < 0)
    {
        free(buf);
        return (0);
    }
    free(buf);
    return (y);
}

CodePudding user response：

I don't know why you think it's necessary, but assuming the read() reads all letters (i == letters), buf[i] = '\0' overruns your buffer. And your input file is left open if the malloc() fails.

CodePudding user response：

You do not have to and you do not need it in this code.,

This line makes a C string (adding null terminating character). As you use write you do not need it, but if you use any function taking string as a parameter you will need it.

CodePudding user response：

Given that the only use of buf after the "termination" is the write call and that writes buf[0] to buf[i-1], then the character at buf[i] is ignored. As such the "termination" serves no purpose at all.

Not only does it serve no purpose, it is in fact also a bug. If the file contains at least letters bytes, then i == letters and buf[i] will be out of bounds.