How to get newline character from scanf even if it's the only input

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一生所求
一生所求 2020-12-11 04:52

I\'m doing homework that asks me to read an integer n representing the size of a loop and then read a line of characters n times and print it right

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  • 2020-12-11 05:26

    Using fgets to read in a line is simpler and more robust:

    if (!fgets(buff, 28, stdin))
    {
        // reading failed, do appropriate error handling
        // we're just exiting here
        exit(EXIT_FAILURE);
    }
    // We have successfully read in a line, or at least the first 27
    // characters of the line. Check whether a full line was read,
    // if it was, whether the line was empty
    size_t l = strlen(buff);    // <string.h> must be included
    if (buff[l-1] == '\n')
    {
        // a full line was read, remove trailing newline unless
        // the line was empty
        if (l > 1)
        {
            buff[l-1] = 0;
        }
    }
    else
    {
        // the input was too long, what now?
        // leave the remaining input for the next iteration or
        // empty the input buffer?
    }
    printf("%s\n",buff);
    

    It doesn't work with scanf("%s",buff) because most scanf conversions ignore leading white space:

    Input white-space characters (as specified by the isspace function) are skipped, unless the specification includes a [, c, or n specifier.

    So if the user inputs an empty line, scanf ignores that input unless its format is one of the exceptional.

    You can use scanf with a character set format instead,

    scanf("%27[^\n]%*c", buff);
    

    to read all characters until a newline (but limited to 28 - 1 here to avoid buffer overruns), and then consume a newline without storing it (the * in the %*c conversion specifier suppresses assignment), that would handle non-empty lines consisting entirely of whitespace, which the %s conversion would not. But if the first character of the input is a newline, the %27[^\n] conversion fails (thanks to chux for drawing attention to that), the newline is left in the input buffer, and subsequent scans with that format would also fail if the newline isn't removed from the input buffer.

    A somewhat robust (but ugly; and not dealing with too long input) loop using scanf would, as far as I can see, need to check for a newline before scanning, e.g.

    for(int ct = 0; ct < i; ++ct)
    {
        int ch = getchar();
        if (ch == EOF)
        {
            // something bad happened; we quit
            exit(EXIT_FAILURE);
        }
        if (ch == '\n')
        {
            // we had an empty line
            printf("\n\n");
        }
        else
        {
            // The first character was not a newline, scanning
            // with the character set format would have succeeded.
            // But we don't know what comes next, so we put the
            // character back first.
            // Although one character of pushback is guaranteed,
            if (ungetc(ch,stdin) == EOF)
            {
                // pushback failed
                exit(EXIT_FAILURE);
            }
            scanf("%27[^\n]%*c",buff);
            printf("%s\n",buff);
        }
    }
    

    Use fgets, really. It's better.

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  • 2020-12-11 05:26

    Yes, scanf() can do it if you use the right format-specifiers:
    --To accept a string of no more than 27 chars and
    --To accept all chars for that string (including space) except newline '\n', and
    --To see if scanf() made any conversions, write:

    scanRet = scanf("%27[^\n]s", buf);  // fills buf[28] with a string. 
    if(scanRet < 1) printf("Sorry, scanf didn't change buf[] array.\n");
    

    CAUTION! Repeated scanf() calls or conversions often get spoiled by unexpected inputs from users and/or leftover 'junk' in stdin. You can clear stdin between each scanf() call like this:

    char junk; 
    do {
      junk = getc(stdin);
      printf("junk: %c (int %d)\n", junk, (int)junk); // show me the junk we cleared
    }
    while(junk != '\n' || junk != EOF);
    

    Here's my best 'clean' solution using only scanf():

    char buf[28] = "";      // sets buf[0] = '\0';
    char junk;
    int i, iMax, scanRet;
    
      printf("how many lines? (integer>0):\n");
      scanRet = scanf(" %d", &iMax);
      if(scanRet < 1) {
        printf("Sorry, I couldn't read your integer. Bye.\n");
        return 1;    // error exit.
      }
      printf("please enter %d lines of text (<=27 chars/line):\n",iMax);
      for(i=0; i<iMax; i++) {
        // but first, clear any leftover junk in stdin;
        do {
          junk = getc(stdin);
          printf("junk: %c (int %d)\n", junk, (int)junk); // VERBOSE.
        }
        while (junk != '\n' && junk != EOF);
        // THEN try to read user's latest text string:
        scanRet = scanf("%27[^\n]",buf); // [^\n]== accept ALL chars except newline
        if(scanRet < 1) {  // format conversion failed. Nothing new in buf.
          printf("(empty line)\n");
        }
        else {
          printf("%s\n", buf); // non-empty user string 
        }
      }
      return 0;  // normal exit.
    }
    
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  • 2020-12-11 05:27

    I have two solutions for this:

    • Use fgets instead of scanf.
    • Append a \n at the end of the string, because you know the use will end the input with \n.

    First solution:

    ...
    char buf[28];
    fgets(buf, 28, stdin);
    ...
    

    Second solution:

    #include <string.h>
    
    ...
    char buf[28];
    scanf("%s", buf);
    strcat(buf, "\n"); // add the newline to the string
    ...
    
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  • 2020-12-11 05:28

    One option is to read one line at a time using gets and then parse each line using sscanf.

    EDIT based on comments: Using fgets is more appropriate since you can avoid buffer overrun.

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