C I/O

Program input

There are 3 main sources of input for programs:

1. from the command line

   • you get access to data on the command line by using argc and argv[][]

2. from standard input (also called stdin)

   • stdin can be the keyboard, a data file, or the output of another program

3. from an 'internally-defined' file

   • open a file, use fscanf(), and don't forget to close the file

1. command line

To read from the command line:

• include argc and argv in your parameter list for main().

• use sscanf() to read the arguments (it stands for 'string scanf()')

  • the first argument of a sscanf() is a string

Here is a program that counts from 1 to num, where num is provided by the user on the command line

   1 // countc.c
   2 // reads an integer from the command line and counts
   3 #include <stdio.h>
   4 #include <stdlib.h>
   5 
   6 int main(int argc, char *argv[]) {
   7    int num = 0;
   8    if (argc < 2 || sscanf(argv[1], "%d", &num) != 1) { // num is defined here
   9       fprintf(stderr,"Usage: %s integer\n", argv[0]);
  10       return EXIT_FAILURE;
  11    }
  12    for (int i=1; i<=num; i++) {
  13       printf("%d ", i);
  14    }
  15    printf("\n");
  16    return EXIT_SUCCESS;
  17 }

Notice the program prints a 'Usage' message if an integer argument is missing (discussed in next session)

To execute the program:

prompt$ dcc -o countc countc.c

prompt$ ./count
Usage: ./countc integer

prompt$ ./countc !t#q
Usage: ./countc integer

prompt$ ./countc 10
1 2 3 4 5 6 7 8 9 10

2. standard input

To read from standard input (usually called simply stdin)

• a scanf() is used (instead of sscanf())

  • a scanf() misses the string argument of a sscanf()

• so where does num comes from?

  • ... the default 'channel' stdin

   1 //counts.c
   2 // reads an integer from stdin and counts
   3 #include <stdio.h>
   4 #include <stdlib.h>
   5 
   6 int main(void) {
   7    int num;
   8    if (scanf("%d", &num) != 1) {
   9       fprintf(stderr, "Usage: a number expected\n");
  10       return EXIT_FAILURE;
  11    }
  12    // the rest of the program is exactly the same as the command-line version
  13    for (int i=1; i<=num; i++) {
  14       printf("%d ",i);
  15    }
  16    printf("\n");
  17    return EXIT_SUCCESS;
  18 }

Notice the Usage message this time is simpler than the command-line version above

• ... because we did not declare argc and argv, and so cannot use argv[0] this time!!

  • we could have if we wanted to of course

There are many ways to 'test' a program that reads stdin.

1. Using the keyboard

    prompt$ dcc -o counts counts.c
    prompt$ ./counts
    10
    1 2 3 4 5 6 7 8 9 10

   where the integer 10 was typed on the keyboard by the user, and the program generates the count from 1 to 10.

2. Using a data file, input.txt say, which contains the integer 10 (followed by a newline).

    prompt$ more input.txt
    10
   
    prompt$ ./counts < input.txt
    1 2 3 4 5 6 7 8 9 10

3. Using a pipe command. A pipe command joins the stdout of a program to the stdin of another program. If we have a program called write10.c:

      1  // write10.c
      2  // just print the string 10
      3  #include <stdio.h>
      4  #include <stdlib.h>
      5 
      6  int main(void) {
      7     printf("10\n");
      8     return EXIT_SUCCESS;
      9  }
   

   then we can pipe its stdout to the stdin of our counting program

    prompt$ dcc -o write10 write10.c
    prompt$ dcc -o counts counts.c
    prompt$ ./write10 | ./counts
    1 2 3 4 5 6 7 8 9 10

   But you can actually generate a string much more easily in UNIX using echo

    prompt$ echo "10" | ./counts
    1 2 3 4 5 6 7 8 9 10

User prompting

You can still use a 'user prompt' when you use stdin but it messes up the output.

   1 // counts+.c
   2 // reads an integer from stdin and counts
   3 // prompts the user
   4 #include <stdio.h>
   5 #include <stdlib.h>
   6 
   7 int main(void) {
   8    int num;
   9    printf("Please input a number: "); // this line added to counts.c
  10    if (scanf("%d", &num) != 1) {
  11       fprintf(stderr, "Usage: a number expected\n");
  12       return EXIT_FAILURE;
  13    }
  14    for (int i=1; i<=num; i++) {
  15       printf("%d ",i);
  16    }
  17    printf("\n");
  18    return EXIT_SUCCESS;
  19 }

results in

prompt$ ./counts+
Please input a number: 10
1 2 3 4 5 6 7 8 9 10

where the program prints the user prompt, the user types in 10, and the program then outputs the count to 10.

• that looks fine

If you instead use a pipe as input, then you do not see what the input is

prompt$ echo "10" | ./counts+
Please input a number: 1 2 3 4 5 6 7 8 9 10

You see here that the 10 generated by the echo does not appear on the screen: you just see the output of the program

• which is sort-of messed up

3. A file

A program can open and close, and read from, and write to, a file that is 'internally' defined.

This is generally done when you have large volumes of stored data, or complex data (such as structs) or non-printable data.

• these don't happen often

   1 // files.c
   2 // read a number 'num' from a file input.txt
   3 // write a count from 1 to 'num' to the file OUT
   4 
   5 #define IN  "input.txt"
   6 #define OUT "output.txt"
   7 
   8 #include <stdio.h>
   9 #include <stdlib.h>
  10 
  11 #define NUMDIG 6
  12 
  13 int main(void) {
  14    FILE *fpi, *fpo; // these are file pointers
  15    char s[NUMDIG];
  16 
  17    fpi = fopen(IN, "r");
  18    if (fpi == NULL) { // an important check
  19        fprintf(stderr, "Can't open %s\n", IN);
  20        return EXIT_FAILURE;
  21    }
  22    else {
  23        int num;
  24        if (fscanf(fpi, "%d", &num) != 1) {
  25            fprintf(stderr, "No number found in %s\n", IN);
  26            return EXIT_FAILURE;
  27        }
  28        else {
  29            fpo = fopen(OUT, "w");
  30            if (fpo == NULL) { // an important check
  31                fprintf(stderr, "Can't create %s!\n", OUT);
  32                return EXIT_FAILURE;
  33            }
  34            else {
  35                fprintf(fpo, "%s", "Counts\n");
  36                for (int i=1; i<=num; i++) {
  37                    sprintf(s, "%d", i);
  38                    fprintf(fpo, "%s\n", s);
  39                }
  40                fclose(fpo);
  41                printf("file %s created\n", OUT);
  42                return EXIT_SUCCESS;
  43            }
  44        }
  45    }
  46 }

Program output

There are two standard output 'streams', standard output stdout and standard error stderr.

• both are normally defined as the screen

The general form for a print statement is fprintf(stream, ...), where stream can be stdout, stderr or a user-defined file.

• the call printf(...) is the same as fprintf(stdout, ...) and is default to the screen but may be re-directed

• a fprintf(stderr, ...) is usually reserved for serious errors

  • for example, in countc.c above the 'Usage' message went to stderr:

       1   fprintf(stderr,"Usage: %s integer\n", argv[0]);
    

  • you could argue over whether 'bad' usage is a serious error (maybe just a printf would have been enough)

There is 'systematic' naming here:

• standard input is scanf(),

  • if you read from a string then use sscanf(), where the first argument is the string

• standard output is printf(),

  • if you write to a file then use fprintf(), where the first argument is a stream

Like stdin, we can re-direct stdout to a file. For example:

dcc -o counts counts.c
./counts > output.txt
10

(where the integer 10 is input by the user) will result in the count from 1 to 10 going to the file output.txt

If you create a data file input.txt that contains the string 10, then the following will generate the same output text file.

./counts < input.txt > output.txt

As we saw before, you can let echo generate data and use that in a pipe. This also generates the same output text file.

echo "10" | ./counts > output.txt

Input/output design considerations