C复习笔记(3)-6.19

6月19日.

(1)

Write a program to count blanks, tabs, and newlines.

#include <stdio.h>

int main(void)

{

  int blanks, tabs, newlines;

  int c;

  int done = 0;

  int lastchar = 0;

  blanks = 0;

  tabs = 0;

  newlines = 0;

  while(done == 0)

  {

    c = getchar();

    if(c == ' ')

      ++blanks;

    if(c == '/t')

      ++tabs;

    if(c == '/n')

      ++newlines;

    if(c == EOF)

    {

      if(lastchar != '/n')

      {

        ++newlines; /* this is a bit of a semantic stretch, but it copes

                     * with implementations where a text file might not

                     * end with a newline. 

                     */

      }

      done = 1;

    }

    lastchar = c;

  }

  printf("Blanks: %d/nTabs: %d/nLines: %d/n", blanks, tabs, newlines);

  return 0;

} 

Note: if all the inputs are from keyboard, then every line ended with a newline.

(2)

Write a program to copy its input to its output, replacing each string of one or more blanks by a single blank.

Besides the answer in the C Answer Book, I uses the Continue Statement.

#include <stdio.h>

 

int main(void)

{

    int c;

    int lastchar;

 

    while ((c = getchar())!= EOF)

    {

       if (c == ' ')

       {

           if (lastchar == ' ')

              continue;

       }

       putchar(c);

       lastchar = c;

    }

}

(3)

Write a program to copy its input to its output, replacing each tab by /t , each backspace by /b , and each backslash by // . This makes tabs and backspaces visible in an unambiguous way.

#include <stdio.h>

 

int main(void)

{

    int c;

   

    while ((c = getchar())!= EOF)

    {

       if (c == '/b')

       {

           printf("//b");

           continue;

       }

       if (c == '/t')

       {

           printf("//t");

           continue;

       }

       if (c == '//')

       {

           printf("////");

           continue;

       }

       putchar(c);

    }

 

}

Why I copy the code here? I don’t know how to input a Backspace from keyboard.

 

#include <stdio.h>

#define IN 1 /*inside a word*/

#define OUT 0 /*outside a word*/

 

/*word counting*/

int main()

{

    int c;

    int state = OUT;

    int count = 0;

 

    while ((c=getchar()) != EOF) {

    if ((c != ' ') && (c != '/t') && (c != '/n'))

    {

        if(state == OUT)

            ++count;

        state = IN;

        }

        else

        {

        state = OUT;

        }

    }

    printf("%d",count);

    return 0;

}

You'll also find that it's easier to make extensive changes in programs where magic numbers appear only as symbolic constants.

 

Expressions connected by && or || are evaluated left to right, and it is guaranteed that evaluation will stop as soon as the truth or falsehood is known.

 

static char *str = "a b c d e f g h i j k l m "

                       "a b c d e f g h i j k l m "

                      "n/n";

The string can be defined like this, have you seen this before, aha!