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The system will copy characters from tin into buffer until a white space character is encountered. The user must ensure that buffer is defined to be a character string long enough to hold the input. @The functions declared in the cstring.h> header file may be used to manipulate strings. These include the string length function s trlen ( > , the string copying functions s trcpy ( > and s trncpy ( > , the string concatenating functions s trca t ( ) and strncat ( > , the string comparing functions strcmp ( > and strncmp ( > , and the token extracting function s trtok ( ) . These functions are described in Section 7.8.

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6.4 Proving a circle problem stated in words Prove that if a radius bisects a chord, then it is perpendicular to the chord.

I is appended to the string:

main0 { char s[] = ABCD ; for (int i = 0; i c 5; i++) tout CC "s[" cc i CC "1 =

To Prove: OC ' AB Plan: Prove ^ AOD > ^ BOD to show j 1 > j 2. Also, j 1 and j 2 are supplementary.

"1 CC s[i] CC "'\n";

&Geography ....&Continents ........&Africa ........An&tarctica ........As&ia ........A&ustralia ........&Europe ........&North America ........&South America ....&Oceans ........&Arctic ........A&tlantic ........&Indian ........&Pacific ....&Seas ........&Baltic ........B&ering ........B&lack ........&Caribbean ........&Mediterranean ........&Persian Gulf ........&Red ........&South China

PROOF:

When the NUL character is sent to cou t, nothing is printed- not even a blank. This is seen by printing one apostrophe immediately before the character and another apostrophe immediately after the character.

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Statements 1. Draw OA and OB. 2. 3. 4. 5. 6. 7. 8. OA > OB OC bisects AB. AD > DB OD > OD ^ AOD > ^ BOD j1 > j2 j 1 is the supplement of j 2.

7.4 STRING I/O Input and output of strings is done in several ways in C++ programs. The best way is by means of string class operators as described in 10. More straightforward methods are described here.

Reasons 1. A straight line segment may be drawn between any two points. 2. Radii of a circle are congruent. 3. Given 4. To bisect is to divide into two > parts. 5. Reflexive property 6. SSS s 7. Corresponding parts of congruent ^ are >. 8. Adjacent are supplementary if exterior sides lie in a straight line. 9. Congruent supplementary angles are right angles. 10. Rt. are formed by perpendiculars.

main0 char word[80]; do 1 tin >> word; if (*word) tout CC "\t\"" } while (*word);

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The length of a tangent from a point to a circle is the length of the segment of the tangent from the given point to the point of tangency. Thus, PA is the length of the tangent from P to circle O in Fig. 6-16.

buffer:

CC word cc "\"\n";

Fig. 6-16

In this Iun, the whi le loop iterated 10 times: once for each word entered (including the Control-D that stopped the loop). Each word in the input stream c in is echoed to the output stream cou t. Note that the output stream is not flushed until the input stream encounters the end of the line. Each string is printed with a double quotation mark on each side. This character must be designated by the character pair \ I inside a string literal. The expression *word controls the loop. It is the initial character in the string. It will be nonzero (i.e., true ) as long as the string word contains a string of length greater than 0. The string of length 0, called the empty string, contains the NUL character I\ 0 I in its first element. Pressing Control-D on a UNIX or Macintosh computer (Control-Z on a DOS PC or in VAX/VMS) sends the end-of-file character in from tin. This loads the empty string into word, setting *word (which is the same as word [ 0 ] ) to and stopping the loop. The last line of output shows only the Control-D echo. Note that punctuation marks (apostrophes, commas, periods, etc.) are included in the strings, but whitespace characters (blanks, tabs, newlines, etc.) are not. The do loop in Example 7.3 could be replaced with:

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