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the expression - flags &= -mask clears all the flags in the mask, leaving all other flags unchanged. The effect of this function is to clear all the flags in the format mask and to return their previous settings:

patron patron #

EXAMPLE 12.6 Using unset f ( ) to Clear a Stream s Format Flags main0 { char buffer[80]; cin.unsetf(ios::skipws); // clear default skipws flag tin >> buffer; tout << " [" cc buffer CC t']\n"; tin >> buffer; tout << " [" CC buffer X-C rl]\nII; tin >> buffer; tout -c-c " [" XC buffer XX "]\n"; int n = 234; cout.setf( ios::hex 1 ios::uppercase 1 ios: :showbase ); tout C-C n K-C endl; cout.unsetf( ios::basefield ); // clear all radix flags tout -C-C n -CC endl;

The skipws flag is set by default, which means that ordinarily white space is skipped before reading a value into a variable. But the call to tin. unsetf clears this flag so that white space will not be ignored. The first three characters input are two blanks and the H, so the first read into buf f er reads only a single blank. That is because the second blank signals an end to that input. The second read into buf f er reads the second blank and the entire six-character string He1 lo, ". Again, input is terminated by the first

blank encountered after the reading begins. The call to tout . unset f clears all the flags that are in the base f i eld group. This includes the hex flag that was set with the cou t . set f call. m a y n o t w o rof tout .unsetf o l d e r c o m p i l e r s ( e . g . , v e r s i o n 4 . 5 o f Warning: This use k o n s o m e Microsoft Turbo C++).

The set f ( > function that takes two arguments uses a format mask in its second argument:

Fig. 11-8

long setf(long f, long mask) ( long t = -flags; -flags = (-flags&-mask)1 (f&mask); return t; )

The mask specifies which group of flags is to be changed. The function clears all the flags in the group specified by mask and then sets the flag f if it is in that group.

On most UNIX workstations, the int type ranges from -2,147,483,648 How many bytes will an object of this type occupy in memory

to 2,147,483,647.

Use existence dependency to determine the association in the student section problem of Example 11.10. The object model developed in Example 11.10 does not satisfy the existence dependency rules, since section cannot be existence dependent on student or vice versa. Thus, an additional object called enrollment must be used. See Fig. 11-9.

The range from -2,147,483,648 to 2,147,483,647 covers 4,294,967,296 values. This number is exactly 232, so each int requires 32 bits which is 4 bytes of memory.

How do the following two statements differ:

char ch = 'A'; char ch = 65;

Both statements have the same effect: declare ch to be a char and initialize it with the value 65. Since this is the ASCII code for I A I , that character constant can also be used to initialize ch to 65.

Multiplicities are restrictions on the associations between instances of objects. The multiplicity is speci ed by an expression at the end of the association. The expression can be a single value, a range of values, or a list of ranges or single values. In the range, the two values are separated by two periods. The problem domain often has restrictions on how many relationships an instance of an object can have with instances of other objects.

char c = 100; tout << c;

Run a program like that in Example 1.14 and compare the constants CHAR -MAX, SCHAR-MAX, and UCHARJWX:

Solved Programming Problems 1.25 Write a program that prints the first sentence of the Gettysburg Address.

Essentially, all that we need to do here is use a sequence of output statements, sending pieces of the sentence to the cou t object: