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tout << x <<
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<< y << " < M << z CC endl;
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a. tin CC count;
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CHAP. 21
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CONDITIONAL STATEMENTS AND INTEGER TYPES
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tout << "Enter n: 'I; tin x= n; if (n < 0) tout << "That is negative. Try again.\n"; tin >> n; else tout << "o.k. n = ' -CC n << endl;
Solved Problems
Construct a logical expression to represent each of the following conditions: a. score is greater than or equal to 80 but less than 90; 6. answer is either IN I or I n '; c. n is even but not 8; d. ch is a capital letter.
a. (score >= ZA (answer == h (n%2 80 SC& score < 90); N answer == 'n');
== 0 && n != 8); 'A' &SC ch <= 'z');
b. (ch >= 2.14 .
What is wrong with the following code:
if (x == 0) if (y == 0) tout << "x and y are both zero.\n"; else tout << "x is not zero.\n"; The programmer clearly intended for the second output I x i s no t z ero . \ n I to be printed if the first condition (x = = 0 ) is false, regardless of the second condition ( y = = 0 ) . That is, the el se was intended to be matched with the first i f . But the e 1 se matching rule causes it to be matched it with the second condition, which means that the output I x i s not z ero . \ n I will be printed only when x is zero and y is not zero. The e 1 se matching rule can be overridden with braces: if (x == 0) { if (y == 0) tout << "x and y are both zero.\n"; else tout << 'lx is not zero.\n"; the way the programmer had intended it to be.
Now the e 1 se will be matched with the first i f,
What is the difference between the following two statements:
if (n > 2) { if (n < 6) tout C-C "OK"; > else tout << "N&I; }
if (n > 2) { if (n < 6) tout C-K "OK"; else tout << "NG";
CONDITIONAL STATEMENTS AND INTEGER TYPES
[CHAP. 2
In the first statement, the el se is matched with the first i f . In the second statement, the e 1 se is matched with the second i f. If y1 < 2, the first statement will print NG while the second statement will do nothing. If 2 < y2 < 6, both statements will print OK. If y2 > 6, the first statement will do nothing while the second statement will print NG. N o t e that this code is difficult to read because it does not follow standard i ndentation conventions. The first statement should be written if (n>2) { if (n c 6) tout << "OK";
else tout CC "NG"; The braces are needed here to override the else matching rule. This else is intended to match the first i f. The second statement should be written if (n > 2) if (n c 6) tout << "OK"; else tout CC "NG"; Here the braces are not needed because the el se is intended to be matched with the second i f.
Solved Programming Problems
Write and run a program that reads the user s age and then prints You are a child. if the age < 18, You are an adult. if 18 I age < 65, and You are a senior citizen. if age 2 65.
Here weusedthe else if construct because the three outcomes depend upon age being in one of three disjoint intervals: main0 int age; tout << "Enter your age: "; tin >> age; if (age < 18) tout CC "You are a child.\n"; else if (age c 65) tout CC "You are an adult.\n"; else tout CC "you are a senior citizen.\&';
If control reaches the second condition ( age c 6 5 ) , then the first condition must be false so in fact 18 < age < 65. Similarly, if control reaches the second el se, then both conditions must be false so in factage 2 65. 2.17
Write and run a program that reads two integers and then uses the conditional expression oper-
ator to print either multiple or not according to whether one of the integers is a multiple of the other.
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