2d barcode generator vb.net b. These two boolean expressions are equivalent: p T F !pp T F p T F p T F in Software

Generator EAN-13 in Software b. These two boolean expressions are equivalent: p T F !pp T F p T F p T F

b. These two boolean expressions are equivalent: p T F !pp T F p T F p T F
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c. These two boolean expressions are not equivalent: p T T F F q T F T F !p || q T F T T p T T F F q T F T F p || !q T T F T
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d. These two boolean expressions are equivalent: p T T T T F F F F q T T F F T T F F r T F T F T F T F p && (q&&r) T F F F F F F F p T T T T F F F F q T T F F T T F F r T F T F T F T F (p&&q) && r T F F F F F F F
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e. These two boolean expressions are not equivalent: p T T T T F F F F 3.8 q T T F F T T F F r T F T F T F T F p || (q&&r) T T T T T F F F p T T T T F F F F q T T F F T T F F r T F T F T F T F (p||q) && r T F T F T F F F
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The term short-circuiting is used to describe the way C++ evaluates compound logical expressions like (x > 2 || y > 5) and (x > 2 && y > 5). If x is greater than 2 in the first expression, then y will not be evaluated. If x is less than or equal to 2 in the second expression, then y will not be evaluated. In these cases only the first part of the compound expression is evaluated because that value alone determines the truth value of the compound expression. The programmer probably intended to test the condition (x == 0). But by using assignment operator = instead of the equality operator == the result will be radically different from what was intended. For example, if x has the value 22 prior to the if statement, then the if statement will change the value of x to 0. Moreover, the assignment expression (x = 0) will be evaluated to 0 which means false, so the else part of the selection statement will execute, reporting that x is not zero! The programmer probably intended to test the condition (x < y && y < z). The code as written will compile and run, but not as intended. For example, if the prior values of x, y, and z are 44, 66, and 22, respectively, then the algebraic condition x < y < z is false. But as written, the code will be evaluated from left to right, as (x < y) < z. First the condition x < y will be evaluated as true. But this has the numeric value 1, so the expression (x < y) is evaluated to 1. Then the combined expression (x < y) < z is evaluated as (1) < 66 which is also true. So the output statement will execute, erroneously reporting that 44 < 66 < 22. a. (score >= 80 && score < 90) b. (answer == 'N' || answer == 'n') c. (n%2 == 0 && n != 8) d. (ch >= 'A' && ch <= 'Z') a. (n > 0 && n < 7 && n != 3) b. (n > 0 && n < 7 && n%2 != 0) c. ((ch >= 'A' && ch <= 'Z') || (ch >= 'a' && ch <= 'z')) The programmer clearly intended for the second output "x is not zero." to be printed if the first condition (x == 0) is false, regardless of the second condition (y == 0). That is, the else was intended to be matched with the first if. But the else matching rule causes it to be matched with the second condition, which means that the output "x is not zero." will be printed only when x is zero and y is not zero. The else matching rule can be overridden with braces: if (x == 0) { if (y == 0) cout << "x and y are both zero." << endl; } else cout << "x is not zero." << endl; Now the else will be matched with the first if, the way the programmer had intended it to be. In the first statement, the else is matched with the first if. In the second statement, the else is matched with the second if. If n 2, the first statement will print NG while the second statement will do nothing. If 2 < n < 6, both statements will print OK. If n 6, the first statement will do nothing while the second statement will print NG. Note that this code is difficult to read because it does not follow standard indentation conventions. The first statement should be written
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