ssrs ean 128 Here is part of Table C.l in Appendix C. in Software

Generator QR in Software Here is part of Table C.l in Appendix C.

Here is part of Table C.l in Appendix C.
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CHAP. l]
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INTRODUCTION TO PROGRAMMING IN C++
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Table 1.2 Some C++ Operators Operator
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Description Negate Multiply Divide Remainder, modulo Add Subtract Bit shift left, output Simple assignment
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Precedence Associativity 15 13 13 13 12 12 11 2 Right Left Left Left Left Left Left Right
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Arity Unary Binary Binary Binary Binary Binary Binary Binary
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-n m*n m/n m%n m + n m - n tout << n m = n
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It lists 8 operators that apply to integer variables. They fall into five distinct precedence levels. For example, the unary negate operator - has precedence level 15, and the binary multiply operator * has precedence level 13, so negative is evaluated before multiply. Thus the expression m* - n is evaluated as m* ( -n) . Assignment operators have lower precedence than nearly all other operators, so they are usually performed last. The column labeled Associativity tells what happens when several different operators with the same precedence level appear in the same expression. For example + and - both have precedence level 12 and are left associative, so the operators are evaluated from left to right. For example, in the expression
8 - 5 + 4
first 5 is subtracted from 8, and then 4 is added to that sum:
(8 -5)+4=3+4=7
The column labeled A&y lists whether the operator is unary or binary. Chary means that the operator takes only one operand. For example, the post-increment operator + + is unary: n + + operates on the single variable n. Binary means that the operator takes two operands. For example, the add operator + is binary: m + n operates on the two variables m and n.
1.15 THE INCREMENT AND DECREMENT OPERATORS
Of the many features C++ inherited from C, some of the most useful are the increment operator ++ and decrement operator - -. These operators transform a variable into a statement expression that abbreviates a special form of assignment.
INTRODUCTION
PROGRAMMING
[CHAP. 1
EXAMPLE 1.17 Increment and Decrement Operators
This shows how the increment and decrement operators work:
#include ciost ream.h> // Tests the i ncremen t main0 and decrement operators:
int m = 44, n = 66; tout CC Nrn = ' CC m <<
endl;
++m; IIw
tout CC "m = U CC m << endl;
n I- ;
tout << flrn return 0; = u CC m << endl;
Both the pre-increment operator + +m and the post-increment operator m+ + have the same effect here: they add 1 to the value of m. Similarly, both the pre-decrement operator - -n and the post-decrement operator n- - have the same effect here: they subtract 1 from the value of n.
When used as a stand-alone expression statement, ++m and m++ are both equivalent to the assignment
m = m + 1;
they simply increase the value of m by 1. Similarly, the expression statements --n and n-are both equivalent to the assignment
n = n - 1 ;
they simply decrease the value of n by 1. (The increment operator ++ was used in the name C++ because it increments the original C programming language; it has everything that C has, and more.) However, when used as subexpressions (i.e., expressions within expressions), the pre-increment operation ++m is different from the post-increment operation m+ +. The pre-increment increases the variable first before using it in the larger expression, whereas the post-increment increases the value of the variable only after using the prior value of the variable within the larger expression. Since the incrementing process is equivalent to a separate assignment, there are really two statements to be executed when the increment operation is used as a subexpression: the incrementing assignment and the larger enclosing statement. The difference between the pre-increment and the post-increment is simply the difference between executing the assignment before or after the enclosing statement.
CHAP. l]
INTRODUCTION TO PROGRAMMING IN C++
EXAMPLE 1.18 Pre-Increment and Post-Increment Operators
This shows the difference between the pre-increment and the post-increment:
#include <iostream.h> // Tests the increment main0 int n = tout n = tout tout
decrement
operators:
m = 66, n; ++m; << -x m << '1, n = C-C n C-C endl; m++; << I m = m n = cc n << endl; << m cc endl; tout -x I'm = '1 C-K ++m << endl; return 0;
In the first assignment, m is pre-incremented, increasing its value to 67, which is then assigned to n. In the second assignment, m is post-incremented, so 67 is assigned to n and then m is increased to 68. In the third output statement, m is post-incremented, so the current valueof m (68) is dropped into the output stream and then m is increased to 69. In the last output statement, m is pre-incremented, so m is increased to 70 firstandthen that value is dropped into the output stream.
Use of the increment and decrement operators as subexpressions can be tricky and should be used with caution. For example, the order of evaluations of expressions that involve them is not defined by the language and consequently can be unpredictable:
EXAMPLE 1.19 The Unpredictablilty of the Order of Evaluation of Subexpressions #include main0 int n = x = ++n tout << tout << 5, x; * --n; "n = ' << n <=c ', x = ' C-K x -CC endl; ++n << u '1 << ++n << 'I ' << ++n << endl; <iostream.h>
In the assignment to x, n is first increment to 6 and then decremented back to 5 before the multiply operator is evaluated, computing 5 * 5. In the last line, the three subexpressions are evaluated from right to
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