THEORY AND PROBLEMS OF PROGRAMMING WITH Second Edition phần 1 pot

... 11 0 n 15 SI 47 I 79 0 11 1 0 16 DLE 48 0 80 P 11 2 P 17 DC 1 49 1 81 Q 11 3 q 18 DC2 50 2 82 R 11 4 r 19 DC3 51 3 83 S 11 5 S 20 DC4 52 4 84 T 11 6 t 21 ... 71 G 10 3 g 8 BS 40 ( 72 H 10 4 h 9 HT 41 1 73 I 10 5 i * 10 LF 42 74 J 10 6 j 11 VT 43 + 75 K 10 7 k 12 FF 44 1 76 L 10 8 1 13 CR 45 -77 M 10 9 m 14 ... 11 . Generating Consecutive Integer Quantities -Examples 6.8,6 .11 ,6 .14 ,6 .15 12 . Averaging a List of Numbers -Examples 6 .10 , 6 .13 ,6 .17 ,6.3 1 13 . Repeated Averaging of a List of...

THEORY AND PROBLEMS OF PROGRAMMING WITH Second Edition phần 2 pptx

... -3.3e+00: :12 3 12 -3.3e+00 : : +12 3 +12 -3,3e+00: : +12 3 +12 -3.3e+00 : 12 12 . -3.3 -3.30000: 88 DATA INPUT AND OUTPUT [CHAP. 4 The first line illustrates how integer and floating-point ... encountered the use of this function in Chaps. 1 and 2, and in Example 4 .1. Let us now examine it more thorough 1 y. The getchar function is a part of the standard C I/O library. ... 1 Execution of this program results in the following output. (The colons indicate the beginning of the first field and the end of the last field in each line.) : 12 34 17 77 a08c: : 12 34...

THEORY AND PROBLEMS OF PROGRAMMING WITH Second Edition phần 3 doc

... 1. 3 510 9 Iteration number : 9 X= 1. 35238 Iteration number: 10 X= 1. 3 51 75 Iteration number: 11 X= I.35206 Iteration number : 12 X= 1. 3 519 1 Iteration number: 13 X= 1. 3 519 8 ... Iteration number: 14 X= 1. 3 519 6 Iteration number: 15 X= 1 .3 519 6 Iteration number: 16 X= 1. 3 519 5 Root= 1. 3 519 5 No. of iterations= 16 Now suppose that a value of x = 10 had been ... resulting in x = [ 10 -3( 1 .47577)2]0.2= 1. 28225 Continuing this procedure, we obtain x = [ 10 -3( 1. 213 225)~IO.~ = 1. 38344 x = [10 -3 (1. 38344)2]0*2= 1. 33 613 and so on. Notice...

THEORY AND PROBLEMS OF PROGRAMMING WITH Second Edition phần 4 doc

... 47.7 9 0 .12 8 .1 10 0 .12 63.2 (b) Calculate the cumulative product of a list of n numbers. Test your program with the following six data items: 6.2, 12 .3, 5.0, 18 .8, 7 .1, 12 .8. (c) ... version of this problem). 17 0 CONTROL STATEMENTS [CHAP. 6 1 ’ i= 1 f = 0.06 x = 27.5 2 0.08 13 .4 3 0.08 53.8 4 0 .10 29.2 0 .10 74.55 6 0 .10 87.0 7 0 .12 39.9 8 0 .12 47.7 ... Depreciation: 11 63.64 Current Value: 4072.73 End of Year 4 Depreciation: 10 18 .18 Current Value: 3054.55 End of Year 5 Depreciation: 872.73 Current Value: 218 1.82 End of Year 6...

THEORY AND PROBLEMS OF PROGRAMMING WITH Second Edition phần 5 ppt

... the limits x = 1 and x = 4. Solve this problem first with 16 evenly spaced points, then with 61 points, and finally with 3 01 points. Note that the accuracy of the solution will ... the number of random variates, the mean and the standard deviation be input quantities to the program. Generate each random variate within a function that accepts the mean and standard deviation ... { 11 , 2J 3J 4)J 15 J 6J 7J 8), (9, 10 , 11 , 12 ) This definition results in the same initial assignments as in the last example. Thus, the four values in the first inner pair of braces...

THEORY AND PROBLEMS OF PROGRAMMING WITH Second Edition phần 6 ppsx

... -6A= X= 45678 910 11 5 5678 9 10 111 2 -4 6 7 8 9 10 11 12 13 7 -2 (6) Suppose A is a table of floating-point numbers having k rows and m columns, and B is a table of floating- ... 0 - 1 1 0 0 3 - 917 617 413 912 317 -3 3 4 -11 2 0 314 Display the elements of A, B and C. Be sure that everything is clearly labeled. (c) Read in the first m elements of a one-dimensional ... POINTERS [CHAP. 10 Values: i=l f=0.300000 d=0.005000 c=* Addresses : &i=ll7E &f =11 80 &d=ll86 &c=l18E Pointer values: px =11 7E px + 1= 118 0 px + 2 =11 82 px + 3 =11 84 The first...

THEORY AND PROBLEMS OF PROGRAMMING WITH Second Edition phần 7 pps

... CHAP. 10 1 PONTERS 33 1 10 .54 A C program contains the following declaration. static float table[2][3) = { tl .1, 1. 2, 1. 3), {2 .1, 2.2, 2.3) 1; What is the meaning of table? ... of (table + 1 )? What is the meaning of *(table + 1 )? What is the meaning of (*(table + 1 ) + 1 )? What is the meaning of (*(table) + 1 )? Whatisthevalueof*(*(table ... Pass f 1 and f2 to table-gen as arguments. Test the program using the values a =2, b =-0 .1, c =0.5 where the values oft are 1, 2,3, . . . ,60. 355 CHAP. 11 1 STRUCTURES AND UNIONS...

THEORY AND PROBLEMS OF PROGRAMMING WITH Second Edition phần 8 doc

... processed? 11 .11 What is the precedence of the period (.)operator? What is its associativity? 11 .12 Can the period operator be used with an array of structures? Explain. 11 .13 What is ... assigned? 11 .8 How is an array of structures initialized? 11 .9 What is the scope of a member name? What does this imply with respect to the naming of members within different structures? 11 .10 ... Lauderdale, FL 2 21 9 C 13 5.00 0.00 13 5.00 12 12 91 1998 Martin Peterson 17 87 Pacific Parkway San Diego, CA 8452 0 387.42 352.42 35.00 1 2 12 9 11 998 Phyllis Smith 10 00 Great White Way...

THEORY AND PROBLEMS OF PROGRAMMING WITH Second Edition phần 9 ppt

... 0 01 1 3 3 4 010 0 4 4 5 010 1 5 5 6 01 10 6 6 7 011 1 7 7 8 10 00 10 8 9 10 01 11 9 10 10 10 12 A 11 10 1 1 13 B 12 11 00 14 C 13 11 01 15 D 14 11 10 16 E 15 ... of the second word. Fig. 13 .2 shows the layout of the bit fields within the two 16 -bit words. word 2 word 1 bitno. 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 15 14 13 12 11 ... patterns. I lost bits I a = 0 710 11 01 IMI 011 1 a >> 6 = 0000 0007 7 011 0 710 = 0xlb6 I 0s I 453 1 CHAP. 14 1 SOME ADDITIONALFEATURES OF C case white: foreground...

THEORY AND PROBLEMS OF PROGRAMMING WITH Second Edition phần 10 ppsx

... 10 11 110 0 0 011 (4) 3a00 0 011 10 10 0000 0000 (r) 5b3c 010 1 10 11 0 011 11 00 (s) fbc3 11 11 10 11 110 0 0 011 (t) fbOO 11 11 10 11 0000 0000 (U) fbff 11 11 10 11 11 11 11 11 13 .40 ... 0 01 0 1 100 0 01 1 (a) 5d3c 010 1 11 01 0 011 11 00 (b) 2202 0 010 0 010 0000 0 010 (c) 9dc5 10 01 110 1 11 00 010 1 (4 bfc7 10 11 111 1 11 00 011 1 (e) 80cl 10 00 0000 11 00 00 01 v) 623a 011 0 ... PROBLEMS (I) ffff 11 11 11 11 11 11 11 11 (validforanyvalueofa) (m) a000 10 10 0000 0000 0000 (n) c100 11 00 00 01 0000 0000 (0) aOc3 10 10 0000 11 00 0 011 (p) 5bc3 010 1 10 11...

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