{ "metadata": { "name": "chapter11" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "markdown", "metadata": {}, "source": "

Chapter 11: Pointers

" }, { "cell_type": "markdown", "metadata": {}, "source": "

Example 11.1, Page 11.2

" }, { "cell_type": "code", "collapsed": false, "input": "# realtionship between two variables, their address and associated pointers\n\n\nu,v=[3],[]\npu=u\nv=pu\npv=v\n\nprint \"u=\",u[0],\nprint \"&u=\",id(u),\nprint \"pu=\",id(u),\nprint \"*pu\",pu[0]\n\nprint \"v=\",v[0],\nprint \"&v=\",id(v),\nprint \"pv=\",id(v),\nprint \"*pv\",pv[0]\n", "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": "u= 3 &u= 56575840 pu= 56575840 *pu 3\nv= 3 &v= 56575840 pv= 56575840 *pv 3\n" } ], "prompt_number": 11 }, { "cell_type": "markdown", "metadata": {}, "source": "

Example 11.2, Page number: 11.3

" }, { "cell_type": "code", "collapsed": false, "input": "# a simple program using pointers\n\n\nv=[3]\nu1=2*(v[0]+5)\npv=v\nu2=2*(pv[0]+5)\n\nprint \"u1=%d u2=%d\" %(u1,u2)", "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": "u1=16 u2=16\n" } ], "prompt_number": 12 }, { "cell_type": "markdown", "metadata": {}, "source": "

Example 11.3, Page munber: 11.4

" }, { "cell_type": "code", "collapsed": false, "input": "# a simple programme using pointers\n\nv=[3]\npv=v\nprint \"*pv=%d v=%d\" %(pv[0],v[0])\n\npv[0]=0\nprint \"*pv=%d v=%d\" %(pv[0],v[0])", "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": "*pv=3 v=3\n*pv=0 v=0\n" } ], "prompt_number": 13 }, { "cell_type": "markdown", "metadata": {}, "source": "

Example 11.7, Page number: 11.6

" }, { "cell_type": "code", "collapsed": false, "input": "# difference between by pass by reference and pass by value\n\n\ndef funct1(u,v):\n u=0\n v=0\n print \"Within funct1 : u=%d v=%d\" %(u,v)\n return\n\ndef funct2(u,v):\n u[0]=0\n v[0]=0\n print \"Within funct2 : *pu=%d *pv=%d\" %(u[0],v[0])\n return\n\nu=[1]\nv=[3]\n\nprint \"Before calling funct1: u=%d v=%d\" %(u[0],v[0])\nfunct1(u[0],v[0])\nprint \"After calling funct1 : u=%d v=%d\" %(u[0],v[0])\n\nprint \"Before calling funct2: u=%d v=%d\" %(u[0],v[0])\nfunct2(u,v)\nprint \"After calling funct2 : u=%d v=%d\" %(u[0],v[0])", "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": "Before calling funct1: u=1 v=3\nWithin funct1 : u=0 v=0\nAfter calling funct1 : u=1 v=3\nBefore calling funct2: u=1 v=3\nWithin funct2 : *pu=0 *pv=0\nAfter calling funct2 : u=0 v=0\n" } ], "prompt_number": 14 }, { "cell_type": "markdown", "metadata": {}, "source": "

Example 11.8, Page number: 11.8

" }, { "cell_type": "code", "collapsed": false, "input": "# Analyzing a line of text\n\n\n\ndef scan_line(line,pv,pc,pd,pw,po):\n\n for c in line:\n\n if c=='A' or c=='E' or c=='I' or c=='O' or c=='U':\n pv[0]+=1\n elif c>='A' and c<='Z':\n pc[0]+=1\n elif c>='0' and c<='9':\n pd[0]+=1\n elif c==' ' or c=='\\t':\n pw[0]+=1\n else:\n po[0]+=1\n\n return\n\n\nvowel,consonants,digits,whitespc,other=[0],[0],[0],[0],[0]\nline=\"Personal computers with memories in excess of 4096 KB are now quite common.\"\nline=line.upper()\nscan_line(line,vowel,consonants,digits,whitespc,other)\nprint \"\\n\\n\"\nprint \"No. of vowels : \",vowel[0]\nprint \"No. of consonants : \",consonants[0]\nprint \"No. of digits : \",digits[0]\nprint \"No. of whitespace characters : \",whitespc[0]\nprint \"No. of other characters : \",other[0]", "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": "\n\n\nNo. of vowels : 23\nNo. of consonants : 35\nNo. of digits : 4\nNo. of whitespace characters : 12\nNo. of other characters : 1\n" } ], "prompt_number": 15 }, { "cell_type": "markdown", "metadata": {}, "source": "

Example 11.12, Page number: 11.15

" }, { "cell_type": "code", "collapsed": false, "input": "# unlike C in python the elements of a list are not arranged in memory uniformly\n\nx=[10,11,12,13,14,15,16,17,18,19]\n\nfor i in range(0,10):\n print \"i=%d x[i]=%d *(x+1)=%d\" %(i,x[i],x[i]),\n print \" &x[i]=\",id(x[i]),\n print \" x+i=\",id(x[i])", "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": "i=0 x[i]=10 *(x+1)=10 &x[i]= 30696620 x+i= 30696620\ni=1 x[i]=11 *(x+1)=11 &x[i]= 30696608 x+i= 30696608\ni=2 x[i]=12 *(x+1)=12 &x[i]= 30696596 x+i= 30696596\ni=3 x[i]=13 *(x+1)=13 &x[i]= 30696584 x+i= 30696584\ni=4 x[i]=14 *(x+1)=14 &x[i]= 30696572 x+i= 30696572\ni=5 x[i]=15 *(x+1)=15 &x[i]= 30696560 x+i= 30696560\ni=6 x[i]=16 *(x+1)=16 &x[i]= 30696548 x+i= 30696548\ni=7 x[i]=17 *(x+1)=17 &x[i]= 30696536 x+i= 30696536\ni=8 x[i]=18 *(x+1)=18 &x[i]= 30696524 x+i= 30696524\ni=9 x[i]=19 *(x+1)=19 &x[i]= 30696512 x+i= 30696512\n" } ], "prompt_number": 16 }, { "cell_type": "markdown", "metadata": {}, "source": "

Example 11.14, Page number: 11.17

" }, { "cell_type": "code", "collapsed": false, "input": "# string as one dimensional character array\n\ndef main():\n global x\n y=\"This string is declared within main\"\n y=list(y)\n x=list(x)\n print x\n print y\n\n return\n\nx=\"This string is declared externally\"\nmain()\n", "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": "['T', 'h', 'i', 's', ' ', 's', 't', 'r', 'i', 'n', 'g', ' ', 'i', 's', ' ', 'd', 'e', 'c', 'l', 'a', 'r', 'e', 'd', ' ', 'e', 'x', 't', 'e', 'r', 'n', 'a', 'l', 'l', 'y']\n['T', 'h', 'i', 's', ' ', 's', 't', 'r', 'i', 'n', 'g', ' ', 'i', 's', ' ', 'd', 'e', 'c', 'l', 'a', 'r', 'e', 'd', ' ', 'w', 'i', 't', 'h', 'i', 'n', ' ', 'm', 'a', 'i', 'n']\n" } ], "prompt_number": 17 }, { "cell_type": "markdown", "metadata": {}, "source": "

Example 11.16, Page number: 11.19

" }, { "cell_type": "code", "collapsed": false, "input": "# Reoedering a list of numbers\n\n\ndef reorder(x):\n n=len(x)\n for item in range(0,n-1):\n for i in range(item+1,n):\n if x[i]Example 11.17, Page number: 11.21

" }, { "cell_type": "code", "collapsed": false, "input": "# Pointer arithmetic doesnt exixt in python\n\ni=[4]\ni[0]=1\nf=0.3\nd=0.005\nc='*'\n\npx=i\nprint 'Values: i=%d f=%f d=%f c=%c \\n' %(i[0],f,d,c)\nprint 'Addresses: &i=%X &f=%X &d=%X &c=%X \\n' %(id(i[0]),id(f),id(d),id(c))\nprint 'Pointer Values: px=%X px+1=%X px+2=%X px+3=%X' %(id(px[0]),id(px[0]+1),id(px[0]+2),id(px[0]+3))\n\n", "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": "Values: i=1 f=0.300000 d=0.005000 c=* \n\nAddresses: &i=1D46518 &f=32BD940 &d=32BD950 &c=1D9F158 \n\nPointer Values: px=1D46518 px+1=1D4650C px+2=1D46500 px+3=1D464F4\n" } ], "prompt_number": 19 }, { "cell_type": "markdown", "metadata": {}, "source": "

Example 11.18, Page number: 11.22

" }, { "cell_type": "code", "collapsed": false, "input": "# two different pointer variable points to the first and last element of an integer array\n\na=[1,2,3,4,5,6]\n\npx=a[0]\npy=a[5]\n\nprint 'px=%X py=%X \\n' %(id(px),id(py))\nprint 'py - px = %X' %(id(py)-id(px))", "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": "px=1D46518 py=1D464DC \n\npy - px = -3C\n" } ], "prompt_number": 20 }, { "cell_type": "markdown", "metadata": {}, "source": "

Example 11.22, Page number: 11.26

" }, { "cell_type": "code", "collapsed": false, "input": "# adding two tables of numbers\n\ndef readinput(m,n,i=0):\n\n at=[]\n for row in range(0,m):\n temp=[]\n for col in range(0,n):\n t=i\n i+=2\n temp.append(t)\n at.append(temp)\n\n\n return at\n \ndef computesum(a,b,m,n):\n\n c=[]\n\n for row in range(0,m):\n temp=[]\n for col in range(0,n):\n t=a[row][col]+b[row][col]\n temp.append(t)\n c.append(temp)\n\n return c\n\ndef writeoutput(c,m,n):\n\n for row in range(0,m):\n for col in range(0,n):\n print \"%4d\" %(c[row][col]),\n print\n\n return\n\n\n\nprint \"\\n FIRST TABLE : \\n\"\na=readinput(5,5,1)\nwriteoutput(a,5,5)\n\nprint \"\\n SECOND TABLE : \\n\"\nb=readinput(5,5,50)\nwriteoutput(b,5,5)\n\nc=computesum(a,b,5,5)\nprint \"Sums of the elements : \\n\"\nwriteoutput(c,5,5)\n", "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": "\n FIRST TABLE : \n\n 1 3 5 7 9\n 11 13 15 17 19\n 21 23 25 27 29\n 31 33 35 37 39\n 41 43 45 47 49\n\n SECOND TABLE : \n\n 50 52 54 56 58\n 60 62 64 66 68\n 70 72 74 76 78\n 80 82 84 86 88\n 90 92 94 96 98\nSums of the elements : \n\n 51 55 59 63 67\n 71 75 79 83 87\n 91 95 99 103 107\n 111 115 119 123 127\n 131 135 139 143 147\n" } ], "prompt_number": 21 }, { "cell_type": "markdown", "metadata": {}, "source": "

Example 11.24, Page number: 11.31

" }, { "cell_type": "code", "collapsed": false, "input": "# adding two tables of numbers\n\ndef readinput(m,n,i=0):\n\n at=[]\n for row in range(0,m):\n temp=[]\n for col in range(0,n):\n t=i\n i+=2\n temp.append(t)\n at.append(temp)\n\n\n return at\n \ndef computesum(a,b,m,n):\n\n c=[]\n\n for row in range(0,m):\n temp=[]\n for col in range(0,n):\n t=a[row][col]+b[row][col]\n temp.append(t)\n c.append(temp)\n\n return c\n\ndef writeoutput(c,m,n):\n\n for row in range(0,m):\n for col in range(0,n):\n print \"%4d\" %(c[row][col]),\n print\n\n return\n\n\n\nprint \"\\n FIRST TABLE : \\n\"\na=readinput(5,5,1)\nwriteoutput(a,5,5)\n\nprint \"\\n SECOND TABLE : \\n\"\nb=readinput(5,5,50)\nwriteoutput(b,5,5)\n\nc=computesum(a,b,5,5)\nprint \"Sums of the elements : \\n\"\nwriteoutput(c,5,5)\n", "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": "\n FIRST TABLE : \n\n 1 3 5 7 9\n 11 13 15 17 19\n 21 23 25 27 29\n 31 33 35 37 39\n 41 43 45 47 49\n\n SECOND TABLE : \n\n 50 52 54 56 58\n 60 62 64 66 68\n 70 72 74 76 78\n 80 82 84 86 88\n 90 92 94 96 98\nSums of the elements : \n\n 51 55 59 63 67\n 71 75 79 83 87\n 91 95 99 103 107\n 111 115 119 123 127\n 131 135 139 143 147\n" } ], "prompt_number": 22 }, { "cell_type": "markdown", "metadata": {}, "source": "

Example 11.26, Page number: 11.34

" }, { "cell_type": "code", "collapsed": false, "input": "# Reordering a list of strings\n\n\ndef reorder(x):\n\n n=len(x)\n for item in range(0,n-1):\n for i in range(item+1,n):\n if x[item]>x[i]:\n temp=x[item]\n x[item]=x[i]\n x[i]=temp\n\n\n return\n\nx=['PACIFIC','ATLANTIC','INDIAN','CARIBBEAN','BERING','BLACK','RED','NORTH','BALTIC','CASPIAN']\nprint 'Original list of strings :\\n\\n'\n\nfor i in x:\n print \"String : \",i\n\nreorder(x)\n\nprint \"\\nReodered list of strings : \\n\\n\"\n\nfor i in x:\n print \"String : \",i", "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": "Original list of strings :\n\n\nString : PACIFIC\nString : ATLANTIC\nString : INDIAN\nString : CARIBBEAN\nString : BERING\nString : BLACK\nString : RED\nString : NORTH\nString : BALTIC\nString : CASPIAN\n\nReodered list of strings : \n\n\nString : ATLANTIC\nString : BALTIC\nString : BERING\nString : BLACK\nString : CARIBBEAN\nString : CASPIAN\nString : INDIAN\nString : NORTH\nString : PACIFIC\nString : RED\n" } ], "prompt_number": 23 }, { "cell_type": "markdown", "metadata": {}, "source": "

Example 11.28, Page number: 11.37

" }, { "cell_type": "code", "collapsed": false, "input": "# Display the day of the year\n\n\ndef convert(mm,dd,yy):\n yy-=1900\n ndays=long(30.42*(mm-1)+dd)\n if mm==2:\n ndays+=1\n if mm>2 and mm<8:\n ndays-=1\n if yy%4==0 and mm<2:\n ndays+=1\n\n ncycles=yy/4\n ndays+=ncycles*1461\n\n nyears=yy%4\n if nyears>0:\n ndays+=365*nyears+1\n if ndays>59:\n ndays-=1\n\n day=ndays%7\n\n return day\n\ndef main(mm,dd,yy):\n day_of_week=convert(mm,dd,yy)\n print \"%s, %s %d %d\" %(weekday[day_of_week],month[mm-1],dd,yy)\n return\n \nweekday=['Sunday','Monday','Tuesday','Wednesday','Thursday','Friday','Saturday']\nmonth=['January','February','March','April','May','June','July','August',\\\n 'September','October','November','December']\n\nmain(10,29,1929)\nprint\nmain(8,15,1945)\nprint\nmain(7,20,1969)\nprint\nmain(5,24,1997)\nprint\nmain(8,30,2010)\nprint\nmain(4,12,2069)\n", "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": "Tuesday, October 29 1929\n\nWednesday, August 15 1945\n\nSunday, July 20 1969\n\nSaturday, May 24 1997\n\nMonday, August 30 2010\n\nFriday, April 12 2069\n" } ], "prompt_number": 24 }, { "cell_type": "markdown", "metadata": {}, "source": "

Example 11.30, Page number: 11.44

" }, { "cell_type": "code", "collapsed": false, "input": "# Future value of monthly deposits (compound interest calculations)\n\n\ndef ratio_choose(freq,a,m,n):\n\n if freq=='C':\n ratio=md3(a,m,n)\n elif freq=='D':\n ratio=md2(a,m,n)\n else:\n ratio=md1(a,m,n)\n\n return ratio\n\n\ndef table(freq,a,m,n):\n\n\n print \"Interest Rate Future Amount\\n\\n\"\n for count in range(1,21):\n i=0.01*count\n f=a*ratio_choose(freq,i,m,n)\n print \" %2d %.2f\" %(count,f)\n\n return\n\ndef md1(i,m,n):\n \n factor=1+i/m\n ratio=12*(factor**(m*n)-1)/i\n return ratio\n\ndef md2(i,m,n):\n\n factor=1+i/m\n ratio=(factor**(m*n)-1)/(factor**(m/12)-1)\n return ratio\n\ndef md3(i,dummy,n):\n ratio=(10**(i*n)-1)/(10**(i/12)-1)\n return ratio\n\ndef main(freq):\n m=0\n freq=freq.upper()\n if freq=='A':\n m=1\n print \"\\nAnual Compounding\\n\"\n \n elif freq=='S':\n m=2\n print \"\\nSemiannual Compounding\\n\"\n \n elif freq=='Q':\n print \"\\nQuaterly Compounding\\n\"\n m=4\n \n elif freq=='M':\n m=12\n print \"\\nMonthly Compounding\\n\"\n \n elif freq=='D':\n m=360\n print \"\\nDaily Compounding\\n\"\n\n elif freq=='C':\n m=0\n print \"\\nContinuous Compounding\\n\"\n \n else:\n print \"\\nERROR!!! Please Repeat\\n\\n\"\n return\n \n return m\n\n\na,n,freq=100,3,'m'\nprint \"FUTURE VALUE OF A SERIES OF MONTHLY DEPOSITS\\n\\n\"\nprint \"Frequency of Compunding (A,S,Q,M,D,C): \",freq\nprint \"Amount of Each Monthly Payement : \",a\nprint \"Number of years: \",n\nm=main(freq)\n\ntable(freq,a,m,n)\n\n", "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": "FUTURE VALUE OF A SERIES OF MONTHLY DEPOSITS\n\n\nFrequency of Compunding (A,S,Q,M,D,C): m\nAmount of Each Monthly Payement : 100\nNumber of years: 3\n\nMonthly Compounding\n\nInterest Rate Future Amount\n\n\n 1 3653.00\n 2 3707.01\n 3 3762.06\n 4 3818.16\n 5 3875.33\n 6 3933.61\n 7 3993.01\n 8 4053.56\n 9 4115.27\n 10 4178.18\n 11 4242.31\n 12 4307.69\n 13 4374.33\n 14 4442.28\n 15 4511.55\n 16 4582.17\n 17 4654.18\n 18 4727.60\n 19 4802.45\n 20 4878.78\n" } ], "prompt_number": 25 }, { "cell_type": "code", "collapsed": false, "input": "", "language": "python", "metadata": {}, "outputs": [], "prompt_number": 25 } ], "metadata": {} } ] }