Steady Hand Fun With The Raspberry Pi - Adrirobot

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Pointers can also be used together with arrays,#i ncl u de i nt mai n() {i nt i , * p, arr[4] = {6, 2, 4, 7} ;p = &arr[0] ; /* As s i gn the addres s of the fi rst el ement to p * /for( i =0; i
fpri ntf( fi l ePtr, " Wri ti ng ou t s ome i mportant data. . . %d, %d, %d\n" , 2, 3, 4) ;When the file access has been completed it has to be closed, flushing any data in memory to the disk.fcl os e( fi l ePtr) ; /* Cl os e textFi l e. txt, fl u s hi ng data to di s k * /The final example program uses many of the concepts discussed so far, to provide a simple encryption algorithm,#i ncl u de i nt mai n(i nt argc, char * argv[] ) {i nt mas k = 163; /* Decl are an i nt and as s i gn i t wi th a val u e l es s than 256. * /char c; /* Decl are a character ( whi ch i s one byte, maxi mu m val u e 255. ) * /FI L E * i npu tFi l e = 0, * ou tpu tFi l e = 0; /* decl are two fi l e poi nters * /i f( argc! =3) { /* Check the number of argu ments * /pri ntf( " Us age: %s \n" , argv[0] ) ;retu rn 1; /* Report an error * /}i npu tFi l e = fopen(argv[1] , " r" ) ; /* Open the i npu t fi l e. * /i f( ! i npu tFi l e) retu rn 2; /* I f fi l e poi nter i s nul l retu rn an error. * /ou tpu tFi l e = fopen(argv[2] , " w" ) ; /* Open the ou tpu t fi l e. * /i f( ! ou tpu tFi l e) retu rn 3; /* I f the fi l e poi nter i s nul l retu rn an error * /c = fgetc( i npu tFi l e) ; /* Get the fi rst character. * /whi l e( c ! = E OF) { /* L oop u nti l end- of- fi l e i s reached. * /c ^= mas k; /* E xcl u s i ve- OR wi th the mas k. * /fpu tc( c, ou tpu tFi l e) ; /* Wri te to the ou tpu t fi l e. * /c = fgetc( i npu tFi l e) ; /* Get another character. * /}fcl os e( i npu tFi l e) ; /* Cl os e the i npu t fi l e. * /fcl os e( ou tpu tFi l e) ; /* Cl os e the ou tpu t fi l e. * /}retu rn 0; /* Retu rn s u cces s to the operati ng s ys tem * /The program uses an exclusive-OR to encrypt and decrypt files. An exclusive-OR is true if either bit is set and false ifboth or no bits are set. For example, 2 ^ 3 = 1 in decimal or in binary 10 ^ 11 = 01. Type in a text file withnano and then try to encode the file. Then run the program again to decode the file.Challenge problemImprove the encryption algorithm by using a simple random number generator, e.g:#i ncl u de i nt newMas k( ) {i nt mas k = ( dou bl e) rand( ) /RAND_MAX* 254+1; /* Nu mber between 1 and 255 * /retu rn mas k;}i nt mai n() {s rand( 1234567) ; /* Set the s eed once * /i nt mas k = newMas k( ) ; /* Get a new mas k each ti me i t i s u s ed. * /retu rn 0;}The solution to the problem will be given next time.Article by W. H. Bell & D. Shepley23
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