Write an interactive c program that will encode or decode a line of text. To encode a line of text, proceed as follows: Convert each character, including blank spaces, to its ASCII equivalent. Generate a positive random integer. Add this integer to the ASCII equivalent of each character. The same random integer will be used for the entire line of text. Suppose that N1 represents the lowest permissible value in the ASCII code, and N2 represents the highest permissible value. If the number obtained in step 2 above exceeds N2, then subtract the largest possible multiple of N2 from this number, and add the remainder to N1. Hence the encoded number will always fall between N1 and N2, and will therefore always represent some ASCII character. Display the characters that correspond to the encoded ASCII values. The procedure is reversed when decoding a line of text. Be certain, however, that the same random number is used in decoding as was used in encoding.

1 Answers   Amazon, CSJM, HCL, Microsoft, TCS, Wipro,

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What is a far pointer?What is the utility?

4 Answers  

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List the different types of c tokens?

0 Answers  

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what is a static function

10 Answers   Satyam,

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Write a program of advanced Fibonacci series.

0 Answers   Aspiring Minds,

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Why do we need a structure?

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Tell us something about keyword 'auto'.

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main() { intj; while9j<=10) { printf("\n%d",j); j=j+1; } }

4 Answers  

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struct abc { unsigned int a; char b; float r; }; struct xyz { int u; struct abc tt; }ww; ww = (struct xyz*)malloc(sizeof(struct xyz)); will the memory be allocated for the inner structure also?

1 Answers   Wipro,

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Can we change the value of #define in c?

0 Answers  

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what do you mean by defining a variable in our c code?

2 Answers  

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18)struct base {int a,b; base(); int virtual function1(); } struct derv1:base{ int b,c,d; derv1() int virtual function1(); } struct derv2 : base {int a,e; } base::base() { a=2;b=3; } derv1::derv1(){ b=5; c=10;d=11;} base::function1() {return(100); } derv1::function1() { return(200); } main() base ba; derv1 d1,d2; printf("%d %d",d1.a,d1.b) o/p is a)a=2;b=3; b)a=3; b=2; c)a=5; b=10; d)none 19) for the above program answer the following q's main() base da; derv1 d1; derv2 d2; printf("%d %d %d",da.function1(),d1.function1(),d2.function1 ()); o/p is a)100,200,200; b)200,100,200; c)200,200,100; d)none 20)struct { int x; int y; }abc; you can not access x by the following 1)abc-->x; 2)abc[0]-->x; abc.x; (abc)-->x; a)1,2,3 b)2&3 c)1&2 d)1,3,4

1 Answers  

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