* Translate the IF statement to assembly language* All values are unsignedстр ebx,есхif( ebx < есх )јa next{mov eax,5еах 3 5;mov edx,6edx = 6;next:}There can be multiple correct solutions

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Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

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Use c programming for the following question. Complete the coding and plz don't use any other libraries
Using the following description, please replace the return line in C code; /* * addOK: determine if x+y succeeds without overflow * Examples: * addOK(0x40000000,0x20000000) = 1 * addOK(0x40000000,0x40000000) = 0 * addOK(0x55555555,0x2AAAAAAA) = 1 * addOK(0x55555555,0x2AAAAAAB) = 0 * addOK(0x80000000,0x40000000) = 1 * addOK(0x80000000,0x80000000) = 0 * Legal ops: ! ~ & ^ | + << >> * Max ops: 20 * */ int addOK(int x, int y) { return 2 } You cannot: 1. Use any control constructs such as if, do, while, for, switch, etc. 2. Define or use any macros. 3. Define any additional functions in this file. 4. Call any functions. 5. Use any other operations, such as &&, ||, -, or ?:, not listed in the the "Legal ops" list for the function. 6. Use any form of casting. 7. Use any data type other than int.
bitcount: Given x, set y to the number of bits in x that are 1. Write the code in Java. Provided input(s): x Permitted: 40 operations (may use !, ~, +, -, <<, >>, &, ^, |) Hint: The obvious solution would be something like ans=0; for(int i=0; i<32; i+=1) { a += x&1; x >>=1; } We don’t allow for loops, but even if you replace it with 32 copies that’s still 96 operations, and we only allow 40 for this task. The trick is to do things in parallel, treating a number like a vector of smaller numbers. Suppose I wanted to count the bits of an 8-bit number with bits abcdefgh. With a little shifting and masking I could make three numbers 0b00e00h 0a00d00g 0000c00f and add them to get xx0yy0zz where xx = a+b, yy = c+d+e, and zz = f+g+h. Extending this trick to several rounds on 32 bits will solve this problem.

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Translate the IF statement to assembly language * All values are unsigned стр еbx,eсх if( ebx <3 есх ) ja next { mov eax,5 eах 3 5; mov edx,6 edx = 6; next: } There can be multiple correct solutions