Write multiword addition for the given data and give the content of the related memory locations after the execution of the code. DS: 0710 ;data segment datal dw ed09ah data2 dw 302bh result dw 2dup (0) ;code segment
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- 2. function 2: series_generator2(int n);// n is the number of elements to display// series: 19683 6561 2187 729 243 81 27 9 3 1. Convert the following pseudo code into assembly codo- (//' represents-comments). // Each element is 1 Byte long // Each element is 1 Byte long arrayl 13h, 14h, 15h, 16h array2 12h, 13h, 14h, 15h // lengthl is a Symbolic constant // length2 is a Symbolic constant // this variable is 1 byte long initialized with 30 lengthl = number of items in Arrayl. length2 = number of items in Array2 samplel = 30h sample2 = 5h. // this variable is 1 byte long 1 byte long variable maxlength = max of lengthl and length2. // maxlength is // this is a variable initialized with 0 index = 0. While ( If (arrayl [index] > array2[index] index < maxlength ){ * samplel) / (array2[index] expl = (arrayl[index] only store the quotient of the division in expl and expl is 7/16 bit long variable Sample2) else expl = 0 index = index + 1 }Generate intermediate code for the code segment: (a>b) or (b
- IN C PROGRAMMING LANGUAGE: Please write a pointer version of strncmp() named pstr_ncmp(char *s, char *t, int n) which compares the first n characters.Please design a data structure to allow the following three operations: 1. push (insert a new number) 2. find_Kmin (return the value of the K-th smallest number without removing it). 3. pop All operations should be done in 0(log n) time or better.Define an array of ten pointers to floating point numbers. Then read ten numbers into individual locations referenced by pointers. Add all of the numbers and store the result in the pointer-referenced location. Display the contents of all locations.
- The following numbers are inserted into an empty LLRBT in the given order: 11, 2, 14, 25, 15, 13, 16. Please draw the resulting LLRBT.Question 1 Assume that arrX array is already defined and allocated in memory, its base address 0xabcd3000 and already stored in $a2 and its number of elements in $a3. Answer each of the next questions as required. arrX: .word 0x99,0x20,0x73,0x40,0x50,0x69, ... Please write question numbers and answer parts in this question in order. Q1) The fourth element of arrX is located at HEX address ___________ Q2) Write one MIPS instruction to copy into $t9 the value of the fourth element of arrX: ___________ Q3) Write one MIPS instruction to calculate $t0 = 32 * fourth element: __________ Q4) Write NO more than 3 MIPS instructions to decrement the fourth element of arrX: __________ Q5) Write no more than 6 instructions to calculate $v0= first element - last element in arrX.Assume that arrx array is already defined and allocated in memory, its base address Oxabcd3000 and already stored in Sa2 and its number of elements in Sa3. Answer each of the next questions as required. arrX: .word 0x99,0x20,0x73,0x40,0x50,0x69, ... Please write question numbers and answer parts in this question in order. Q1) The fourth element of arrX is located at HEX address Q2) Write one MIPS instruction to copy into $t9 the value of the fourth element of arrx: Q3) Write one MIPS instruction to calculate Sto = 32 * fourth element: Q4) Write NO more than 3 MIPS instructions to decrement the fourth element of arrX: Q5) Write no more than 6 instructions to calculate Svo= first element - last element in arrx.Segmentation: Select all of the following statements that are true. In segmentation, a logical address always has a length of 32 bit. In order to translate logical into physical addresses, the memory management unit uses the segment part of the logical address to determine the start address in the segment table and adds the offset to this to get the physical address. In segmentation, the logical address consists of a segment part and an offset. The segment length is limited by the maximum possible segment number. When applying segmentation, processes are only allowed to access the memory within their segments. Segments can be assigned access rights and privilege levels.SEE MORE QUESTIONS