Jump to level 1Convert the C to assembly. y's base address is in DM[7040].y[3] = y[5];addi v $t01w1w1w<SW$t2$t3$t3addi v $t3$t3'I$zero ✓0 ($t10 ($t20($t2$zero ✓0 ( $t2"70400Szero$to$t1$12$13Registers00000Data memory0995000501250207040

Answered: Image /qna-images/answer/98d38025-b9aa-46a8-bcac-a69965a9b141.jpg

Convert the C to assembly. y's base address is in DM[7040]. y[3] = y[5]; addi v 1w 1w 1w 3 < SW < addi v $t0 $t2 V $t3 $t3 $t3 $t3 > I I $zero 0($t1 0 ($t2 0 ( $t2 $zero ✓ 0 ( $t2 7040 0 Szero $t0 $t1 $12 $13 Registers 0 0 0 0 0 Data men 5012 C 5020 99 7040 500

Convert the C to assembly. y's base address is in DM[7040]. y[3] = y[5]; addi v 1w 1w 1w 3 < SW < addi v $t0 $t2 V $t3 $t3 $t3 $t3 > I I $zero 0($t1 0 ($t2 0 ( $t2 $zero ✓ 0 ( $t2 7040 0 Szero $t0 $t1 $12 $13 Registers 0 0 0 0 0 Data men 5012 C 5020 99 7040 500

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

See similar textbooks

Related questions

Q: Evaluate the C code statements in MIPS assembly code for(i = 0; i < a; i++) for(j = 0; j < b;…

A: The MIPS code for the given loop for(i = 0; i < a; i++) for(j = 0; j < b; j++)…

Q: Assume that variables i and k are in registers s0 and s1, respectively and that B is an integer…

A: variables i and k are in registers s0 and s1, respectively and that B is an integer array whose base…

Q: In C, both long x and int *y are 8-byte values. How can we can tell which one is which in the…

A: Here is the explanation regarding pointers and integers.

Q: 3- What is the difference between: a- MOV BX,[1234H] and LEA BX,[1234H] b- LDS AX,[200H] and LES…

A: - According to the guideline we are allowed to answer only the first question. - The given…

Q: Convert the following C code to native hardware MIPS assembly. The variables a, b and x are signed…

A: sge Rdest, $s0, $s1 # if (a >= b), if a>=b, it stores value 1 in Rdest beq Rdest, $0,…

Q: What addresses in memory contain array A?

A: Given: What addresses in memory contain array A?

Q: gorithm (logic) to process reverse polish notat he operators +, -, and * are will always returni…

A: Given Line: 2 3 4 + + Result: 9 Line: 12 6 / 4 * 4 + Result: 12 Line: 2 2 3 3 * * * Result: 36 Line:…

Q: ere function f returns the results through address passing. Assuming c is a global complex number…

A: Complex increase is a more troublesome activity to comprehend from either an arithmetical or a…

Q: described each line and write the PA, also specify the type of addressing mode for the following…

A: In the realm of low-level programming, the efficient manipulation of data within a computer's memory…

Q: g[3] = h[1] + A[2]; Base address of g in $s1, base address of h in $s2, base address of A in $s3…

A: Here's the MIPS assembly code for the given line of code:

Q: Assuming variable b is stored in $s2 and that the base address of array A is in $s3, what is the…

A: Actions to be performed for an instructionlet's take a load instruction at address 0x800:0x800 lw…

Q: Need to be done in masm x86

A: Here's a program that meets the requirements you specified:```cpp#include <iostream>#include…

Q: Write out how to solve the following code using any addressing mode for (3,3) GPR. a[] is address…

A: Answer: I have given answered in the handwritten format in brief explanation.

Q: Quiz-1: Refer to Lecture-3 Module 1 swap (int* V, int k) { temp = V[k]; /* temp in $t0 */…

A: Lets use a temporary variable to create a program for registers.

Q: Consider two C++ int arrays A and B. • The base address of A is held by $s0, and B $s1. • Variable i…

A: An instruction is a set of operators and operands used to perform some computation. The instructions…

Q: 12.3 LAB: Array of squares - Iw/sw Given an array of four integers, write a program that populates a…

A: According to the information given:- We have to create array of square with the help of Store Word…

Q: Extend Example 7-36 to write (store) the string below of 28 characters into EEPROM locations…

A: Answer is given below-

Concept explainers

Computer Fundamentals

The computer is termed computation. For calculating or computing something the device that has been used is known as the computer. Or we can say that for performing a fast arithmetic operation the device that has been used is a computer. Storing, processing, and retrieving are the main features of a computer. 

Question

Jump to level 1
Convert the C to assembly. y's base address is in DM[7040].
y[3] = y[5];
addi v $t0
1w
1w
1w
<
SW
$t2
$t3
$t3
addi v $t3
$t3
'
I
$zero ✓
0 ($t1
0 ($t2
0($t2
$zero ✓
0 ( $t2
"
7040
0
Szero
$to
$t1
$12
$13
Registers
0
0
0
0
0
Data memory
0
99
5000
5012
5020
7040

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, computer-science and related others by exploring similar questions and additional content below.

Similar questions

Implement nibble_swap() function. Take in 4 byte integer and returns integer with swapped nibbles in first two bytes. Use a 3 bit register. The template below is for your reference. The code in the Assembly code portion needs to be completed in Assembly x86-64 #include extern "C" int nibble_swap(int var); int main(int avgc, char** avgs) { int v = nibble_swap(0xCAFEBABA); //answer: 0xACEFABAB printf("Swapped Variable is : 0%X \n", v); return 0; } ----------------Assembly Code --------------------------- ;yasm -f elf64 var.asm section .text global start : ;put ur code here ret
Using C Language In this function, b has the address of a memory buffer that is num_bytes long. The function should repeatedly copy the 16 byte pattern that pattern16 points at into the memory buffer until num_bytes have been written. If num_bytes is not a multple of 16, the final write of the 16 byte pattern should be truncated to finish filling the buffer. void memset16(void *b, int num_bytes, void *pattern16) For example if the 16 bytes that pattern16 points at is 00 11 22 33 44 55 66 77 88 99 aa bb cc dd ee ff, then memset(b, 20, pattern16) should write to the buffer pointed at by p the 20 bytes 00 11 22 33 44 55 66 77 88 99 aa bb cc dd ee ff 00 11 22 33. Use SSE instructions to improve efficiency. Here's pseudocode. x = SSE unaligned load from pattern16while (num_bytes >= 16) SSE unaligned store x to p advance p by 16 bytes decrement num_bytes by 16while (num_bytes > 0) store 1 byte from pattern16 to p advance p by 1 byte advance pattern16 by 1 byte…
How many hexadecimal digits are used when representing a nibble in hex form? 0 1 O 3 O 2 O O Patti the Programmer decides to create a function with three int16 parameters, sending all three by-reference. Will she need to pad her functions activation record to make it 32-bit aligned? O The answer will depend on how many registers get preserved on the stack O Yes O No O The answer will depend on the value of the passed parameters How many bits of stack memory are used when passing an int16 variable by-value to a function? O 32 O 64 O 16 O 8
The memory location at address of 0×003FB01 contains 1-byte memory variable J (0010_0001), and the memory location at the address of 0X003FB02 contains 1-byte memory variable K (0001_0010), see figure. If we declared char *Ptr; char J; what is the information hold by Ptr in hexadecimal form after executing C programming statement Ptr = &J? 7 Address in Data in Hex. FormatHex. Format 0X003FBF04 1110 0101 M 0X003FBF03 0000 1111 0×003FBF02 0001 0010 0×003FBF01 0010 0001 K J The information hold by Ptr is 0X003FBF04. а. O b. The information hold by Ptr is 0×003FBF02. The information hold by Ptr is 0×003FBF03. The information hold by Ptr is 0×003FBF01. O d.
The memory location at address of 0X003FB01 contains 1-byte memory variable J (0010_0001), and the memory location at the address of 0X003FB02 contains 1-byte memory variable K (0001_0010), see figure. If we declared char *Ptr; char J; what is the information hold by Ptr in hexadecimal form after executing C programming statement Ptr = &J? 7 Address in Data in Hex. Format Hex. Format OX003FBF04 1110 0101 0X003FBF03 0000 1111 0X003FBF02 0001 0010 K 0X003FBF01 0010 0001 J The information hold by Ptr is 0X003FBF03. a. The information hold by Ptr is OX003FBF04. Ob. The information hold by Ptr is 0×003FBF02. Oc. The information hold by Ptr is 0X003FBF01. d.
Need answer for D,E,F,G
E In the following code block(Reference:Q11), you will a set of assembly instructions with corresponding line numbers (line numbers are for informational purpose only and they are not part of the source code). 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 mov edx, 5 dec ecx jmp LABEL1 mov eax, 1 LABEL1: mul edx jmp ecx mov edx, 0678h sub edx, eax jmp DWORD PTR [edx] neg ebx add ecx, ebx mov eax, 0 For each of the conditions/scenario listed below, indicate the corresponding line number (that cause or is associated with the condition/scenario). Enter 0 (Zero) if the condition is not caused by the block of code. 1) Memory indirect jump: type your answer... type your answer... type your answer... type your answer... 2) Register indirect jump: 3) Relative short jump: type your answer... 5) Two's complement type your answer... 4) Relative near jump: 6) Unreachable code