Part 1: he base address is 0x10010000. What is the address of the memory location that holds Ox56789018? char char_arr[] = {0x23, 0x24, 0x25); //+0,+1, +2 it wordNum1 = 0x78; //+4 har varl = 'a';//Hint: varl: .byte 'a' //+8 it int_arr[] = {0x01, 0x003, 0x56789018}; //+12, +16, +20 dec...hex ar *str = "Hi";//Hint: str: .asciiz "Hi" Part 2: The base address is 0x10010000. What is the address of the memory location that holds 'H'? char char_arr[] = {0x23, 0x24, 0x25); //+0,+1, +2 int wordNum1 = 0x78; //+4 char varl 'a';//Hint: varl: .byte 'a' //+8 int int_arr[] = (0x01, 0x003, 0x56789018); //+12, +16, +20dechex char *str= "Hi";//Hint: str: .asciiz base+24 de hex char myChar '0'; "Hi" 'H' or ASCII cod2 =48 in addre * Part 3 is on second attachment. Thank you.
Part 1: he base address is 0x10010000. What is the address of the memory location that holds Ox56789018? char char_arr[] = {0x23, 0x24, 0x25); //+0,+1, +2 it wordNum1 = 0x78; //+4 har varl = 'a';//Hint: varl: .byte 'a' //+8 it int_arr[] = {0x01, 0x003, 0x56789018}; //+12, +16, +20 dec...hex ar *str = "Hi";//Hint: str: .asciiz "Hi" Part 2: The base address is 0x10010000. What is the address of the memory location that holds 'H'? char char_arr[] = {0x23, 0x24, 0x25); //+0,+1, +2 int wordNum1 = 0x78; //+4 char varl 'a';//Hint: varl: .byte 'a' //+8 int int_arr[] = (0x01, 0x003, 0x56789018); //+12, +16, +20dechex char *str= "Hi";//Hint: str: .asciiz base+24 de hex char myChar '0'; "Hi" 'H' or ASCII cod2 =48 in addre * Part 3 is on second attachment. Thank you.
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
Related questions
Question
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Hello. Please answer the attached MIPS computer architecture question and all its parts correctly. You do not have to give a lengthy explanation for each part.
*If you answer the question and all its parts correctly, I will give you a thumbs up. Thanks.
![Part 1:
The base address is 0x10010000.
What is the address of the memory location that holds Ox56789018?
char char_arr[] = {0x23,0x24,0x25}; //+0, +1, +2
int wordNum1 = 0x78; //+4
char varl = 'a'; //Hint: varl:
.byte
'a ' //+8
int int_arr[] = {0x01, 0x003, 0x56789018}; //+12, +16, +20 dec=...hex
char *str = "Hi";//Hint: str:
"Hi"
.asciiz
Part 2:
The base address is Ox10010000.
What is the address of the memory location that holds 'H'?
char char_arr[] = {0x23, 0x24, 0x25}; //+0, +1, +2
int wordNum1 = 0x78; //+4
char varl = 'a'; //Hint: varl:
'a' //+8
int int_arr[] = {0x01, 0x003,0x56789018); //+12, +16,+20 dec=...hex
char *str = "Hi";//Hint: str:
asciiz
"Hi"
base+24 dec...hex
char mychar = '0';
.byte
'H' or ASCII cod2 = 48 in address
* Part 3 is on second attachment.
Thank you.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1223c4c6-4ebb-4911-bc0d-edb37c26385f%2Fa57b0e8e-caea-486e-b620-b2a0545f3dbd%2Fkckb1np_processed.png&w=3840&q=75)
Transcribed Image Text:Part 1:
The base address is 0x10010000.
What is the address of the memory location that holds Ox56789018?
char char_arr[] = {0x23,0x24,0x25}; //+0, +1, +2
int wordNum1 = 0x78; //+4
char varl = 'a'; //Hint: varl:
.byte
'a ' //+8
int int_arr[] = {0x01, 0x003, 0x56789018}; //+12, +16, +20 dec=...hex
char *str = "Hi";//Hint: str:
"Hi"
.asciiz
Part 2:
The base address is Ox10010000.
What is the address of the memory location that holds 'H'?
char char_arr[] = {0x23, 0x24, 0x25}; //+0, +1, +2
int wordNum1 = 0x78; //+4
char varl = 'a'; //Hint: varl:
'a' //+8
int int_arr[] = {0x01, 0x003,0x56789018); //+12, +16,+20 dec=...hex
char *str = "Hi";//Hint: str:
asciiz
"Hi"
base+24 dec...hex
char mychar = '0';
.byte
'H' or ASCII cod2 = 48 in address
* Part 3 is on second attachment.
Thank you.
![Part 3:
The base address is 0x10010000.
What is the address of the memory location that holds 'o'?
char char_arr[]
int wordNum1 = 0x78; //+4
char varl = 'a'; //Hint: varl:
.byte
'a' //+8
int int_arr[] {0x01, 0x003, 0x56789018}; //+12, +16, +20 dec=...hex
char *str = "Hi"; //Hint: str:
.asciiz
"Hi"
=
=
{0x23,0x24,0x25}; //+0,+1, +2
'H' or ASCII cod2 =48 in address base+24dec=...hex
//'i' will be at base+28 dec=19hex .asciiz puts a null character after the string. Therefore, you see 00 at address Ox1001001A.
char myChar = '0'; //'0' will be save in the next byte.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1223c4c6-4ebb-4911-bc0d-edb37c26385f%2Fa57b0e8e-caea-486e-b620-b2a0545f3dbd%2F0dkn6dw_processed.png&w=3840&q=75)
Transcribed Image Text:Part 3:
The base address is 0x10010000.
What is the address of the memory location that holds 'o'?
char char_arr[]
int wordNum1 = 0x78; //+4
char varl = 'a'; //Hint: varl:
.byte
'a' //+8
int int_arr[] {0x01, 0x003, 0x56789018}; //+12, +16, +20 dec=...hex
char *str = "Hi"; //Hint: str:
.asciiz
"Hi"
=
=
{0x23,0x24,0x25}; //+0,+1, +2
'H' or ASCII cod2 =48 in address base+24dec=...hex
//'i' will be at base+28 dec=19hex .asciiz puts a null character after the string. Therefore, you see 00 at address Ox1001001A.
char myChar = '0'; //'0' will be save in the next byte.
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