Database System Concepts
Database System Concepts
7th Edition
ISBN: 9780078022159
Author: Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher: McGraw-Hill Education
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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.
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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.
expand button
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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