# Answered Essay: b) What message will printf print for part a) (3 pts)

Question 3 (10 pts): Memory Map Practice

a) In order to debug an embedded application or to understand an embedded application developed by someone else, it is often necessary to understand how data memory is modified by C code statements. Fill in the given memory map after the C fragment below has been executed. Note: The TM4C123 uses Little Endian ordering for memory (i.e. lower significant bytes of an element are stored at lower addresses). (7 pts)

char msg[4] = “Sal”;

short age[1] = {0x2070};

int speed[2];

msg[3] = 0x61;

msg[1] = 0x6F;

age[0] = age[0] + 2;

speed[0] = 0x696C6345;

age[3] = 0x7370;

age[4] = 0x65;

printf(“%s”, msg);

 Memory Location 0xFF00 0xFF01 0xFF02 0xFF03 0xFF04 0xFF05 0xFF06 0xFF07 0xFF08 0xFF09 0xFF0A 0xFF0B 0xFF0C 0xFF0D Value Array msg age speed Index 0 1 2 3 0 0 1

b) What message will printf print for part a) (3 pts)

Question 3 (10 pts): Memory Map Practice a) In order to debug an embedded application or to understand an embedded application developed by someone else, it is often necessary to understand how data memory is modified by C code statements Fill in the given memory map after the C fragment below has been executed. Note: The TM4C123 uses Little Endian ordering for memory (i.e. lower significant bytes of an element are stored at lower addresses).(7 pts) char msq[4] “Sal”; short age [1] {0x2070}; int, speed [2]: msg [3] msg [1] 0x61; 0x6F; age [0] = age [0] + 2; speed [0] 0x696C6345; age [3] age [4] = 0x7370; = 0x65; printf(“%s”, msg); Mem Location 0xFF00lOxFF01 0xFF02 0xFF03 0xFF04 0xFF05 0xFF06 0xFF07 0xFF08 0xFF09 0xFFOAlOxFFOB 0xFFOCIOxFFOD Value ms age speed Index0 0 0 1 b) What message will printf print for part a)(3 pts)

Editable code:

a)

Using C Code:

#include<stdio.h>

int main()

{

char msg[4] = “Sal”;

short age[1] = {0x2070};

int speed[2];

msg[3] = 0x61;

msg[1] = 0x6F;

age[0] = age[0] + 2;

speed[0] = 0x696C6345;

age[3] = 0x7370;

age[4] = 0x65;

printf(“For msg[0]: %02xn”, msg[0]);

printf(“For msg[1]: %02xn”, msg[1]);

printf(“For msg[2]: %02xn”, msg[2]);

printf(“For msg[3]: %02xn”, msg[3]);

printf(“For age[0]: %04xn”, age[0]);

printf(“For speed[0]: %08xn”, speed[0]);

printf(“For speed[1]: %08xn”, speed[1]);

}

b)

Using C Code:

#include<stdio.h>

int main()

{

//declare char variable for print

char print[14];

//each value from the part a

print[0]=0x53;

print[1]=0x6F;

print[2]=0x6C;

print[3]=0x61;

print[4]=0x72;

print[5]=0x20;

print[6]=0x45;

print[7]=0x63;

print[8]=0x6C;

print[9]=0x69;

print[10]=0x70;

print[11]=0x73;

print[12]=0x65;

print[13]=0x00;

//display the message for printf from part a)

printf(“%c%c%c%c%c%c%c%c%c%c%c%c%c%c”, print[0],print[1],print[2],print[3],print[4],print[5],print[6],

print[7],print[8],print[9],print[10],print[11],print[12],print[13]);

}

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