Consider the following C code snippet:

// Global Variable

uint8_t ipt1[4];

uint8_t ipt1[4]; // Input data 1

uint16_t ipt2[4]; // Input data 2

uint32_t ipt3[4]; // Input data 3

int main(void) {

for (int i = 0; i

ipt1[i] = i

ipt2[i] = i

ipt3[i] = i

}

}

Assuming the addresses of ipt1, ipt2, andipt3, are 0x2000_0000,0x2000_0010, and0x2000_0020, respectively. Draw the memory map (contents in decimal plus addresses in hexadecimal) as shown in the example below:

Draw the memory map of the first 4 addresses starting from ipt1.

Draw the memory map of the first 8 addresses starting from ipt2.

Draw the memory map of the first 16 addresses starting from ipt3.

8 bits High Address ex20000007 01110010 By grouping bits together we can store more values 8 bits = | byte 16 bits = 2 bytes = | halfword 32 bits = 4 bytes = I word From software perspective, memory is an addressable array of bytes. The byte stored at the memory address Ox20000004 is Ob10000100 00100101 ex20000006 0x20000005 11100010 @x20000004 10000100 ex20000003 01100001 ex20000002 10001111 ex20000001 00010010 @x20000000 10010100 Ob10000100 0x84 - 132 Binary Hexadecimal Decimal Low Address 8 bits High Address ex20000007 01110010 By grouping bits together we can store more values 8 bits = | byte 16 bits = 2 bytes = | halfword 32 bits = 4 bytes = I word From software perspective, memory is an addressable array of bytes. The byte stored at the memory address Ox20000004 is Ob10000100 00100101 ex20000006 0x20000005 11100010 @x20000004 10000100 ex20000003 01100001 ex20000002 10001111 ex20000001 00010010 @x20000000 10010100 Ob10000100 0x84 - 132 Binary Hexadecimal Decimal Low Address