What are the values of PBO-7 and PD0-7 after the following code segment executes? unsigned char w = 0x35; DDRB = OX0F; DDRD PORTB = w& OX0F; OXOF %3! PORTD - W >>4; PB7 P86 PB5 P84 P83 P82 PB1 PBO PD7 PD6 PD5 PD4 PD3 PD2 PD1 PDO
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- array([[ 1960, 54211], [ 1961, 55438], [ 1962, 56225], [ 1963, 56695], [ 1964, 57032], [ 1965, 57360], [ 1966, 57715], [ 1967, 58055], [ 1968, 58386], [ 1969, 58726], [ 1970, 59063], [ 1971, 59440], [ 1972, 59840], [ 1973, 60243], [ 1974, 60528], [ 1975, 60657], [ 1976, 60586], [ 1977, 60366], [ 1978, 60103], [ 1979, 59980], [ 1980, 60096], [ 1981, 60567], [ 1982, 61345], [ 1983, 62201], [ 1984, 62836], [ 1985, 63026], [ 1986, 62644], [ 1987, 61833], [ 1988, 61079], [ 1989, 61032], [ 1990, 62149], [ 1991, 64622], [ 1992, 68235], [ 1993, 72504], [ 1994, 76700], [ 1995, 80324], [ 1996, 83200], [ 1997, 85451], [ 1998, 87277], [ 1999, 89005], [ 2000, 90853], [ 2001, 92898], [ 2002, 94992], [ 2003, 97017], [ 2004, 98737], [ 2005, 100031], [ 2006, 100832], [ 2007, 101220], [ 2008, 101353], [ 2009, 101453], [ 2010, 101669], [ 2011, 102053], [ 2012, 102577], [ 2013, 103187], [ 2014, 103795], [ 2015, 104341], [ 2016, 104822], [ 2017, 105264]]) Question 2 Now that we have have our…In c language unsigned int data [ 3 ] = { 0x11121314, 0x23242526, 0x35363738 } ; How would print the individual bytes using a pointer pointing to data array ? Your output can be 0x11 0x12, 0x13 0x14 0x23 0x24 0x25 0x26 0x35 0x36 0x37 0x38STRING-ARRAY C program that scans a single char and prints out only the index/indexes of the character. If there are several matches, separate them by a single line. INPUT- scans a single char SAMPLE:Sample String: TheQuickBrownFoxJumpsOverTheLazyDogInput= TOutput= 0 26
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- If you want to define integer far pointer *n to address (Ox2000;0x200) O int far *n=(int far*)MK_FP(0x2000,0x200); O int far *n=(far int*)MK_FP(0x2000,0x200), int far *n-(int far*)MK FP(0x2000:0x200), O int far *n=(far int*)MK_FP(0x2000:0x200);void show_byte(byte_pointer start, int len) { Q2 int i; for(i=0; iA-1D-9 Given the code segment below, what will be the sum of ALL elements of dCData after running the codes? Limit your answer to one (1) decimal place only if answer is a floating point number. int main() int i; double dAData[3] = {1.e, 2.0, 3.0}; double dBData[3] = {4.e, 5.0, 6.0}; double dCData[3]; for (i = e; i < 3; i++) dCData[i] = dAData[i] + dBData[2-i] +i * i; return e; Assume the necessary libraries and functions are included and implemented. Write INVALID (in all CAPITAL LETTERS) if it is an invalid access, or if the task will cause a syntax or logical error.Please help me in c++ program declare two characker arrays and takes thier input and add /0 at the end of both then store value of first char array in third array which terminates terminates array when reads /0 at end of first char array and start printing 2nd char array in same 3rd string and print /0 at the end of third string. Output: Enter first string: this is /0 Enter seond string: cat /0 third string: this is cat /0c programming language The program below uses pointer arithmetic to determine the size of a 'char'variable. By using pointer arithmetic we can find out the value of 'cp' and thevalue of 'cp+1'. Since cp is a pointer, this addition involves pointer arithmetic:adding one to a pointer makes the pointer point to the next element of the sametype.For a pointer to a char, adding 1 really just means adding 1 to the address, butthis is only because each char is 1 byte.1. Compile and run the program and see what it does.2. Write some code that does pointer arithmetic with a pointer to an int anddetermine how big an int is.3. Same idea – figure out how big a double is, by using pointer arithmetic andprinting out the value of the pointer before and after adding 1.4. What should happen if you added 2 to the pointers from exercises 1through 3, instead of 1? Use your program to verify your answer.#include <stdio.h>int main( ){ char c = 'Z'; char *cp = &c; printf("cp is %p\n", cp);…int X[900]; int Y[600]; int sum, sum1, sum2, sum3; //parallelism : dividing outer loop in three parts //i = 1 to 300 for(i=1;i<=300;i++) { for(j=1;j<600;j++) { sum1 = X[i] + Y[j]; } } //i = 301 to 600 for(i=301;i<=600;i++) { for(j=1;j<600;j++) { sum1 = X[i] + Y[j]; } } //i = 601 to 900 for(i=601;i<=900;i++) { for(j=1;j<600;j++) { sum1 = X[i] + Y[j]; } } sum = sum1 + sum2 + sum3;} another way to solve the question that send in the picSEE MORE QUESTIONS