In each of the following code snippets, data is copied from x to y. How many bytes of data are copied? The answer should be a C expression. (e) int x=10, y=x; (f) int x[10]; int *y = x; (g) int x[10] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}; void f(int y[5]) { ... } int main() { f(x); } (h) int x[10], y[10]; ... memcpy(y, x, 5*sizeof(int))
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In each of the following code snippets, data is copied from x to y. How many bytes of data are
copied? The answer should be a C expression.
(e) int x=10, y=x;
(f) int x[10]; int *y = x;
(g) int x[10] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
void f(int y[5]) { ... }
int main() { f(x); }
(h) int x[10], y[10];
...
memcpy(y, x, 5*sizeof(int));
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- . What will be the output of the following code, consider the memory address of variable x is 0x0044c and pointer variable ptr is 0x0066f, and consider int data type take 2 byte in the memory void main() { int x[5] = { 1,2,3,4,5 }; int *ptr = x; ptr = ptr + 1; cout << ptr << endl; cout << *ptr << endl; cout << &ptr << endl; cout << &x[0] << endl; cout << *x << endl; x += 2; cout << x << endl; }Describe what problem occurs in the following code. What modifications should be made to it to eliminate the problem? int[] sums = {6, 12, 3, 32, 12, 10, 9, 6}; for (int index = 1; index <= sums.length; index += 1) { System.out.println(sums[index]); }Consider the following function: int secret(int m, int n) { int temp = n; for (int i = 1; i < abs(m); i++) temp = temp + n; if (m < 0) temp = -temp; return temp; } What is the output of the following C++ statements? i.cout << secret(5, 4) << endl; ii.cout << secret(-3, 20) << endl;
- Computer Science Write a program that processes a data sequence according to the following specifications: The sequence has three kinds of positive numbers: red, black and white. Each red number is preceded by the letter r, each black number is preceded by the letter b and each white number is preceded by the letter w. The number 0 indicates the end of data. If a number is preceded by any character other than r, b or w, the number is ignored. For instance, if the input to the program is: r 23 b 15 b 11 w 17 c 13 r 19 b 0, then we have two red numbers, 23 and 19, two black numbers 15 and 11, and one white number, 17. The number 13 is ignored. (The terminating 0 is also a black number but it is not processed.) The program produces the following output: The sum of red numbers is 42The sum of black numbers is 26The sum of white numbers is 17 Assume that input is from the keyboard and is in the prescribed format. (a). In 2-3 sentences, describe your strategy for solving this…Consider the following C++ code statements. How could you assign the address of i to iPtr? int i = 3; int *iPtr = NULL; A) *iPtr = i; B) *iPtr = &i; C) iPtr = &i; D) iPtr = *i;Consider the following declarations. In each case, write the English name of the type of x.Example: double *x;Answer: pointer-to-double (a) int *x[];(b) int (*x)[];(c) double **x;(d) unsigned long int (*x[])[n];
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- in c++ You are given an array A of non-negative integers of size m. Your task is to sort the array in increasing order and print out the original indices of the new sorted array. Example: A= {12, 2, 6, 10, 9, 24,23, 33} After sorting the new array becomes A= {2,6,9,10,12,23,24,33}. The required output should be sorted array and original indexes "1 2 4 3 0 6 7", index of 2 in original array was 1 , index for 6 was 2 and so on.Consider the following function: int secret(int m, int n){int temp = n;for (int i = 1; i < abs(m); i++)temp = temp + n;if (m < 0)temp = -temp;return temp;} a. What is the output of the following C++ statements? i. cout << secret(18, 4) << endl; ii. cout << secret(-10, 20) << endl;b. What does the function secret do?The topic is Booth's algorithm - Explain this C programming code - i have a seminar. //header files #include #include #include int a=0,b=0,c=0,a1=0,b1=0,com[5]={1,0,0,0,0}; int anum[5]={0},anumcp[5] ={0},bnum[5]={0}; int acomp[5]={0},bcomp[5]={0},pro[5]={0},res[5]={0}; // binary function void binary(){ a1 = fabs(a); b1 = fabs(b); int r, r2, i, temp; //for loop for binary for(i = 0; i < 5; i++){ r = a1 % 2; a1 = a1 / 2; r2 = b1 % 2; b1 = b1 / 2; anum[i] = r; anumcp[i] = r; bnum[i] = r2; if(r2 == 0){ bcomp[i] = 1; } if(r == 0){ acomp[i] =1; } } //2's complimanting part for the program c = 0; for( i = 0; i < 5; i++){ res[i] = com[i]+ bcomp[i] + c; if(res[i]>=2){ c = 1; } else c = 0; res[i] =…