The output of the following code int a[]={10,20,30,40}, *p; p = &a[2]; p++; printf(“%d”,*p);
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The output of the following code
int a[]={10,20,30,40}, *p;
p = &a[2];
p++;
printf(“%d”,*p);
Step by step
Solved in 2 steps with 1 images
- int* p: int a[3]{1, 2, 3}; p = a; What is the value of *(p+2)?#include using namespace std; int find(int arr[], int value, int left, int right) { int midpt = (left+right)/2; if (left > right) return -1; if ( arr[midpt] return midpt; else if (arr[midpt] < value) == value) return else return find(arr,value, left,midpt-1); } void main(void) { int arr[] ={4,5,6,9,11}; cout<#include <stdio.h> struct Single { int num; }; void printSingle(int f) { int binaryNum[33]; int i = 0; while(f>0) { binaryNum[i] = f % 2; f = f/2; i++; } for (int j=i-1; j>= 0; j--) { printf("%d",binaryNum[j]); } } int main() { struct Single single; single.num = 33; printf("Number: %d\n",single.num); printSingle(single.num); return 0; }#include <stdio.h> struct dna { int number; char text; char stringvalue[30]; }; int main() { struct dna dnavalue[5]; int i; for(i=0; i<4; i++) { printf("Sample %d\n",i+1); printf("Enter Number:\n"); scanf("%d", &dnavalue[i].number); printf("Enter Text :\n"); scanf("%c",&dnavalue[i].text); printf("Enter String :\n"); scanf("%s",dnavalue[i].stringvalue); } printf("Sample DNA Number Text String\n"); for(i=0; i<4; i++) { printf("%d\t\t", i+1); printf("%d\t\t", dnavalue[i].number); printf("%c\t\t", dnavalue[i].text); printf("%s", dnavalue[i].stringvalue); printf("\n"); } return 0; } Write comments on every line of this program explaining what's happening briefly#include <stdio.h> int main(){ int arr[10]; int i; for (i=0; i<10; i++){ arr[i] = i; } for (i=0; i<10; i++){ printf("arr[%d] = %d\n", i, arr[i]); } return 0;} Copy the code and modify it as follows: Add another separate array of 10 integers. The following code parts are to be added (in the same order as specified) after the printing of the values of the first array:(1) Copy the content of the first array to the second array. Add 10 to each of the values of the second array. Make use of a single loop to achieve this part.(2) Print each element of the second array. Make use of another loop for this partز١C++ n is the size of arr1, m is the size of arr2CFG: Example 1 • Draw the CFG for the following code: int f(int n){ } int m = n* n; if (n < 0) else return 0; return m;#include <stdio.h>#include <string.h> int findRepeat(char* s){int p , i , j; p = -1; for (i = 0 ; i < strlen(s) ; i++){for (j = i + 1; j < strlen(s); j++){if (s[i] == s[j]){p = i;break;}}if (p != -1)break;}return p;} int main(){char str[] = "Hello World";int pos = findRepeat(str);if (pos == -1)printf("Not found");elseprintf("%c", str[pos]);return 0;}can you please explain this code for meint X[10]={2,0,6,11,4,5,9,11,-2,-1); From the code above, what is the value of X[8] ?#include using namespace std; main() { int num array[] = (1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; for (int n-1; n<-5;++n) { cout <« n<< * "; } return 0; Out put : 1 23 4 567 89 10 O Out put : 3 4 56789 10 O Out put : 2 3 45 67 89 10 11 O Out put : 1 2 3 456 Out put : 1 2 345 Not listed Oedit code to print letters not numbers #include<stdio.h> void main(){ //Required matrices and variables int matrix1[6][6]; int matrix2[6][6]; int matrix3[6][6]; int i,j,k,n,m,l; //Reads number from user printf("Enter a number : "); scanf("%d", &n); //Value for diagonal m = 2*n; //Temperary variable of 'n' l = n; //Creates matrix 1 for(i = 0; i< 6; i++){ k = l; for(j = 0; j<6; j++){ if(j == (6-i-1)) matrix1[i][j] = m; else if(j > (6-i-1)) matrix1[i][j] = (k++ + 1); else matrix1[i][j] = k--; } l--; } //Prints matrix 1 for(i = 0; i< 6; i++){ printf("["); for(j = 0; j<6; j++) printf("%d ", matrix1[i][j]); printf("]\n"); } printf("\n\n"); //Creates matrix 2 l = n; for(i = 0; i<6 ; i++) for(j = 0; j<6; j++){ if(i < 3 && j < 3) matrix2[i][j] = l;…SEE MORE QUESTIONS
