he following program main() { static char a[3][4] = {"abcd", "mnop", "fghi"}; putchar(**a); }
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A: Required:
The following
main()
{
static char a[3][4] = {"abcd", "mnop", "fghi"};
putchar(**a);
}
A. |
will not compile successfully |
B. |
results in run-time error |
C. |
prints garbage |
D. |
none of the above |
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- #include <iostream> using namespace std; class test{ public: test(){ cout<<"alpha\n"; } ~test(){ cout<<"beta\n"; } }; int main(int argc, char const *argv[]) { test *a = new test[5]; delete[] a; return 0; } Give output.#include <iostream>using namespace std;class test{public:test(){cout<<"alpha\n";}~test(){cout<<"beta\n";}};int main(int argc, char const *argv[]){test *a = new test[5];delete[] a;return 0;} Give output.#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; }
- int fun(int k){ return ( ); void main(){ int n; cin >> n; n = n * fun(n); <-- 1 Fill in the appropriate statement and expression in fun, so that when it is called in main, and after the execution of the statement marked 1, the value of n would always be n3.#include <iostream> using namespace std; class Test { static int x; public: Test() { x++; } static int getX() {return x;} }; int Test::x = 0; int main() { cout << Test::getX() << " "; Test t[5]; cout << Test::getX(); } Is the given code is correct or not.#include <stdio.h>void cubeByReference( int *nPtr ); // function prototypeint main( void ){ int number = 5; // initialize number printf("The original value of number is %d", number ); // pass address of number to cubeByReference cubeByReference( &number ); printf("\nThe new value of number is %d\n", number );} // end main void cubeByReference( int *nPtr ){ *nPtr = *nPtr* *nPtr* *nPtr;} passing argument by reference - We modify the code above 1- define a second argument (example "int number2 = 9") and a pointer to it 2- define a second function (addByReference) that adds number2 to number - passing both arguments by reference 3- print-out the result (that is in number) Upload the output and .c code
- Editable source code: #include<stdio.h> int isEven(int); int main(void) { // TODO: Write your code here return 0;} int isEven(int n) { return n % 2 == 0;} The instructions is on the photo as well as the expected output#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 partSEE MORE QUESTIONSRecommended textbooks for youMicrosoft Visual C#Computer ScienceISBN:9781337102100Author:Joyce, Farrell.Publisher:Cengage Learning,Microsoft Visual C#Computer ScienceISBN:9781337102100Author:Joyce, Farrell.Publisher:Cengage Learning,