The code below is written in C. In C, for each integer variable 4 Byte, for each char variable 1 Byte, and for each double variable 8 Byte are allocated from the memory. After the execution of the code below, how many bytes are leaked? int main (){ int *a;
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The code below is written in C. In C, for each integer variable 4 Byte, for each char variable 1 Byte, and for each double variable 8 Byte are allocated from the memory. After the execution of the code below, how many bytes are leaked?
int main (){
int *a;
a = (int*) malloc (60*sizeof(int));
a[0] = 60;
a = (double*) malloc (a[0]*sizeof(double));
char *d,*c = (char*) malloc (10*sizeof(char));
double b = 5;
free(a);
free(d);
return 0;
}
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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; }In C, write a function that gets 3 pointers int* a, int* b, int* c, and rotates the values in their addresses to the left. That is, a gets the value of b, b gets the value of c, and c gets the value of a. void rotate3 (int* a, int* b, int* c); For example, if we have int x=1, y=2, z=3, then after calling rotate3 (&x, &y, &z) we should have x==2, y==3, and z==1. if we have int x=7, y=1, z=6, then after calling rotate3 (&x, &y, &z) we should have x==1, y==6, and z==7.Write a C program that uses the following: a main() to read two integer values from the user, val1 and val2, and prints the returned value from swap().a swap() that uses call by reference (takes the addresses into pointers) to swap values, and prints their values after the swap "num1 = # and num2 = #". This function returns the largest of the two values. If these are equal, it returns their sum.
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