Determine the output generated by the following code. #include int main() { float x=1.0, *px; float y=1.5, *py; px=&x; printf("%.1f\n", *px); py=&y; printf("%.1f\n", *py); printf("%.1f\n", *px +*py); return 0;

Change this code into a program that receives a number and prints its divisors.( C language)   #include <stdio.h>   int main() { unsigned guess; /* current guess for prime */ unsigned factor; /* possible factor of guess */ unsigned limit; /* find primes up to this value */   printf("Find primes up to: "); scanf("%u", &limit);   printf("2\n"); /* treat first two primes as special case */ printf("3\n");   guess = 5; /* initial guess */ while ( guess <= limit ) { /* look for a factor of guess */ factor = 3; while ( factor*factor < guess && guess % factor != 0 ) factor += 2;if ( guess % factor != 0 ) printf("%d\n", guess); guess += 2; /* only look at odd numbers */ } return 0; }

int a = 5; int b = 6; int nain(void){ printf("%d Xd \n", a, b); int a; printf("d Xd \n", a, b); return e; When this program runs it prints (at least using gcc -std=c99 program.c -o program on Vera Virtual). 5 6; e 6; This shows that the compiler implements the second int a; by initializing the local variable a to zero. Now consider this program: int a = 5; int b = 6; int nain(void){ int a-a + b; printf("d Xd \n", a, b); return e; When this second program runs it prints, perhaps surprisingly: 6 6; Based on the first program, which a (local or global) is being used to compute a + b ? Finally, consider this third program: int a = 6; int b = 5; int nain(void){ b = a + a; a = b + b; int a = a + b; int b= a - b; printf("d\n", a + b); return e; Activate Windows What integer does this third program print to the console? (at least using gcc -std=c99 program.c -o