COMP1911 23T2 Introduction to Programming
  1. Given these declarations 
    int     n;
int    *p, *q;
    What will happen when each of the following statements are executed (in order)? 
    p = &n;
*p =  5;
*q = 17;
q =  p;
*q =  8;

    p = &n;   // p will point to n
*p =  5;  // 5 will be stored into n
*q = 17;  // the program will attempt to store 17 into the address given by q,
          // which could cause an error because q has not been initialized.
          // dcc will give us a warning about this
q =  p;   // q will point to the same address that p does, ie n
*q =  8;  // 8 will be stored into n
  2. We haven't used scanf to it's full potential yet. What do you think the following program will output? 
    int main(void) {
    int scanfResult;
    int a, b, c;
    scanfResult = scanf("%d %d %d", &a, &b, &c);
    printf("%d %d %d %d", scanfResult, a, b, c);
}
    scanf returns the number of items that were successfully read in. If we want to test if "more" than expected numbers were passed in, we need to create a while loop.
  3. Consider the following code 
    void swap (int x, int y);

int main(void) {
    int x = 10;
    int y = 99;
    
    printf("%d %d\n",x,y);
    
    swap(x,y);
    
    printf("%d %d\n",x,y);
    

    return EXIT_SUCCESS;
}

void swap (int x, int y) {
   int tmp;
   tmp = x;
   x = y;
   y = tmp;
}
    1. What will the output be and why? 
      The program passes copies of the values in x and y into the function, 
so any changes made in the function are only made on the local copies. 
x and y in the main function remain unchanged. So the output is.
10 99
10 99
    2. Modify the code so that the function prototype is void swap (int *x, int *y); 
      void swap ( int *x, int *y );

int main(void){
    int x = 10;
    int y = 99;
    
    printf("%d %d\n",x,y);
    
    swap(&x,&y);
    
    printf("%d %d\n",x,y);
    

    return EXIT_SUCCESS;
}


void swap (int *x, int *y){
   int tmp;
   tmp = *x;
   *x = *y;
   *y = tmp;
}
    3. What will the output of the modified code be and why? 
      The program passes addresses of the variables in x and y into the function, 
so any changes made in the function are made by going to the addresses of  the original x and y variables and modifying the values in them. 
This means that the values x and y in the main function are changed. 
10 99
99 10
  4. Consider: 

    float ff[] = {1.1, 2.2, 3.3, 4.4, 5.5, 6.6};
float *fp = ff;

    What are the similarities between ff and fp? What are the differences?

    They are both pointers to float. They both point to the beginning of the float array ff. They can both be used to access the array.
    The difference is that ff is a constant pointer, defined to always point to the beginning of the array, while fp is variable and can be made to point somewhere else.
  5. What would be the output of the following code? 
    int x = -9;
int *p1 = &x;
int *p2;

p2 = p1;
printf("%d\n", *p2);
*p2 = 10;
printf("%d\n",x);

    -9
10
  6. Write a function with the prototype below that calculates the sum and the product of all the integers in the given array. 
    void sumProd(int nums[], int len, int *sum, int *product);
    Sample solution (whole program) 
    #include <stdio.h>

void sumProd(int nums[], int len, int *sum, int *product);

int main(int argc, char *argv[]){
   int nums[] = {3,4,1,5,6,1};
   int prod;
   int sum;

   //Pass in the address of the sum and product variables
   sumProd(nums, 6, &sum, &prod);

   printf("The sum is %d and prod is %d\n",sum,prod);
   return 0;
}


// Calculates the sum and product of the array nums.
// Actually modifies the  variables that *sum and *product are pointing to
void sumProd(int nums[], int len, int *sum, int *product) {
    int i;
    *sum = 0;
    *product = 1;
    for (i = 0; i < len; i = i + 1) {
        *sum = *sum + nums[i];
        *product = *product * nums[i];
    }
}