LAB #2 (01/22/2025, 01/24/2025 and 01/27/2025)

Assignment (due 01/29/2025)

  1. (Variable Type) It is very important to use the correct type (int, float, char, double) for each variable defined/declared. If you mix different types of variables, the precedence is double > float > int=char.
    Compose the following program and run it. Discuss (a) the result (why it did not work as expected) and (b) how you can modify the code so that it prints the expected value. 
    #include <stdio.h>
int main()
{
int a=10, b=20;
float c;
c=a/b;
printf("Answer = %f \n",c);
return 0;
}
  2. (If statement) The syntax of the if statement is: 
    
if (.....) {statement 1; statement 2; statement 3;}
 else statement4;
    Note that if you want to execute multiple statements after if, you have to enclose them by { and }. else is optional above.
    Write a program to tell whether an integer entered from the keyboard is an even number or an odd number. Use the following as a template: 
    #include <stdio.h>
int main()
{
int a;

printf("Enter an integer ="); scanf("%d", &a);

if (a%3==0) printf("The number you entered is a multiple of 3.\n");
 else printf("The number you entered is NOT a multiple of 3.\n");

return 0;
}
    Note: a%m returns the remainder of a divided by m.
  3. A for loop is useful when you know in advance the number of iterations to be executed. The for loop consists of three parts each of which is separated by a semicolon (;). The first part is initialization of the variable for iteration, the second part is a test of the iteration variable and if this test fails, the loop is finished. The third part is a counter for the iteration variable. A block can be followed after the for statement (no semicolon !!) to execute multiple statements for each iteration. 
    for (initial value; condition ; counter increment) statement;
    The following program prints an integer value from 15 to 20. 
    #include <stdio.h>
int main()
{
int i;

for (i=5;i<11;i++) 
 printf("%d\n", i+10);

return 0;
}
    Write a code to print odd integers starting from 1 to 21.
    Hint: 
    int i;
for (i=0;i<=10;i++) print (2 times i + 1) .....
  4. It is possible to approximate the sum of an infinite series by using the for statement.
    The following code computes
    ∞
    ∑
    i=0 
    		 1
    2i
    		 = 1 + 1
    2
    		 + 1
    4
    		 + 1
    8
    		 + 1
    16
    		 + …
    which converges to 2. (proof) . Here is a code to do the above. 
    #include <stdio.h>
#include <math.h>
int main()
{
int i, n;
float  sum=0;

printf("Enter # of terms ="); scanf("%d", &n);

for (i=0; i<= n ; i++) sum = sum + 1/pow(2,i);

printf("Sum = %f \n",sum);
return 0;
}
    Write a code to compute the sum (should be close to 1.2)
    1 + 1
    23
    		 + 1
    33
    		 + 1
    43
    		 + 1
    53
    		 + …+ 1
    1003
    Note: pow(a,b) in math.h returns ab.


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On 15 Jan 2025, 20:21.