Which of the following protocols allows seemingly ill-matched computers to communicate with one another?
1 answer:
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Answer:
C code :
#include<stdio.h>
int main()
{
int j;
float k,l, x; //taking float to give the real results.
printf("Enter your two operands: "); //entering the numbers on which //the operation to be performed.
scanf("%f %f", &k, &l);
printf("\n Now enter the operation you want to do: ");
printf("1 for Addition, 2 for Subtraction, 3 for Multiplication, 4 for Division ");
scanf("%d", &j); //j takes the input for opearation.
switch(j)
{
case 1:
x=k+l;
printf("%.2f+%.2f=%.2f",k,l,x); //we write %.2f to get the result //upto 2 decimal point.
break;
case 2:
x=k-l;
printf("%.2f-%.2f=%.2f",k,l,x);
break;
case 3:
x=k*l;
printf("%.2f*%.2f=%.2f",k,l,x);
break;
case 4:
if(l!=0) //division is not possible if the denominator is 0.
{
x=k/l;
printf("%.2f/%.2f=%.2f",k,l,x);
}
else
printf("Division result is undefined");
default:
printf("\n invalid operation");
}
}
Output is in image.
Explanation:
At first we take two numbers.
Then we take integers from 1 to 4 for Addition, subtraction, multiplication, division respectively.
Then accordingly the case is followed and the operation is performed.
