Lab 3 (C Programming) Solution
Feb 16, 2019programs written here are compiled and run in gcc (Ubuntu 7.3.0-27ubuntu1~18.04) 7.3.0
3) Given the three numbers a(=8), b(=4),c and constant value PI=3.1415, calculate and display the following result using macros (preprocessor directives).
a) c = PI * mult(a,b) /* the macro mult(a,b) perform the multiplication of a & b*/
b) c = PI * sum(a,b) /* the macro sum(a,b) perform the addition of a & b*/
c) c = PI * sub(a,b) /* the macro sub(a,b) perform the subtraction of a & b*/
d) c = PI * div(a,b) /* the macro div(a,b) perform the division of a & b*/Source Code:
#include <stdio.h>
#define PI 3.1416
#define mult(a,b) a * b
#define sum(a,b) a + b
#define sub(a,b) a - b
#define div(a,b) a / b
int main()
{
int a = 8, b = 4;
float mul = PI * mult(a, b);
float summ = PI * sum(a, b);
float subb = PI * sub(a, b);
float divv = PI * div(a, b);
printf("mul = %f\n", mul);
printf("sum = %f\n", summ);
printf("sub = %f\n", subb);
printf("div = %f\n", divv);
return 0;
}
Output:
mul = 100.531197 sum = 29.132799 sub = 21.132799 div = 6.283200
4) Write a program to add two numbers (5 and 7) and display its sum.
Source Code(using ASCII):
#include <stdio.h>
int main()
{
char ch;
printf("Enter the character :");
scanf(" %ch", &ch);
if(ch >= 48 && ch <=57){
printf("It is Number.");
}else if((ch >= 65 && ch <= 90) || (ch <= 97 && ch >= 122){
printf("It is Alphabet.");
}else{
printf("It is special.");
}
return 0;
}
Output:
Enter the character :D It is Alphabet.
Source Code(using builtin functions):
#include <stdio.h>
#include <ctype.h>
int main()
{
char ch;
printf("Enter the character:");
scanf(" %c", &ch);
if(isalnum(ch)){
printf("It is alphanumberic.\n");
}
if(isblank(ch)){
printf("Its is blank character.\n");
}
if(isalpha(ch)){
printf("It is alphabet.\n");
}
if(iscntrl(ch)){
printf("It is control character.\n");
}
if(isdigit(ch)){
printf("It is digit.\n");
}
if(isupper(ch)){
printf("It is Uppercase.\n");
}
return 0;
}
Output:
Enter the character:D It is alphanumberic. It is alphabet. It is Uppercase.