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.