C Exercises: Display multiple variables
C Basic Declarations and Expressions: Exercise-7 with Solution
Write a C program to display following variables.
a+ c, x + c,dx + x, ((int) dx) + ax, a + x, s + b, ax + b, s + c, ax + c, ax + ux
Variable declaration :
int a = 125, b = 12345;
long ax = 1234567890;
short s = 4043;
float x = 2.13459;
double dx = 1.1415927;
char c = 'W';
unsigned long ux = 2541567890;
C Code:
#include <stdio.h>
int main()
{
int a = 125, b = 12345; /* Declare and initialize integer variables */
long ax = 1234567890; /* Declare and initialize long integer variable */
short s = 4043; /* Declare and initialize short integer variable */
float x = 2.13459; /* Declare and initialize floating-point variable */
double dx = 1.1415927; /* Declare and initialize double precision variable */
char c = 'W'; /* Declare and initialize character variable */
unsigned long ux = 2541567890; /* Declare and initialize unsigned long integer variable */
/* Various arithmetic operations and type conversions */
printf("a + c = %d\n", a + c);
printf("x + c = %f\n", x + c);
printf("dx + x = %f\n", dx + x);
printf("((int) dx) + ax = %ld\n", ((int) dx) + ax);
printf("a + x = %f\n", a + x);
printf("s + b = %d\n", s + b);
printf("ax + b = %ld\n", ax + b);
printf("s + c = %hd\n", s + c);
printf("ax + c = %ld\n", ax + c);
printf("ax + ux = %lu\n", ax + ux);
return 0;
}
Sample Output:
a + c = 212 x + c = 89.134590 dx + x = 3.276183 ((int) dx) + ax = 1234567891 a + x = 127.134590 s + b = 16388 ax + b = 1234580235 s + c = 4130 ax + c = 1234567977 ax + ux = 3776135780
Flowchart:
C Programming Code Editor:
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