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How to solve the expression (a/b)+(c/d) using typecasting in c

C-and-Cpp Increment-And-Decrement-Operators 11-12 min 20-30 sec 16-10-20, 11:25 p.m. Ammu15

Answers:

abc

17-10-20, 12:02 p.m. VarshaChaprana

Means what

17-10-20, 10:31 p.m. Ammu15

That's only assignment.

20-06-21, 10:18 p.m. Sujan_A_T

#include <stdio.h>

int main() {

int a, b, c, d;

printf("Enter a and b: ");

scanf("%d %d", &a, &b);

printf("Enter c and d: ");

scanf("%d %d", &c, &d);

float ans = (a/b)+(c/d);

printf("Answer(without typecast) = %f\n", ans);

ans = ((float)a/b)+((float)c/d);

printf("Answer(with typecast) = %f\n", ans);

return 0;

}

Tell if there is wrong.

20-06-21, 10:19 p.m. Sujan_A_T

<font size="4">Program in c to solve the expression (a/b)+(c/d):</font>

#include <stdio.h>

int main()

{

int a,b,c,d;

float x,y,z;

printf("Enter the values of a,b,c and d:");

scanf("%d%d%d%d",&a,&b,&c,&d);

x=a/b;

y=c/d;

z=x+y;

printf("Value of x is %.2f\n",z);

x=(float)a/b;

y=(float)c/d;

z=(float)x+y;

printf("Value of x is %.2f\n",z);

return 0;

}

output:

Enter the values of a,b,c and d:3

Value of x is 1.00

Value of x is 1.75

22-01-24, 4:20 p.m. Vaishnavi2885

#include<stdio.h>

#include<conio.h>

void main()

{
int a,b,c,d;

float x,y,z;

printf("enter the a,b,c,d value");

scanf("%d%d%d%d",&a&b&c&d);

x=a/b;

y=c/d;

z=x+y;

printf("value of z is %2f",z);

getch();

}

12-08-24, 3:01 p.m. 238T1A05D8

#include <stdio.h>

int main()

{

int a,b,c,d;

float x,y,z;

printf("Enter the values of a,b,c and d:");

scanf("%d%d%d%d",&a,&b,&c,&d);

x=a/b;

y=c/d;

z=x+y;

printf("Value of x is %.2f\n",z);

x=(float)a/b;

y=(float)c/d;

z=(float)x+y;

printf("Value of x is %.2f\n",z);

return 0;

}

output:

Enter the values of a,b,c and d:3

Value of x is 1.00

Value of x is 1.75

12-08-24, 3:01 p.m. 238T1A05D3

#include<stdio.h>

void main()

{

int a,b,c,d,x,y;

float ans;

printf("Enter a and b values:");

scanf("%d %d",&a,&b);

printf("Enter c and d value:");

scanf(%d %d",&c,&d);

x=(float)a/b;

y=(float)c/d;

ans=x+y;

printf("The final answer is %d",ans);

getch();

}

12-08-24, 4:12 p.m. ajayrahulpulli@gmail.com

#include <stdio.h>

int main()

{

int a,b,c,d;

float x,y,z;

printf("Enter the values of a,b,c and d:");

scanf("%d%d%d%d",&a,&b,&c,&d);

x=a/b;

y=c/d;

z=x+y;

printf("Value of x is %.2f\n",z);

x=(float)a/b;

y=(float)c/d;

z=(float)x+y;

printf("Value of x is %.2f\n",z);

return 0;

}

16-08-24, 4:53 p.m. 238T1A05F5

/* program to solve the following expression: (a/b) + (c/d) The values of a, b, c and d are taken as input from the user.using typecasting to perform real division*/

#include <stdio.h>

int main()

{

int a,b,c,d;

float x,y,z;

printf("Enter the values of a,b,c and d:");

scanf("%d%d%d%d",&a,&b,&c,&d);

x=a/b;

y=c/d;

z=x+y;

printf("Value of x is %.2f\n",z);

x=(float)a/b;

y=(float)c/d;

z=(float)x+y;

printf("Value of x is %.2f\n",z);

return 0;

}

19-08-24, 2:49 p.m. 238T1A05G1

#include <stdio.h>

int main()

{

int a,b,c,d;

float x,y,z;

printf("Enter the values of a,b,c and d:");

scanf("%d%d%d%d",&a,&b,&c,&d);

x=a/b;

y=c/d;

z=x+y;

printf("Value of x is %.2f\n",z);

x=(float)a/b;

y=(float)c/d;

z=(float)x+y;

printf("Value of x is %.2f\n",z);

return 0;

}

19-08-24, 3:04 p.m. nasreensk2005@gmail.com

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