Write a C programme for arithmetic operations using menu driven?

Answer 1

#include<stdio.h>

void main()

{

int num1, num2,res;

char ch;

printf("\n Enter the 1st operand");

scanf("%d",&num1);

printf("\n Enter the operator");

scanf("%c",&ch);

printf("\n Enter the 2nd operand");

scanf("%d",&num2);

switch(ch)

{

case '+' :res=num1+num2; break;

case '-' :res=num1-num2; break;

case '*' :res=num1*num2; break;

case '/' :res=num1/num2; break;

}

printf("\n The result of %d %d %d is =%d",num1,ch,num2,res);

}

Answer 2

main()

{

int a,b,ch;

float result;

do

{

clrscr();

printf("select the operation from list");

printf("\n 1.ADDITION \n 2.SUBSTRACTION");

printf("\n 3.MULTIPLICATION \n 4.DIVISION");

printf("\n Enter your choice:");

scanf("%d",&ch);

}

while((ch>4)(ch<1));

switch(ch)

{

case1: printf("\n enter two numbers:");

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

result=a+b;

printf("\n result=%f",result);

break;

case2: printf("\n enter two numbers:");

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

result=a-b;

printf("\n result=%f",result);

break;

case3: printf("\n enter two numbers:");

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

result=a*b;

printf("\n result=%f",result);

break;

case4: printf("\n enter two numbers:");

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

result=a/b;

printf("\n result=%f",result);

break;

}

getch();

}

Answer 3

#include<stdio.h>

#include<conio.h>

void add()

{

int a=10,b=10,sum;

sum=a+b;

printf("%d",sum);

}

void main()

{

int ch;

clrscr();

printf("Enter your choice:-");

scanf("%d",&ch);

switch(ch)

{

case 1:

add();

break;

}

getch();

}

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