This is the result after I run the code
it’s only appear the n2 instead of the sum of n1+n2
<code>#include <stdio.h>
int main()
{
char op,n1,n2;
printf("Enter the operation: ");
scanf("%c",&op);
switch(op)
{
case '+':
printf("Enter the number: ");
scanf("%d %d",&n1,&n2);
printf("result = %d ",n1+n2);
break;
case '-':
printf("result = %d ",n1-n2);
break;
case '*':
printf("result = %d ",n1*n2);
break;
case '/':
printf("result = %d ",n1/n2);
break;
default:
printf("Error: unknown operator!");
break;
}
return(0);
}
</code>
<code>#include <stdio.h>
int main()
{
char op,n1,n2;
printf("Enter the operation: ");
scanf("%c",&op);
switch(op)
{
case '+':
printf("Enter the number: ");
scanf("%d %d",&n1,&n2);
printf("result = %d ",n1+n2);
break;
case '-':
printf("result = %d ",n1-n2);
break;
case '*':
printf("result = %d ",n1*n2);
break;
case '/':
printf("result = %d ",n1/n2);
break;
default:
printf("Error: unknown operator!");
break;
}
return(0);
}
</code>
#include <stdio.h>
int main()
{
char op,n1,n2;
printf("Enter the operation: ");
scanf("%c",&op);
switch(op)
{
case '+':
printf("Enter the number: ");
scanf("%d %d",&n1,&n2);
printf("result = %d ",n1+n2);
break;
case '-':
printf("result = %d ",n1-n2);
break;
case '*':
printf("result = %d ",n1*n2);
break;
case '/':
printf("result = %d ",n1/n2);
break;
default:
printf("Error: unknown operator!");
break;
}
return(0);
}
Is there a way to fix this issue or alternative way?
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