Sum of 2 numbers with 9 digits

[Divit]

Hi! I'm fairly new to C. I wrote a simple c program, a calculator. But when I tried to add 123456789+123456789 it prints the answer as 246913584. I'm not sure what went wrong. Could someone help me?

Written in Notepad++ compiled in gcc

#include<stdio.h>
main()
{
float n1,n2,sum,sub,div,prod;
char sym,choice;
system("cls");
printf("This is a calculator:\n");
printf("\nEnter the expression:\n\n");
cal:
	scanf("%f%c%f",&n1,&sym,&n2);
	sum=n1+n2;
	sub=n1-n2;
	prod=n1*n2;
	div=n1/n2;
	if(sym=='+')
		{
		printf("\nSum = %.2f\n\n",sum);
		goto cal;
		}
	if(sym=='-')
		{
		printf("\nDifference = %.2f\n\n",sub);
		goto cal;
		}
	if(sym=='*')
		{
		printf("\nProduct = %.2f\n\n",prod);
		goto cal;
		}
	if(sym=='/')
		{
		printf("\nQuotient = %.2f\n\n",div);
		goto cal;
		}
	if(sym=='?')
		{
		getch();
		}
}

https://www.dropbox.com/s/z38tq1ylbz7qlpd/calc.png

[dmh2000]

you are using the type 'float' which has only 24 bits of precision. the numbers you are entering need more than that, so they get approximated which results in the inaccuracy. try your program using 'double' instead of 'float'.

[Adak]

If you are adding or subtracting integers, you should use an integral data type, instead of a floating point type. Also, you should check your MAX_INT in limits.h, and limit input to something that is within the range of the data type you're using. See what you have for extended range integers, also. You should have both signed and unsigned long long int's that will considerably extend the range for the int's.