c-program to implement a simple calculator

C-Program to implement a Simple Calculator

Developed by: Sendash Pangambam

For suggestions and queries be free to contact me at sendashpangambam@live.com

This program can be used as a simple calculator.

Just type the arithmetic expression you want to compute in the following format:

<Operand_1><Operator><Operand_2>

e.g. 1900+20145

Operators:

ADDITION + e.g. 1234+4321

SUBTRACTION - e.g. 9712-3421

MULTIPLICATION * e.g. 5621*245

DIVISION / e.g. 915/56

EXPONENTIAL ^ e.g. 1231^64

***Terms and conditions:

This program is written for calculations involving integers only.

Inconvenience is regretted.

For suggestions and queries be free to contact me at sendashpangambam@live.com

HOW MY PROGRAM WORKS?

1. A string of the format given above is supplied as input, e.g. 21900*914990

2. The two integers on the both sides of the operator, „*‟ in this case, are

extracted from the input string.

3. According to the operator in the input string, calculation is done.

VARIABLES USED IN THIS PROGRAM:

1. ch – To read any character from the keyboard.

2. exp – Input expression.

3. op – Operator.

4. fop, sop – 1st and 2nd operand.

5. tfop, tsop – Temporary 1st and 2nd operand, used during division.

6. ddif – Difference in digits of extracted integers and length of input string.

7. And some other variables that you have already known are also used.

For suggestions and queries be free to contact me at sendashpangambam@live.com

“LET‟S GO TO THE PROGRAM”

For suggestions and queries be free to contact me at sendashpangambam@live.com

/*Simple Calculator*/ #include<stdio.h> #include<conio.h> #include<math.h> #include<stdlib.h> #include<string.h> long int appzero(long int n,int d) /*Function to append zeroes to the right of a number*/ { int i; for(i=0;i<d;i++) n=n*10; return(n); } int ndgts(long int n) /*Function to count the number of digits in a given integer*/

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{ int i; for(i=1;n!=0;i++) n=n/10; return(i-1); } long int reverse(long int n) /*Function to find reverse of an integer*/ { long int rev=0; while(n>0) { rev=rev*10+(n%10); n=n/10; } return(rev); }

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void main() { char ch; clrscr(); /*SOME INSTRUCTION ABOUT THIS PROGRAM*/ printf("READ ME:\n\n\tThis program can be used as a simple calculator.\n"); printf("\n\tJust type the arithmetic expression you want to compute. \n \t

Syntax for input expression:\n \t\t\t\t < Operand_1 >< Operator >< Operand_2 >\n\t\t\t\t eg. 1900+20145");

printf("\nOperators:\n\tADDITION \t+\teg. 1234+4321 \n\tSUBTRACTION \t-\teg. 9712-3421 \n\tMULTIPLICATION \t*\teg. 5621*245 \n\tDIVISION \t/\teg. 915/56 \n\tEXPONENTIAL \t^\teg. 1231^64\n\n");

printf("=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+= +=+=+=+=+=+=+=+="); printf("\n\n***Terms and conditions:\n\t\tThis program is written for

calculations involving integers.\n\t\tDecimal calculations will be updated in the next versions.\n\t\tInconvinience is regretted.\n\n");

printf("=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=

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+=+=+=+=+=+=+=+=\n"); getch(); clrscr(); printf("DO YOU WISH TO CONTINUE ? (Y/N) "); ch=getch(); if(ch=='N'||ch=='n') { clrscr(); printf("PRESS ANY KEY TO EXIT "); getch(); exit(1); } clrscr(); printf("Let's try my simple calculator program"); getch(); while(1) { char *exp,op;

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long int fop,sop; double tfop,tsop; int ddif=0; clrscr(); printf("Enter the expression: "); gets(exp); fop=atol(exp); strrev(exp); sop=reverse(atol(exp)); ddif=strlen(exp) - ndgts(fop) - ndgts(sop); sop=appzero(sop,ddif-1); strrev(exp); if(exp[ndgts(fop)]=='+') { clrscr(); printf("\n%s = %ld",exp,fop+sop); } if(exp[ndgts(fop)]=='-')

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{ clrscr(); printf("\n%s = %ld",exp,fop-sop); } if(exp[ndgts(fop)]=='*') { clrscr(); printf("\n%s = %ld",exp,fop*sop); } if(exp[ndgts(fop)]=='/') { clrscr(); tfop=fop; tsop=sop; printf("\n%s = %lf",exp,tfop/tsop); } if(exp[ndgts(fop)]=='^') {

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clrscr(); tfop=fop; tsop=sop; printf("\n%s = %lf",exp,pow(tfop,tsop)); } getch(); clrscr();

printf("Do you have another calculation to do? (Y/N)\n\n*** For more help type 'H' \n\n\t\t\t");

ch=getch(); if(ch=='n'||ch=='N') { clrscr(); printf("\tThanks for your co-operation\n\tPress any key to exit"); getch(); exit(1); }

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if(ch=='h'||ch=='H') main(); } }

TRY THIS PROGRAM AND CHECK WHETHER IT WORKS OR NOT.