Simple Command Line Calculator

I've found it helps in places where the integer-only 'eval' command doesn't meet the need.

Cut and paste the following code into a file called 'calc.c' then compile it with something like...

There's lots of improvements and additions you can make to this which can turn it into a very useful tool.

#include <stdio.h>
#include <stdlib.h>

usage()
{
printf("\nHarvey's simple floating point calculator\n");
printf("-----------------------------------------\n\n");
printf("Enter your query on the command line in the form...\n\n");
printf(" calc <num1> <op> <num2> (note the spaces)\n\n");
printf("where num1 and num2 are numbers and 'op' may be...\n");
printf(" + adds the two numbers\n");
printf(" - Subtracts num2 from num1\n");
printf(" x multiplies the two numbers\n");
printf(" / divides num1 by num2\n");
printf(" %% finds the remainder from dividing num1 by num2");
printf(" ^ raises num1 to the power of num2\n\n");
printf("calc will return the result as a floating point number.\n\n");
}

/* =============================================================== */
/* Gets one word (delimited by a space) from the standard input... */
/* =============================================================== */
char *getword()
{
static char word[40];
int wordlength = 0;
int thischar;

/* ------------------------------- */
/* Strip off any leading spaces... */
/* ------------------------------- */
while((thischar = getchar()) != EOF
   && isascii(thischar)
   && isspace(thischar));
/* ------------------- */
/* Now get one word... */
/* ------------------- */
while(thischar != EOF
        && isascii(thischar)
&& !isspace(thischar))
{
if(wordlength < (sizeof(word) - 1))
word[wordlength++] = thischar;
thischar = getchar();
}
word[wordlength] = '0';
return(wordlength == 0 && thischar == EOF ? NULL : word);
}

/* ================================= */
/* This is where it really starts... */
/* ================================= */
main(argc, argv)
int argc;
char *argv[];
{
float operand1, operand2;
float result = 0;
char operation;
int counter;
char *first = (char *)malloc(40);
char *second = (char *)malloc(40);
long longop1, longop2;

/* ----------------------------------------- */
/* Have we got two operands and an operator? */
/* ----------------------------------------- */
if(argc < 4)
{
/* ------------------------------------------------------- */
/* NO: Try reading from stdin in case num1 is in a pipe... */
/* ------------------------------------------------------- */
if(argc > 2)
{
free(first);
if((first = getword()) != NULL)
{
operation = (char)*argv[1];
sprintf(second, "%s", argv[2]);
}
else
{
usage();
exit(1);
}
}
else
{
usage();
exit(1);
}
}
else
{
/* --------------------------------------- */
/* 3 parameters: assume 1st is a number... */
/* --------------------------------------- */
sprintf(first, "%s", argv[1]);
operation = (char)*argv[2];
sprintf(second, "%s", argv[3]);
}

/* ------------------------------------------- */
/* Put the parameters into usable variables... */
/* ------------------------------------------- */
operand1 = atof(first);
operand2 = atof(second);

/* Perform the calculation */
switch(operation)
{
case '+':
result = operand1 + operand2;
break;

case '-':
result = operand1 - operand2;
break;

case 'x':
result = operand1 * operand2;
break;

case '/':
if (operand2 != 0) /* Notice the error-checking code. */
result = operand1 / operand2;
else
printf("Divide by 0 error!\n");
break;

case '%':
if (operand2 != 0)
{
longop1 = (long)operand1;
longop2 = (long)operand2;
result = (float)(longop1 % longop2);
}
else
{
printf("Divide by 0 error!\n");
}
break;

case '^':
result = operand1;
longop2 = (long)operand2;
for (counter = 2; counter < longop2 + 1; counter++)
result = operand1 * result;
break;

default:
printf("Invalid operation!\n");
break;
}

/* Print out the result */
printf("%f\n", result);