is there another way to do this instead of using strcpy?

[jaclynhi]

for (i=1; i<length(formula); i++)
{
strcpy(numString[j], portionofinput-1);
j++;
i++;
lastOp = i;
}

strcpy(numString[j], lastpartofinput);

how to do this instead of using strcpy?
how to get last part of input?
how to get portion of input?

input is: 1+1
should print: 2

 

[olded]

Hi:

I think you'd get more response if you posted all of your variable declarations.

What is &quot;length&quot;? is this a macro replacement for the strlen function?

Regards,

Ed

 

[jaclynhi]

below is my entire code...i cant think of how to change one part in double calArray(char* formula) to something else instead of using string.h.....could someone please help!!! i'm totally lost!!

if user input 1+2+4-3*5
output is 20

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

#include &quot;arithCalculation.h&quot;

#define TRUE 1
#define FALSE 0

void mainarithCalculation()
{
char formula[70];

printf(&quot;\n **********************************************************************\n&quot;
printf(&quot;\n\t\tWELCOME to Task 1: Basic Arithmetic Calculation \n&quot;
printf(&quot;\n **********************************************************************\n&quot;

while(TRUE)
{
/* get user input */
printf(&quot;\n\nPlease enter a formula:&quot;
fgets(formula, 70, stdin);
printf(&quot;Input: \t%s &quot;, formula);

/* if input is over maximum length*/
if (length(formula)>70)
{
printf(&quot;\nEquation length is more then 70!&quot;
continue;
}

/* if input is lesser then 3 in length*/
else if (length(formula)<3)
{
printf(&quot;\nEquation must have two numbers and an operator!&quot;
continue;
}

/* if input does not contain at least 2 values and operator and doesn't end with an operator */
else if (checkInput(formula))
{
printf(&quot;\nInvalid input!&quot;
continue;
}
else
{
returnResult(formula);
clrscr();
break;
}
}
}

/*check for errors
Operators will not be the last and first
There must be at least two numbers*/
int checkInput(char* formula)
{
char one, two;
int i;

/*check if first character is a operator*/
if (checkValidOperator(formula[0]))
{
return FALSE;
}

/*checks for operator errors*/
for(i = 1; i< length(formula); i++)
{
one = formula[i-1];
two = formula;
if (checkValidOperator(one) && checkValidOperator(two))
{
return FALSE;
}
}

/*check for valid characters*/
for (i = 0; i< length(formula); i++)
{
if(!checkValidNumber(formula) && !checkValidOperator(formula))
{
return FALSE;
}
}

/*check if last character is a operator*/
if(checkValidOperator(formula))
{
return FALSE;
}
return TRUE;
}

/*check if input is integer*/
int checkValidNumber(char valid)
{
switch(valid)
{
case '1':
case '2':
case '3':
case '4':
case '5':
case '6':
case '7':
case '8':
case '9':
case '0':
return TRUE;
default:
return FALSE;
}
}

int checkValidOperator(char op)
{
switch(op)
{
case '+':
case '-':
case '*':
case '/':
return TRUE;
default:
return FALSE;
}
}

/*check length of input and returns the length*/
int length(char* formula)
{
int i;
for(i = 0; formula!= '\0'; i++)
{
/* no code */
}
return i;
}

/*calculate the input formula*/
double calArray(char* formula)
{
double num1, num2;
char numString[70][20];
int lastOp = 0;
int i = 0;
int j = 0;
double result;

/*************do this without using string.h************/
for(i = 1; i < length(formula); i++)
{
if(checkValidOperator(formula)
{
/*strcpy(numstring[j], portion of formula - 1);
j++;
i++;
lastOp = i;
*/
}
}
/* strcpy(numstring[j], last part of formula);*/
/*********************************************************/



/* first number */
result = numstring[0];

/* calculate everything */
for(i = 0; i <= j-2; i+=2)
{
result = cal(result, atod(numstring[i+2]), numstring[i+1][0]);
}

return result;
}

double cal(double num1, double num2, char op)
{
double result;
switch(op)
{
case '+':
result = (num1 + num2);
break;
case '-':
result = (num1 - num2);
break;
case '*':
result = (num1 * num2);
break;
case '/':
if (num2 != 0)
{
result = (num1 / num2);
}
else
{
printf(&quot;Divide by 0 error!&quot;
exit(EXIT_FAILURE);
}
default :
printf(&quot;Illegal operator!&quot;
exit(EXIT_FAILURE);
}
return result;
}

/*to return result and print out*/
void returnResult(char* formula)
{
printf(&quot;\nResult: \t%6.2lf&quot;, calArray(formula));
}

void clrscr()
{
system(&quot;cls&quot;
}

/*start of program*/
void mainarithCalculation();

/*get result and return to main*/
void returnResult(char* formula);

/*check is user enter correct input*/
int checkInput(char* formula);

/*check if user entered an integer*/
int checkValidNumber(char valid);

/*check if user entered an operator*/
int checkValidOperator(char op);

/*check the length of the input*/
int length(char* formula);

/*calculate the input formula*/
double calArray(char* numString);
double cal(double num1, double num2, char op);

/*clear screen*/
void clrscr();

 

[olded]

/*
Hi:

Sorry, it's taken so long to get back to you.

First, of all, on my solaris 7 box, I couldn't get your
strcpy method to work. It core dumped.

Anyway, if you have a two-dimensional array like myString[70][20],
you can set individual characters of another string (like formula[] or
*formula) to the multi-dimensional array, but you have to
use both coordinates of the array. Below I'm placing individual
characters of formula to numString[1][0], numString[2][0], etc.

I've hacked up your code to show you what I mean:

Regards,

Ed

*/
#include <stdio.h>

void main()
{
int i;
int j=0;
/*char formula[]=&quot;1+4-6&quot;; */
char *formula=&quot;1+4-6&quot;;
char numString[70][20];

/* I'm starting at 0 and not 1 */
for (i=0; i<strlen(formula); i++)
{
numString[j][0] = formula;
j++;
}

for (i=0; i<strlen(formula); i++)
printf(&quot;%s\n&quot;, numString);
}

 

[Guest_imported]

thanks alot!! it work....