Forming a maths expression that references array elements

Hi guys can someone assist me, I'm quite new in C. I'm trying to write a maths expression of sort where the letters A,B,C,D,E are corresponding elements of an array row. I need to extract it as a string eg If I input y=A*B-C+D*E I wish the ouput string to be
5*3-2+3*4 for the first row. I've trying to do this but it's giving me a tough time.

Code:

---

#include<stdio.h>
#include<stdlib.h>
#define MAXCOLS 80
main()
{
int num_operand =5;
int num_operatr = 4;
int strglen=num_operand+num_operatr;
char operand;
int i, j;
char string[MAXCOLS];
char a[2][5]={{5,3,2,3,4},{2,5,4,6,8}};
printf("Enter a command of the sort A+B*C_D*E:\n");

for(j=0; j<strglen; j++){
operand=getchar();
if(operand >= 'A' && operand <='Z')
string[j]=a[0][(int)operand-(int)('A')];
else{
string[j]=(char)operand;}}

  printf("Expression\n");
  for(i=0; i<strglen; ++i){
  printf("%d \n", string[i]);
  getchar();}
 }

---

Perhaps some indentation would help, as well as not putting multiple closing braces on the same line.

Quote:

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Originally Posted by Salem

Perhaps some indentation would help, as well as not putting multiple closing braces on the same line.

---

Hi, Salem, I've tried it but it's not working, could you help me out by having a go at it?

You realize that saying "it's not working" is pointless. If it was working, you wouldn't be here posting about your broken code. Tell us the following in great detail:

• what you expect the code to do
• what the code is actually doing
• what lines are giving you trouble (you did debug where your problem is, right? ;) )
• whether the code compiles or not and if not, what error messages are being spit out

> I've tried it but it's not working
You have to take the pizza out of the box before putting it in the microwave.

Quote:

---

Originally Posted by MacGyver

You realize that saying "it's not working" is pointless. If it was working, you wouldn't be here posting about your broken code. Tell us the following in great detail:

• what you expect the code to do
• what the code is actually doing
• what lines are giving you trouble (you did debug where your problem is, right? ;) )
• whether the code compiles or not and if not, what error messages are being spit out

---

The code compiles, and when I run it if it encounters an operator it takes replaces it with the ASCII value of the operator. Like I said, I want to have the code perform, the following:
*If I enter A+B-C-D*E as an expression, where A,B,C,D,E references the array columns. This is done by a[j](int)operand-(int)('A')].
*I want to display the expression above as a string 5+3-2-3*4, where the letters have been replaced by their corresponding array elements. That's all I want to achieve.

I don't know whether my approach is wrong cos I'm new in C programming, just getting to know it.

This is what me mean by indentation and layout (red added).

Code:

---

#include<stdio.h>
#include<stdlib.h>
#define MAXCOLS 80

int main()
{
    int num_operand = 5;
    int num_operatr = 4;
    int strglen = num_operand + num_operatr;
    char operand;
    int i, j;
    char string[MAXCOLS];
    char a[2][5] = { {5, 3, 2, 3, 4}, {2, 5, 4, 6, 8} };

    printf("Enter a command of the sort A+B*C_D*E:\n");

    for (j = 0; j < strglen; j++) {
        operand = getchar();
        if (operand >= 'A' && operand <= 'Z')
            string[j] = a[0][(int) operand - (int) ('A')];
        else {
            string[j] = (char) operand;
        }
    }

    printf("Expression\n");
    for (i = 0; i < strglen; ++i) {
        printf("%d \n", string[i]);
        getchar();
    }

    return 0;
}

---

> I want to display the expression above as a string 5+3-2-3*4
Well the problem is, you only print using %d, so rather than getting '+' say, you get 43 instead (in this case, it's ASCII equivalent).
You need some extra code to do something different with the operands, especially when it comes to printing them.

Thanks for the indentation Salem. I've really toiled with this code, I don't know which other approach to adopt, I've tried all I could think of for now but all to no avail. I wish someone could assist.

Perhaps

Code:

---

char a[2][5] = { {'5', '3', '2', '3', '4'}, {'2', '5', '4', '6', '8'} };

---

and replace the final printf format %d with %c

[code]

Code:

---

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

#define MAXCOLS 80

int main()
{
    int num_operand =5;
    int num_operatr = 4;
    int strglen=num_operand+num_operatr;
    char operand;
    int i, j;
    char string[MAXCOLS];
    char getexp[strglen];
    char a[]={'5','3','2','3','4','2','5','4','6','8'};
               

    printf("Enter a command of the sort A+B*C_D*E:\n");
    fgets(getexp, sizeof getexp, stdin);

    for(j=0; getexp[j] != '\0'; j++)
    {
      operand=getexp[j];
     
      if(operand >= 'A' && operand <='Z')
          string[j]=a[operand-(int)('A')];
      else
          string[j]=(char)operand;
    }
    string[j] ='\0';

    printf("Expression\n");
   
    //for(i=0; i<strglen; ++i)
        printf("&#37;s", string);
   
    getchar();
    return 0;
 }
 /* my outout
 Enter a command of the sort A+B*C_D*E:
A+B+C
Expression
5+3+2
*/

---

ssharish2005

I would also change this:

Code:

---

      if(operand >= 'A' && operand <='Z')

---

to

Code:

---

      if(operand >= 'A' && operand <='E')

---

This is because anything larger than E will index outside the array.

If you really need to do this properly (with numbers bigger than single digit) you probably need to implement a better method of converting your "variable in the string" into a number - perhaps a structure like this:

Code:

---

enum {
  OP_PLUS,
  OP_MINUS,
  OP_MUL,
  OP_DIV,
  OP_END
} OPTYPE;

struct operation {
    int operand;
    enum OPTYPE op;
}; 

#define MAXOPERATIONS (some_number)
struct operation oplist[MAXOPERATIONS];

...

---

This sort of structure will allow you to translate a variable to "any" number (integer - but replace int with double in the struct operation and you would be almost unlimited).

--
Mats

Which post is this referring to? If it's mine, then that's quite clear - it's just a few datatypes, you'd need to modify your code to make it use those data-structures.

--
Mats

SSharish2005 thanks for your amendment. It's compiling but not giving any out put.

Hi Mats thanks but don't really understand how to implement it. To be honest I started learning c last week

Ok, let's forget about my suggestion at the moment - you'll probably get to "structures" at some future point in your learning.

I've just tested the code above, and aside from the compiler moaning about some things, I found that your strglen needs to be "one more" to hold the final "nul".

I get the following output:

Code:

---

Enter a command of the sort A+B*C-D*E:
A+B+C+D+E
Expression
5+3+2+3+4

---

from these changes:
add const on the constants in the beginning of the code (that just makes my compiler compile the code in the first place - you obviously have a different compiler than my Visual Studio VC7).

Code:

---

...
    const int num_operand = 5;
    const int num_operatr = 4;
    const int strglen=num_operand+num_operatr;
...
    char getexp[strglen+1];
...

---

--
Mats

Thanks Mats, it works well now for single digits but it doesn't accept double digits. Could you give me more hint of how to make it more general. I'm trying to read up the struct and enum you mentioned about

Thanks Mats it's working now for single digits, I've also added a code to reconvert the array back to numeric. Could you assist me to make it more general so it can accept double digits using the enum and struct you talked about. I'm equally trying to read it up.

structures is a way to "group data together".
for example:

Code:

---

struct employee {
  char name[25];
  char role[20];
  int yearofbirth;
  int salary;
};

 // You can then create an array of "employees":
struct employee staff[100];

---

structs are pretty similar to classes in C++ too - in fact, you can use struct and class ALMOST interchangably in C++ - there are subtle differences that make one better than the other in a particular situation, but that's best left for when you start learning about C++ - and there's much more important things to learn right now. [1]

enum is a way to "make a list of things that you can use to count/identify", such as you could declare a range of colours:

Code:

---

enum { violet, purple, red, yellow, green, blue, white, black } colours;

---

You can then use these in your code to make it much clearer what you mean. Essentially, the above enum is the same as defining constants for the individual colours and using those constants - just an easier way to do it.

[1] In general, which language you use is pretty unimportant. Knowing how to program is a generic skill, and whilst some languages are better for some things than others, most have pretty much the same common ways to operate, and learning how to program in GENERAL is more important than which particular language you do it in.

--
Mats

Ok thanks, I'll try to crack it.