converting numeric members of a string (type char) to type int in C: RPN interpreter

Question

I'm currently trying to create a Reverse-Polish notation interpreter in C, using a stack implemented using struct. It should take all single-digit values (0-9) and the operators +,-,* and /, and reject all others by exiting the program.

I am trying to capture the whole expression as it is typed in as a string, which has the type char, however when I use isdigit() it always returns a non-zero function (IE, it is not a digit) even when it appears to be to the user. I believe this is to do with the fact that the string is of type char, but I don't think I can use anything else, else I would get error messages when entering the operators.

The error is as follows: suppose I enter "11+" into the function. In the debug, I can see this appear in the watches. Advancing the program, I see that isdigit ()has returned 1 instead of 0, so the if statement conditions are fulfilled, and the program exits with exit(1); There is no specific error message provided by the IDE.

Is there a way to implement a conversion of just the "digits" of the string to type int, or do I have to do something else?

Here is the function. although it is still raw and unfinished, this shows the error:

void parseRPN(TopStack *st)
{
char Input[50];
int i;
do{
    printf("please enter an expression in single-digit integers" 
        "using Reverse Polish notation:");
    scanf("%s",&Input);
    if (sizeof(Input)/sizeof(int)>=50)
        {
            printf("that expression was too large for the RPN engine to handle!"
                "please break it down into smaller sub-tasks.\n");
            fflush(stdin);
            continue;
        }
    break;
}while(true);

for (i=0;i<50;i++)
    {
        int ErrorDetect=isdigit(Input[i]);
        if (ErrorDetect==0 && (Input[i]) != '+' || '-' || '*' || '/')
        {
            printf("Error: Invalid operand to RPN\nExiting...");
            exit(1);
        }
        else printf("great success!");
    }
}

Suggestions: use `strlen(Input)` instead of `sizeof(Input)/sizeof(int)` and then also use `for (i=0;i — J. Piquard, Nov 18 '16 at 22:54
thats a nice idea, I am using , so I'll add that in as well. that doesn't fix the problem though... — Thefoilist, Nov 18 '16 at 22:57
Yes, but please answer to my first comment... What error did you get ? — J. Piquard, Nov 18 '16 at 22:59
i've edited it into the question, 3rd paragraph, unless you need something even more specific? — Thefoilist, Nov 18 '16 at 23:01
I am not sure I follow. isDigit is supposed to return non-zero for digits and zero for non-digits. So for the first '1' in "11+", isDigit is supposed to return 1. Or do you mean it returns 1 for the '+'. — sep, Nov 18 '16 at 23:08
As far as I understand, it scans the first element of the string, which is 1. this is then used as an argument to isdigit, which returns 1 (it returns 0 only for digits). I believe it does this due to the fact that the string is really an array of type char, so the 1st element "1" will be interpreted as char 1; instead of int 1. It _should_ scan from left to right... — Thefoilist, Nov 18 '16 at 23:12
isDigit returns 1 (true) for digits http://www.programiz.com/c-programming/library-function/ctype.h/isdigit — sep, Nov 18 '16 at 23:17
In fact, the if-condition `(ErrorDetect==0&&(Input[i])!='+'||'-'||'*'||'/')` doesn't do what you want and ErrorDetect could be any value to have a true condition. Add bracket as follow `((ErrorDetect==0) && ((Input[i])!='+'||'-'||'*'||'/'))`. And work on operators. — J. Piquard, Nov 18 '16 at 23:17

score 2 · Answer 1 · edited May 23 '17 at 11:59

2

Certainly unintended use of ||
@J. Piquard also comments to that

// if (ErrorDetect==0 && (Input[i]) != '+' || '-' || '*' || '/')
if (ErrorDetect==0 && Input[i] != '+' || Input[i] != '-' || 
    Input[i] != '*' || Input[i] != '/')

Or something like that. Suggest adding () for clarity.. I think OP wants different logic.

if (ErrorDetect || (Input[i] != '+' && Input[i] != '-' && 
    Input[i] != '*' && Input[i] != '/'))

Other coding issue may exist, such as:

char Input[50];
// scanf("%s",&Input);
scanf("%49s",&Input);

edited May 23 '17 at 11:59

Community

1
1

answered Nov 18 '16 at 23:18

chux - Reinstate Monica

143,097
13
135
256

Thanks! this fixes the digit interpretation, but it does not fix the operator interpretation, it still exits the program when I input 11+, only this time on the "+". I have tried adding parenthesis for clarity in places, to no avail :/ the new statement is `if ((ErrorDetect==0) && ((Input[i] != '+') || (Input[i] != '-') || (Input[i] != '*') || (Input[i] != '/')))` – Thefoilist Nov 18 '16 at 23:34
@jamesharris Look at `for (i=0;i<50;i++)` why 50, why not stop at `'\0'`? Your code declares the null character an error. – chux - Reinstate Monica Nov 18 '16 at 23:38
I have since changed that code to `for (i=0;i – Thefoilist Nov 18 '16 at 23:43
`'\0'`, not `'/0'`, is an escape sequence that denote a _character constant_ with the value of 0, or the _null character_. All C _strings_ end in a _null character_. – chux - Reinstate Monica Nov 18 '16 at 23:49
Avoid `for (i=0;i – chux - Reinstate Monica Nov 18 '16 at 23:53
so I should be looking for the first NULL value, denoted by `'\0'` and stopping there in order to save processor cycles? – Thefoilist Nov 18 '16 at 23:57
Yes. BTW: In C `NULL` is used in a pointer context. E.g. `int *ptr; ... if (ptr == NULL) ...`. Use _null character_ or `'\0'` to convey that zero at the end of the _string_. – chux - Reinstate Monica Nov 19 '16 at 00:03
using `for (i=0;i=='\0';i++)` now yields 1 iteration of the "great success!", so something is indeed happening now. however it does not yield a 2nd or 3rd iteration for a 3 character input. – Thefoilist Nov 19 '16 at 00:07
Try `for (i=0; Input[i] != '\0';i++)` or the equivalent `for (i=0; Input[i]; i++)` Good luck, TTFN – chux - Reinstate Monica Nov 19 '16 at 00:08

converting numeric members of a string (type char) to type int in C: RPN interpreter

1 Answers1