Reverse Polish converter

Question

I am trying to make a reverse Polish printer which can perform the following operation-

Inputs:

(a+(b*c))
((a+b)*(z+x))
((a+t)*((b+(a+c))^(c+d)))

Outputs:

abc*+
ab+zx+*
at+bac++cd+^*

This is my code:

#include <stdio.h>
#include <string.h>
char pop(int t);
void push(int c, int t);
int main()
{
    int z;
    scanf("%d", &z);
    char *a[100];
    int i = 0;
    int q = z;
    while (q-- > 0)
    {
        char v[400];
        scanf("%s", &v);
        int t;
        for (t = 0; t < strlen(v); t++)    //loop to put the values and signs in the 2 stacks
        {
            if ((v[t] == '*') || (v[t] == '+') || (v[t] == '-') || (v[t] == '^'))
            {
                push(v[t], 2);


            }

            else if (v[t] == ')')
            {
                int y = pop(2);
                push(y, 1);

            }

            else
            {
                push(v[t], 1);

            }
        }
        int k = 0;
        char c;
        while ((c = pop(1)) !='\0')    //loop to put elements in the array v
        {

            if (c != '(')
            {

                v[k++] = c;

            }
        }
        v[k--] = '\0';
        int m;
        for (m=0; m != k; m++, k--)     //for reversing the string
        {
            char t = v[m];
            v[m] = v[k];
            v[k] = t;
        }

        a[i++] =v;
        printf("%s",a[i - 1]);
    }
    int p;
    for (p = 0; p <z ; p++)   //printing the elements
        printf("%s\n",*a[p]);
    return 0;
}
char ac[400];
char as[400];
int ic = 0;
int is = 0;
void push(int c,int t)
{
    if (t == 1 && ic != 400)
        ac[ic++] = c;
    else if (t == 2 && is != 400)
        as[is++] = c;
}
char pop(int t)
{
    if (t == 1 && ic != 0)
        return ac[--ic];
    if (t == 2 && is != 0)
        return as[--is];
    return '\0';
}

But it is not even inputting properly and I am not able to figure out what are the mistakes in this code.Please help to figure out what are the problems.

What do you mean by "not inputting properly"? What's the input? What's the output? What's the desired output? — Tim Čas, Feb 11 '15 at 16:23
`printf("%s\n",*a[p]);` : `*a[p]` is `a char`. and `v` is out of scope. — BLUEPIXY, Feb 11 '15 at 16:38
That 2nd error pointed out by @BLUEPIXY would've been caught by most modern compilers, if warnings were enabled. I **strongly** recommend that you enable warnings (`-Wall` in GCC and CLang; check your compiler's manual if you have some other compiler). — Tim Čas, Feb 11 '15 at 16:48
@Tim Čas From not inputting properly I mean after inputing the no of test cases i.e.int z and first line if input it crashes — Alpa8, Feb 11 '15 at 17:13
@BLUEPIXY I couldn't understand what you mean by-printf("%s\n",*a[p]); : *a[p] is a char. and v is out of scope. — Alpa8, Feb 11 '15 at 17:16
`char *a[100];`: `a` is array of pointer to `char`. `a[p]` is pointer to `char`. `*a[p]` pointer dereference is `char`. — BLUEPIXY, Feb 11 '15 at 17:29
`while (q-- > 0) { char v[400];` : `v` is an effective area in the while block. If you see outside of the block is not valid. — BLUEPIXY, Feb 11 '15 at 17:35

score 0 · Answer 1 · answered Aug 14 '15 at 09:20

after inputing the no of test cases i.e.int z and first line if input it crashes

This is due to the

        printf("%s\n",*a[p]);

as BLUEPIXY noticed, *a[p] is a char; but %s expects a char *, thus you need

        printf("%s\n", a[p]);

and regarding v is out of scope, the crucial factor is not the scope (visibility), but the storage duration (lifetime) of v - its lifetime ends when execution of the block with which it is associated ends, and the value of a pointer a[i] to it becomes indeterminate; by changing

        a[i++] =v;

to

        a[i++] = strdup(v);

you can remedy that.

Reverse Polish converter

1 Answers1