/prog/ - doitfgts

Name: Anonymous 2011-03-19 2:25

Write a function in C that gets passed a pointer to a null-terminated string containing an arithmetic expression such as "3*(4+5)-1" and returns the correct value. It must be able to handle +,-,*,/, and parenthesis. No implied multiplication, no negative numbers, usual order of operations, and the passed string is guaranteed to have matched parenthesis and proper syntax.

Name: Anonymous 2011-03-19 4:48

Write your own eval, Jock-chan.

Name: Anonymous 2011-03-19 5:21

ENTERPRSIE QUALITY SOLUTION



char buf[1024];

sprintf(buf, "echo %s | bc >arse");

system(buf);

FILE* f = fopen("arse","r");

fread(buf, 1, 1024);

fclose(f);

printf("%s\n",buf);

Name: Anonymous 2011-03-19 5:22

-sprintf(buf, "echo %s | bc >arse");
+sprintf(buf, "echo %s | bc >arse", expression);

Name: Anonymous 2011-03-19 6:33

because boredom.



float asdf(const char *inp_t) {

    char *inp = strdup(inp_t);

    float *arr = (float *)malloc(strlen(inp_t) * sizeof(float));

    unsigned pos=0, c=0;



    while (inp[pos] != 0) {

        switch (inp[pos]) {

            case '+':case '-':case '*':case '/': 

                arr[c++] = inp[pos++];

                break;



            case '(': {

                int lpar = pos++;



                for (int t = 1; t > 0; pos++) 

                    t += (inp[pos]=='(')-(inp[pos]==')');



                inp[pos-1] = 0;

                arr[c++] = asdf(&(inp[lpar+1]));

                break;

            }

            default: {

                int len = strcspn(&(inp[pos]), "+-*/()");

                char tmp = inp[pos+len];

                inp[pos+len] = 0;

                arr[c++] = atoi(&(inp[pos]));

                inp[pos=pos+len] = tmp;

            }

        }

    }



    if (--c == 0) return arr[0];



    for (int i=1; i<=c-1; i+=2) {

        if (arr[i] == '-') arr[i+1] = -arr[i+1];



        if (arr[i] == '/' || arr[i] == '*') {

            arr[i+1] = arr[i-1]*((arr[i]=='/')/arr[i+1] + (arr[i]=='*')*arr[i+1]);

            arr[i-1] = 0;

        }

    }



    for (int i=c; i>0; i-=2) arr[i-2] += arr[i];



    return arr[0];

}

Name: Anonymous 2011-03-19 7:02

>>5
>arr[i+1] = arr[i-1]*((arr[i]=='/')/arr[i+1] + (arr[i]=='*')*arr[i+1]);

your code makes my eyes bleed.

Name: Anonymous 2011-03-19 7:45

>>1
Read K&R. It has this exact exercise.

Name: Anonymous 2011-03-19 7:52

>>7
You mean this one?

http://clc-wiki.net/wiki/K%26R2_solutions:Chapter_5:Exercise_10

That's for polish notation, much simpler.

Name: Anonymous 2011-03-19 11:04

>>8
polish my dick

Name: Anonymous 2011-03-19 11:10

>>8
polish notation infix notation >

Name: Anonymous 2011-03-19 11:46

>>10
polish notation < infix notation

Name: Anonymous 2011-03-19 11:52

>>11
Enjoy you're inferior.

Name: Anonymous 2011-03-19 12:08

>>12
you're = you are; (You are a dog -> You're a dog)
your = something that is yours; (Your dog barks)

Name: Anonymous 2011-03-19 12:23

yo'ure = autism; (Yo'ure a faggot)

Name: Anonymous 2011-03-19 12:31

>>13
Y = λf.(λx.f (x x)) (λx.f (x x)); (YHBT -> YHBT Have Been Trolled)

Name: Anonymous 2011-03-19 15:47

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doitfgts

1 Name: Anonymous 2011-03-19 2:25

2 Name: Anonymous 2011-03-19 4:48

3 Name: Anonymous 2011-03-19 5:21

4 Name: Anonymous 2011-03-19 5:22

5 Name: Anonymous 2011-03-19 6:33

6 Name: Anonymous 2011-03-19 7:02

7 Name: Anonymous 2011-03-19 7:45

8 Name: Anonymous 2011-03-19 7:52

9 Name: Anonymous 2011-03-19 11:04

10 Name: Anonymous 2011-03-19 11:10

11 Name: Anonymous 2011-03-19 11:46

12 Name: Anonymous 2011-03-19 11:52

13 Name: Anonymous 2011-03-19 12:08

14 Name: Anonymous 2011-03-19 12:23

15 Name: Anonymous 2011-03-19 12:31

16 Name: Anonymous 2011-03-19 15:47