how to ignore * function

[laxkrzy]

i am creating a basic calculator program for a class and and stuck on the multiplication. when the third argument is * it goes all funky and give a segmentation fault. my only guess is that this is because in the shell * is a wild card.

 
#include<stdio.h>
#include <stdlib.h>
#include<string.h>
int main(int argc, char *argv[]){
  int a;
  int b;
  int c;

  if  (argc  == 4 || (strcmp (argv[4], "dec") == 0 )){

  if (strcmp (argv[2], "+") == 0){
      a = atoi(argv[1]);
      b = atoi(argv[3]);
      c= a+b;
      printf("%d\n", c);}

  if (strcmp (argv[2], "-")==0){
    a = atoi(argv[1]);
    b = atoi(argv[3]);
    c= a-b;
    printf("%d\n", c);}

  if (strcmp (argv[2], "*")==0){
    a = atoi(argv[1]);
    b = atoi(argv[3]);
    c= a*b;
    printf("%d\n", c);}




}
  return 0;
}

our teacher warned us of this and said "Remember what * means in the shell. How do you protect it? " as a tip

how to i protect it

[Bayint Naung]

Use single quote.
./a.out ' 3 * 2 '

[CommonTater]

our teacher warned us of this and said "Remember what * means in the shell. How do you protect it? " as a tip

how to i protect it

Use the standard math symbol X for multiplication.

[KCfromNC]

Depends on the shell. If it's unix-ish, try ./a.out 3 \* 2. Or maybe ./a.out 3 '*' 2. You can't put the whole thing in one set of quotes because then it'll be lumped into argv[1], spaces and all.