C代码中if＆switch语句中的意外输出

Question

I cant figure out why either of the following code fails to operate as expected. There both compiled into execution files.

Outputs:

a.out  , prints 1, expected "no value"
a.out 1, prints 2, expected 1 
a.out 2, prints 2, expected 2

Using a case:

void main(int in)
{
 int a = in ;
 printf("In function if\n");
 if ( in == 1 )
   printf("1\n");
 else
   if ( in == 2)
     printf("2\n");
   else
     printf("wrong value\n");
}

Using a switch:

void main(int in)
{
  switch( in )
    {
    case  1: printf("1\n");                 break;
    case  2: printf("2\n");                 break;
    default: printf("wrong value\n"); break;
    }
};

I'm trying to get the following LISP functionality in C code:

(cond ((= in 1) 1)
      ((= in 2) 2)
      (t        nil))

Thank you for your assistance.

Answer 1

main doesn't accept the input from the command line as direct arguments, you are getting the argument count in there which is 1 if there are no arguments, and 2 if there is one argument, which causes the strange behavior.

main should be defined as int main( int argc, char *argv[] ) or something similar. To get the input, you need to first check if it exists by testing argc (the argument count, plus one for the executable path), and then converting argv[1] to an integer. atoi can be used to convert a string to an integer.

Answer 2

Note that the first parameter to main is argc , the total count of app name + parameters given on the command line. There is a second argument on main , char *argv[] for receiving the command line parameters.

The reason you get the behaviour described is because your parameter in replaces the purpose of argc . ie when you execute the App with no command line params, the argc count is 1, with one param, it will be 2, etc - this is the value passed to in . Since you don't have a second main parameter for the command line parameters ( argv ), you won't receive the actual parameters at all.

To fix this, you'll need to parse the second args array for your in integer parameter:

#include <stdlib.h>

int main(int argc, char *argv[]) {
    if (argc < 2)
       return EXIT_FAILURE; // ... print out the correct command line usage to user

    char *endPointer = NULL;
    long in;

    in = strtol(argv[1], &endPointer, 10);

    if (endPointer != NULL) {
      switch( in ) {
         case  1: 
            // ... same code as above
        }
       return EXIT_SUCCESS;
    }
    return EXIT_FAILURE; // User hasn't provided a number for 1st param
}

Ide One Sample Here