Segmentation fault with strcmp()

Question

if(strcmp(argv[2], NULL) == 0)

I'm passing 3 command line arguments but I also want to run it with only 2 command line arguments with the above statement. But a segmentation fault error is being displayed.

I also tried with

if(argc < 3)

but it also didn't work...same segmentation fault...

Answer 1

Why segmentation fault?

Because of code if(strcmp(argv[2], NULL) == 0) , you are passing NULL as string pointer to strcmp() function; that try to deference at NULL to compare chars codes (eg acsii code) this cause undefined behavior at run time.

You should compare a string pointer with NULL using == as if(argv[2] == NULL)

I'm passing 3 command line arguments but I also want to run it with only 2 command line arguments with the above statement.

You can implement this in two ways:

1. The main syntax is:

    int main(int argc, char* argv[]) 

   The first argument argc is argument counter that is total number of arguments passed to your process including process name.

   So when you pass no extra argument then argc == 1 eg ./exe

   Suppose if you pass three arguments as follows:

    ./exe firstname lastname 

   Then argc == 3 , it looks like you are passing two arguments but including executable name you are actually passing three arguments to process.

   So you can use of argc value to iterate in a loop to print arguments passed (other then executable)

     printf("Process name is: %s", argv[0]); for(i = 1; i < argc; i++ ){ printf("argv[%d] %s\\n", argv[i]); } 

2. Second technique is using second argument: argv[] is NULL terminated array of string strings so argv[argc] is always equals to NULL. You can use this information in loop to iterate and process of arguments passed.

   To understand this suppose you are executing function as:

    ./exe firstname lastname 

   then argv[0] == ./exe , argv[1] == firstname and argv[2] == lastname and argv[3] == NULL , Note this time argc == 3 ( argv[argc] means argv[3] == NULL).

   For example to print all arguments, you can write you code like:

     int i = 1; printf("Process name is: %s", argv[0]); while(argv[i]){// terminates when argv[i] == NULL printf("argv[%d] %s\\n", argv[i]); i++; } 

Do you notice argv[0] is always your executable name! this means whenever you need to print your executable name use argv[0] instead of hard code name of your executable while writing code, so that if you recompile and give new name to your executable then argv[0] always prints correct name. You should write code as follows:

int main(int argc, char* argv[]){
  :
  :// some other code
  if(argc < min_number_of_arguments){
      fprintf(stderr, "Error: wrong number of arguments passed!\n");
      fprintf(stderr, "Usage: %s [first] [second] \n", argv[0]);
      exit(EXIT_FAILURE);
  }
  :
  :// some other code 
   return EXIT_SUCCESS;
}

Answer 2

You can't use strcmp() to compare to NULL . Neither argument can be null. In this situation it doesn't make sense anyway. If the argument isn't present, argc will be < 3, and if it is somehow empty it will be zero length. Never null.

Answer 3

Firstly, you shall always use strcmp(some_string, "") instead of strcmp(some_string, NULL) to check if a string is empty.

However in your problem you shall test

if (argc < 4)

That's because the executable itself is also in the array argv . Consider you're invoking something like ./a.out param0 param1 , then argc would be 3 and argv[0]="./a.out" , argv[1]="param0" , argv[2]="param1" .

EDITED:

Also, never test if(strcmp(argv[2], NULL) == 0) directly. Always test argc first. Since there is no grantee that what value would be stored in argv[argc+n] for n >= 0