C++ comparing c string problem

Question

I have written the following code which will does not work but the second snippet will when I change it.

int main( int argc, char *argv[] )
{
  if( argv[ 1 ] == "-i" )   //This is what does not work
     //Do Something
}

But if I write the code like so this will work.

int main( int argc, char *argv[] )
{
  string opti = "-i";

  if( argv[ 1 ] == opti )   //This is what does work
     //Do Something
}

Is it because the string class has == as an overloaded member and hence can perform this action?

Thanks in advanced.

Answer 1

Is it because the string class has == as an overloaded member and hence can perform this action?

You are correct. Regular values of type char * do not have overloaded operators. To compare C strings,

if (strcmp(argv[1], "-i") == 0) {
    ...
}

By comparing the strings the way you did (with == directly), you are comparing the values of the pointers. Since "-i" is a compile time constant and argv[1] is something else, they will never be equal.

Answer 2

Correct. When you do argv[1] == "-i" (using == on two char* expressions) you're comparing the pointers for equality, not the characters in the string. You have to use the strcmp function to compare the string data.

std::string has overloaded operators to check string==string, char*==string, and string==char*, but it's impossible to overload char*==char* since that already has a defined meaning: comparing the pointers.

Answer 3

This one will work fine in C. Remember to check argc so you won't get seg fault. There is also strncmp() which is a safer way to compare strings as you limit the max length of the strings.

#include <string.h>
#include <stdio.h>

int main( int argc, char *argv[] )
{
  if((argc > 1) && (strcmp(argv[1], "-i") == 0))
  {
      printf("got -i\n");
  }
}

If you are passing multiple params as command options check out getopt()