True and false in C in if statement with getopt() - c

Question

I have a hard time in understatanding how true and false works with "if statement" when I am using argv & getopt.
This is the simple code:
#include <unistd.h>
#include <stdlib.h>
#include <stdio.h>
int main(int argc, char *argv[]) {
int opt;
while ((opt = getopt (argc, argv, "i:l:")) != -1)
switch (opt) {
case 'i':
printf("This is option i");
break;
case 'l':
printf("This is option l");
break;
default:
fprintf(stderr,"Usage: %s here goes usage\n",argv[0]);
}
if (argc == 1) {
printf("Without options");
}
if ((argc == 2) && (argv[1] != "-l") || (argv[1] != "-i")) {
printf("Without option -l or -i but with other argument \n");
printf("argv[1] is %s\n", argv[1]);
}
Usage:
./a.out foo
Output:
Without option -l or -i but with other argument
argv[1] is foo
It's good so far. Now let me check if it works when argv[1] is "-l":
Usage:
./a.out -l
Output:
./a.out: option requires an argument -- 'l'
Usage: ./a.out here goes usage
Without option -l or -i but with other argument
argv[1] is -l
Getopt works fine, but second information occurs even if argv[1] is -l and I set in "if" statement that (argv[1] != "-l"). Why it works like that? I have no clue.
Thanks for any answer.
B.

Answer 1

Do not use:
argv[1] != "-l"
to compare strings in C(a), this compares the addresses of two arrays of char(b) rather than the contents of those two strings. Instaed, you should be using:
strcmp(argv[1], "-l") == 0
(a) This is one of the things fixed in C++ strings, the equality operators there compare the content of the strings.
(b) This may sometimes seem to work with something like "xyzzy" == "xyzzy" but that's only because the compiler is allowed (but not required) to fold identical string constants together (so they take less room). It generally is not possible when one of the strings is not a constant (such as when it's passed on the command line).

Answer 2

You re comparing two char *s using != this compares the addresses.
(argv[1] != "-l")
The way to compare strings is using strcmp. Mind hat the strcmp function is "trinary", and returns either a negative or positive value if the strings are different (to provide order) or 0 if equal.
(strcmp(argv[1], "-l") == 0)

Answer 3

AND (&&) has precedence over OR (||), so I think your code inside the if statement is interpreted like this: if (((argc == 2) && (argv[1] != "-l")) || (argv[1] != "-i")) So the logic inside of it is always true, because even if your option IS -l and it's one and only (number of option != 2) it's also true that the option you typed is NOT -i.