int main (int argc, **argv)
{
if (argv[1] == "-hello")
printf("True\n");
else
printf("False\n");
}
# ./myProg -hello False
Why? I realize strcmp(argv[1], "-hello") == 0 returns true... but why can't I use the equality operator to compare two C strings?
Because argv[1] (for instance) is actually a pointer to the string. So all you're doing is comparing pointers.
You can't compare strings in C with ==, because the C compiler does not really have a clue about strings beyond a string-literal.
The compiler sees a comparison with a char* on either side, so it does a pointer comparison (which compares the addresses stored in the pointers)
In C because, in most contexts, an array "decays into a pointer to its first element".
So, when you have the array "foobar" and use it in most contexts, it decays into a pointer:
if (name == "foobar") /* ... */; /* comparing name with a pointer */
What you want it to compare the contents of the array with something. You can do that manually
if ('p' == *("foobar")) /* ... */; /* false: 'p' != 'f' */
if ('m' == *("foobar"+1)) /* ... */; /* false: 'm' != 'o' */
if ('g' == *("foobar"+2)) /* ... */; /* false: 'g' != 'o' */
or automatically
if (strcmp(name, "foobar")) /* name is not "foobar" */;
Because there is no such thing as a C string.
In C, a string is usually an array of char, or a pointer to char (which is nearly the same). Comparing a pointer/array to a const array won't give the expected results.
UPDATE: what I meant by 'no C string' is, there is no string in C. What's usually referred to as a 'C string' is language independent (as 'Pascal string' is), it's the representation of strings as a null-terminated linear array of characters.
In C, string values (including string literals) are represented as arrays of char followed by a 0 terminator, and you cannot use the == operator to compare array contents; the language simply doesn't define the operation.
Except when it is the operand of either the sizeof or & operators, or when it is a string literal being used to initialize another array in a declaration, an expression with type "N-element array of T" will have its type implicitly converted (decay) to type "pointer to T", and the value of the expression will be the address of the first element of the array.
So when you write
if (argv[1] == "-hello")
the compiler implicitly converts the expression "-hello" from type "7-element array of char" to "pointer to char" (argv[1] is already a pointer type), and the value of the expression is the address of the character '-'. So what == winds up comparing are two pointer values, which are (most likely) never going to be equal since "-hello" and argv[1] (most likely) occupy different regions in memory.
This is why you have to use library functions like strcmp() to compare string values.
Because C strings don't exist as such. They are char arrays ending in a \0.
The equality operator == will test that the pointer to the first element of the array are the same. It won't compare lexicographically.
On the other hand "-hello" == "-hello" may return non zero, but that doesn't mean that the == operator compares lexicographically. That's due to other facts.
If you want to compare lexicographically, you can always do this:
#define STR_EQ(s1,s2) \
strcmp(s1,s2) == 0
I see that you tagged as C++. So you could do this:
std::string arg1 ( argv[1] );
if (arg1 == "-hello"){
// yeahh!!!
}
else{
//awwwww
}
Strings are not native types in C. What you are comparing in that example are two pointers. One to your first argument, and the other is a static character array with the contents of "-hello".
You really want to use strncmp or something similar.
When you're using ==, you're comparing pointers. That is, it will return true if the two operands refer to the same string in memory. Therefore, it's unsuitable for use in comparing strings lexicographically.
Because C strings are array of characters. Arrays are simply pointers to the first element in the array, and when you compare two pointers using == it compares the memory address they point to, not the values that they point to.
