Compare a specific array element to a string with "if" in C?

Question

Rather Trivial Question.

So I tried to do this:

if (array[0]=="some_string")

where array is a list of strings

array[0] = "some_string_1";
array[1] = "some_string_2";
array[2] = "some_string_3";

but obviously it doesn't work... What do I have to do?

N 1.1 · Accepted Answer · 2010-03-20 01:19:26Z

9

if(array[0] == 'n')

"" signifies string
'' signifies char

Note: 'a' is a char and "a" is a string, so 'a' != "a"

char is a single character (actually int), so if you want to compare strings, use strcmp instead.

edited Mar 20, 2010 at 1:19

answered Mar 20, 2010 at 1:09

N 1.1

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3 Comments

Matthew Iselin Over a year ago

" " actually signifies char*, but you're still right otherwise.

user124493 Over a year ago

..But you can only use strcmp() if the string is null-terminated, or if you know the array size of both strings to exceed a length n-1, you can us strncmp().

N 1.1 Over a year ago

@San: i cant explain everything in detail here :)

user124493 · Accepted Answer · 2010-03-20 01:15:36Z

0

in the example you give, you are comparing a single character (array[0]) with the address of a compiled-in string ("some_string").

Because a string literal is treated as a null-terminated character array, and comparisons against a character array with the == operator compare the address of the array.

The example you gave is essentially similar to this:

char* x = "some_string";
char array[10];
if(array[0] == x)
    ...

And you can see from this example that the types simply don't match. As another poster stated, you use the [] operator to obtain a specific character from the offset in the brackets from the start of the array.

answered Mar 20, 2010 at 1:15

user124493

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