Rather Trivial Question.
So I tried to do this:
if(array[0]=="some_string")
but obviously it doesn't work... What do I have to do?
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184answers:
3
+4
A:
if(array[0] == 'n')
""signifiesstring
''signifieschar
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.
N 1.1
2010-03-20 01:09:40
" " actually signifies char*, but you're still right otherwise.
Matthew Iselin
2010-03-20 01:41:53
..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().
San Jacinto
2010-03-20 14:33:53
@San: i cant explain everything in detail here :)
N 1.1
2010-03-20 14:38:03
A:
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.
San Jacinto
2010-03-20 01:15:36
A:
What was the solution to this problem? I have a similar problem, except I am actually comparing a string. I have an array of strings with around 8 elements. I want to check the first element for equality with a string. Eg.
if (array[0] == "ReadMemory") // do stuff
strcmp doesn't work for me here as sometimes there won't be a value in array[0]. I've tried to first check array[0] for a null value, but that doesn't work either. Any suggestions?
ftl25
2010-08-04 10:33:59