Possible Duplicate:
In C arrays why is this true? a[5] == 5[a]
3["zdvnngfgnfg"];
views:
290answers:
3
+8
A:
It's equivalent to
"zdvnngfgnfg"[3];
which is legal and means "take the address of that literal and add 3*sizeof(char) to it". Will have no effect anyway.
Also see this very similar question.
sharptooth
2009-09-21 09:45:37
Doesn't the meaning also include "and then fetch the value at that address"?
unwind
2009-09-21 11:34:09
I suppose no, since the statement is neither on the left nor on the right side of the assignment.
sharptooth
2009-09-21 11:47:56
Since the snippet is a full expression, it does in fact mean "and then fetch the value". But because doing so has no defined side-effects, and the result is discarded, the compiler is allowed to omit it. In fact the compiler will probably omit the pointer addition as well - all it will do is make sure the statement is legal, realise it does nothing, and elide it. Certainly that's what gcc does, even with no optimisation.
Steve Jessop
2009-09-21 11:59:02
Johannes Schaub - litb
2009-09-21 15:05:31
Indexing a constant string makes for especially nice-looking code when used to convert to a hexadecimal digit: char digit = "0123456789ABCDEF"[i];
Heath Hunnicutt
2009-10-16 09:55:50
+6
A:
arr[i] is parsed as *(arr+i) which can be written as *(i+arr) and hence i[arr]
Now "strngjwdgd" is a pointer to a constant character array stored at read only location.
so it works!!
Neeraj
2009-09-21 09:48:02
+2
A:
The string literal(array) decays to a pointer of type char*. Then you take the fourth element:
3["zdvnngfgnfg"] == "zdvnngfgnfg"[3]
Why you can write the subscript infront of the array is another question:
AraK
2009-09-21 09:49:11