Array subscripting operator [ ]
A postfix expression followed by an expression in [ ]
(brackets)
specifies an element of an array. The expression within the brackets
is referred to as a subscript. The first element of an array
has the subscript zero. Array bounds are not checked for built-in
array types.
a[b]
is equivalent
to the expression *((a) + (b))
, and, because addition
is associative, it is also equivalent to b[a]
. Between
expressions a
and b
, one must be
a pointer to a type T
, and the other must have integral
or enumeration type. The result of an array subscript is an lvalue.
The following example demonstrates this: #include <stdio.h>
int main(void) {
int a[3] = { 10, 20, 30 };
printf("a[0] = %d\n", a[0]);
printf("a[1] = %d\n", 1[a]);
printf("a[2] = %d\n", *(2 + a));
return 0;
}
The following is the output of the above example: a[0] = 10
a[1] = 20
a[2] = 30
The
above restrictions on the types of expressions required by the subscript
operator, as well as the relationship between the subscript operator
and pointer arithmetic, do not apply if you overload
operator[]
of
a class.
The first element of each array has the subscript 0
.
The expression contract[35]
refers to the 36th element
in the array contract
.
In a multidimensional array, you can reference each element (in the order of increasing storage locations) by incrementing the right-most subscript most frequently.
100
to
each element in the array code[4][3][6]
: for (first = 0; first < 4; ++first)
{
for (second = 0; second < 3; ++second)
{
for (third = 0; third < 6; ++third)
{
code[first][second][third] = 100;
}
}
}
![C](../gifs/ngc.gif)
struct trio{int a[3];};
struct trio f();
foo (int index)
{
return f().a[index];
}