sizeof an array which is variable

delighteddennis

I want to use arrays with names which are variable. I need to know the size of these variable arrays, but

$size = sizeof($$var_array_name);

isn't working. It always returns 1, even when I there are more elements in the array.

Who knows a solution to solve this problem?

Dennis

junebug

i tried this and what you are trying to do worked fine for me. it sounds like the you may have spelled the $var_array_name variable wrong or some such thing. another thing you could try would be this:

sizeof(${$var_array_name})

give a look though, i think that it might be the spelling.

delighteddennis

Thanks, but I don't think it's th spelling. I'll give you a little example.

When I use this test-program:

$test = "var_name";
$$test[0]="A";
$$test[1]="B";
$$test[2]="C";
$$test[3]="D";
$$test[4]="E";
$$test[5]="F";

$size = sizeof(${$test});
print("$size <BR>");

for ($i=0; $i<6; $i++)
{
print("${$test[$i]} <BR>");
}

I expect this output:
6
A
B
C
D
E
F

But I get this output:
0 // size is 0?
A
F // what happened to "B"?
C
D
E
F

Can anyone tell me what I'm doing wrong?

Thanx,

Dennis

mrskhris

Ambiguity problem.

Here is the excerpt from the manual:

In order to use variable variables with arrays, you have to resolve an ambiguity problem. That is, if you write $$a[1] then the parser needs to know if you meant to use $a[1] as a variable, or if you wanted $$a as the variable and then the [1] index from that variable. The syntax for resolving this ambiguity is: ${$a[1]} for the first case and ${$a}[1] for the second.