PHP extension function array_init throws 'return_value is undeclared'

Question

For the first time I'm trying to create PHP extenstion. I need a function that will return an assoc array. So for the test reasons I created a small function:

PHP_FUNCTION(testing_array) {
    char *firstVal = NULL;
    char *secondVal= NULL;
    int argc = ZEND_NUM_ARGS();
    int firstVal_len;
    int secondVal_len;

    if (zend_parse_parameters(argc TSRMLS_CC, "ss", &firstVal, &firstVal_len, &secondVal, &secondVal_len) == FAILURE)
        return;

    array_init(return_array);
}

But everytime I'm tryimg to compile it, compiler tells me:

/root/php/php-src/ext/hello_world/hello_world.c:87: error: return_array undeclared (first use in this function)
/root/php/php-src/ext/hello_world/hello_world.c:87: error: (Each undeclared identifier is reported only once
/root/php/php-src/ext/hello_world/hello_world.c:87: error: for each function it appears in.)

What I'm doing wrong? In every example I saw, array variable isn't declared.

For a start you never declared return_array as the error saying, should it not be array_init(testing_array)? — Kieran
– Kieran, Commented Aug 18, 2013 at 15:51

TerryE · Accepted Answer · 2013-08-29 23:51:25Z

1

Look at the definition of the PHP_FUNCTION() macro. It declares the argument return_value not return_array. That's why the latter isn't declared.

answered Aug 29, 2013 at 23:51

TerryE

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Stratford · Accepted Answer · 2013-08-18 15:53:11Z

0

The error is quite clear. You have to declare the variable return_array before use it.

answered Aug 18, 2013 at 15:53

Stratford

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1 Comment

Maxim Lanin Over a year ago

I know that I need to declare it, but in every tutorial, in every example I saw, it never declared. You can see it here github.com/php/php-src/blob/master/ext/filter/filter.c#L889 - this is a code of filter_list() -function from the standart php extension "filter".

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