03-22-2008
How to return void function pointer
Hello all
im trying to build function that will return void function pointer
what is mean is ( not working )
the main function
PHP Code:
void * myClass::getFunction(int type){
if(type==1)
return &myClass::Test1;
if(type==2)
return &myClass::Test2;
}
void myClass::Test1(){
printf("nada1");
}
void myClass::Test2(){
printf("nada2");
}
what im doing wrong here ?
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LEARN ABOUT PHP
php_sapi_name
PHP_SAPI_NAME(3) 1 PHP_SAPI_NAME(3)
php_sapi_name - Returns the type of interface between web server and PHP
SYNOPSIS
string php_sapi_name (void )
DESCRIPTION
Returns a lowercase string that describes the type of interface (the Server API, SAPI) that PHP is using. For example, in CLI PHP this
string will be "cli" whereas with Apache it may have several different values depending on the exact SAPI used. Possible values are listed
below.
RETURN VALUES
Returns the interface type, as a lowercase string.
Although not exhaustive, the possible return values include aolserver, apache, apache2filter, apache2handler, caudium, cgi (until PHP
5.3), cgi-fcgi, cli, cli-server, continuity, embed, fpm-fcgi, isapi, litespeed, milter, nsapi, phttpd, pi3web, roxen, thttpd, tux, and web-
james.
EXAMPLES
Example #1
php_sapi_name(3) example
This example checks for the substring cgi because it may also be cgi-fcgi.
<?php
$sapi_type = php_sapi_name();
if (substr($sapi_type, 0, 3) == 'cgi') {
echo "You are using CGI PHP
";
} else {
echo "You are not using CGI PHP
";
}
?>
NOTES
Note
An alternative approach
The PHP constant PHP_SAPI has the same value as php_sapi_name(3).
Tip
A potential gotcha
The defined SAPI may not be obvious, because for example instead of apache it may be defined as apache2handler or apache2filter.
SEE ALSO
PHP_SAPI.
PHP Documentation Group PHP_SAPI_NAME(3)