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Bug #15814 php 4.1.2, apache 1.3.20, parse error near 'zval'
Submitted: 2002-03-01 11:59 UTC Modified: 2002-06-28 02:57 UTC
Votes:1
Avg. Score:5.0 ± 0.0
Reproduced:1 of 1 (100.0%)
Same Version:0 (0.0%)
Same OS:0 (0.0%)
From: glowa at acn dot waw dot pl Assigned:
Status: Not a bug Package: Compile Failure
PHP Version: 4.1.2 OS: sgi irix 6.5
Private report: No CVE-ID: None
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From: glowa at acn dot waw dot pl
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 [2002-03-01 11:59 UTC] glowa at acn dot waw dot pl
When i try to build php with mySql and apxs, configuration:
 ./configure --with-mysql=/usr/freeware --with-apxs=/usr/freeware/apache/sbin/apxs
goes fine but on make I get this:

# make
Making all in Zend
        /bin/sh ../libtool --silent --mode=compile
gcc -DHAVE_CONFIG_H -I. -I. -I../main   -D_XPG_IV -DIRIX -DMOD_SSL=208104 -D
USE_HSREGEX -DEAPI -DUSE_EXPAT -I../TSRM  -g -O2 -prefer-pic -c
zend_operators.c
zend_operators.c: In function `multi_convert_to_long_ex':
zend_operators.c:566: parse error before "zval"
zend_operators.c:566: parse error before "zval"
zend_operators.c:566: parse error before "zval"
zend_operators.c:566: parse error before "zval"
zend_operators.c:566: parse error before "zval"
zend_operators.c:566: parse error before ')' token
zend_operators.c: In function `multi_convert_to_double_ex':
zend_operators.c:581: parse error before "zval"
zend_operators.c:581: parse error before "zval"
zend_operators.c:581: parse error before "zval"
zend_operators.c:581: parse error before "zval"
zend_operators.c:581: parse error before "zval"
zend_operators.c:581: parse error before ')' token
zend_operators.c: In function `multi_convert_to_string_ex':
zend_operators.c:596: parse error before "zval"
zend_operators.c:596: parse error before "zval"
zend_operators.c:596: parse error before "zval"
zend_operators.c:596: parse error before "zval"
zend_operators.c:596: parse error before "zval"
zend_operators.c:596: parse error before ')' token
*** Error code 1 (bu21)
*** Error code 1 (bu21)

Same happens with 4.1.0 and 4.0.6


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 [2002-06-28 02:57 UTC] derick@php.net
Thank you for taking the time to report a problem with PHP.
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