Description
The following code:
<?php
json_decode("\00invalid json");
var_dump(json_last_error());
json_decode("[]", true, 512, JSON_THROW_ON_ERROR);
var_dump(json_last_error());Resulted in this output:
But I expected this output instead:
In json.c file, the function calls JSON_G(error_code) = error_code when I turn off flags of JSON_THROW_ON_ERROR.
but when I turn on flags of JSON_THROW_ON_ERROR, and then JSON_G(error_code) = error_code is not called probably.
PHP_JSON_API zend_result php_json_decode_ex(zval *return_value, const char *str, size_t str_len, zend_long options, zend_long depth) /* {{{ */
{
php_json_parser parser;
php_json_parser_init(&parser, return_value, str, str_len, (int)options, (int)depth);
if (php_json_yyparse(&parser)) {
php_json_error_code error_code = php_json_parser_error_code(&parser);
if (!(options & PHP_JSON_THROW_ON_ERROR)) {
JSON_G(error_code) = error_code; // <--- here
} else {
zend_throw_exception(php_json_exception_ce, php_json_get_error_msg(error_code), error_code);
}
RETVAL_NULL();
return FAILURE;
}
return SUCCESS;
}PHP Version
PHP 8.1
Operating System
mac
Description
The following code:
Resulted in this output:
But I expected this output instead:
In json.c file, the function calls
JSON_G(error_code) = error_codewhen I turn off flags of JSON_THROW_ON_ERROR.but when I turn on flags of JSON_THROW_ON_ERROR, and then
JSON_G(error_code) = error_codeis not called probably.PHP Version
PHP 8.1
Operating System
mac