Currently we fail to notice integer overflow when parsing a
back-reference expression, or when converting the parsed result from
long to int. This changes the result to be int, so no conversion is
needed, and uses the overflow-checking built-ins to detect an
out-of-range back-reference.
libstdc++-v3/ChangeLog:
PR libstdc++/106607
* include/bits/regex_compiler.tcc (_Compiler::_M_cur_int_value):
Use built-ins to check for integer overflow in back-reference
number.
* testsuite/28_regex/basic_regex/106607.cc: New test.
(cherry picked from commit
1b09eea33f2bf9d1eae73b25cc25efb05ea1dc3f)
_Compiler<_TraitsT>::
_M_cur_int_value(int __radix)
{
- long __v = 0;
- for (typename _StringT::size_type __i = 0;
- __i < _M_value.length(); ++__i)
- __v =__v * __radix + _M_traits.value(_M_value[__i], __radix);
+ int __v = 0;
+ for (_CharT __c : _M_value)
+ if (__builtin_mul_overflow(__v, __radix, &__v)
+ || __builtin_add_overflow(__v, _M_traits.value(__c, __radix), &__v))
+ std::__throw_regex_error(regex_constants::error_backref,
+ "invalid back reference");
return __v;
}
--- /dev/null
+// { dg-do run { target c++11 } }
+
+#include <regex>
+#include <string>
+#include <climits>
+#include <testsuite_hooks.h>
+
+// PR libstdc++/106607 - Regex integer overflow on large backreference value
+
+int main()
+{
+ std::regex r("(.)\\1"); // OK
+
+ try
+ {
+ long long n = (unsigned)-1 + 2LL; // 4294967297 for 32-bit int
+ VERIFY( (int)n == 1 ); // 4294967297 % 2^32 == 1
+ std::regex r("(.)\\" + std::to_string(n)); // Invalid back reference.
+ VERIFY(false);
+ }
+ catch (const std::regex_error& e)
+ {
+ VERIFY( e.code() == std::regex_constants::error_backref );
+ }
+}