auditsc: test below 0 on unsigned ino
PATCH 4/9 auditsc: test below 0 on unsigned ino
ino is unsigned so the test didn't work.
---
kernel/auditsc.c +1 -1
@@ -286,7 +286,7 @@ static int audit_match_filetype(struct audit_context *ctx, int which)mode_t mode = which &S_IFMT ;if (index >= ctx->name_count)return 0 ; -if (ctx->names[index].ino ==-1 ) +if (ctx->names[index].ino ==-1ul )return 0 ;if ((ctx->names[index].mode ^ mode) &S_IFMT )return 0 ;
Quoted 1 line · roel kluin
ino is unsigned so the test didn't work.
^^^^^^^^^^^
Kindly report the way to reproduce your remarkable observation. Do not forget to include the compiler version, since the following two lines
kernel/auditsc.c static int audit_match_filetype(struct audit_context *ctx, int which) · line 289
Quoted 2 lines
kernel/auditsc.c +1
-1-if (ctx->names[index].ino ==-1 ) +if (ctx->names[index].ino ==-1ul )
are equivalent. -1 is an expression of type int. ctx->names[index].ino is an expression of type unsigned long. If both operands of a comparison operator have arithmetic types, the usual arithmetic conversions (see 6.3.1.8) are applied to the operands. In this case, both types are integer ones and not modified by integer promotions. One is signed, another is unsigned and the rank of unsigned one is greater or equal to that of the signed one (rank(unsigned long) = rank(signed long) > rank(signed int)). Therefore, the operand with signed integer type (-1) is converted to the type of argument with unsigned integer type. Then they are compared.
In the second case both operands have the same integer type (unsigned long) and comparison is done without any conversions.
Proof that (unsigned long)-1 and -1ul have the same value (namely, the maximal value that can be represented in unsigned long) is left as an exercise for reader.
Assuming that you have indeed observed a case when results of these tests differed, you have found a blatant non-compliance of whatever C compiler you were using. I am sure that maintainers of that compiler would like to see your bug report, especially since you already have a reproducer. So would everybody else, to know which version to avoid.