From 71016baf5524d739d1dd4d9110b111a7c3ddcaf9 Mon Sep 17 00:00:00 2001 From: Denys Vlasenko Date: Fri, 5 Jun 2009 16:24:29 +0200 Subject: printf: accept negative numbers for %x; sh: overflowed numbers are 0 Signed-off-by: Denys Vlasenko --- shell/math.c | 8 ++++++-- 1 file changed, 6 insertions(+), 2 deletions(-) (limited to 'shell/math.c') diff --git a/shell/math.c b/shell/math.c index cc298bd..d75bcae 100644 --- a/shell/math.c +++ b/shell/math.c @@ -562,7 +562,11 @@ arith(const char *expr, int *perrcode, a_e_h_t *math_hooks) } if (isdigit(arithval)) { numstackptr->var = NULL; + errno = 0; + /* call strtoul[l]: */ numstackptr->val = strto_arith_t(expr, (char **) &expr, 0); + if (errno) + numstackptr->val = 0; /* bash compat */ goto num; } for (p = op_tokens; ; p++) { @@ -592,7 +596,7 @@ arith(const char *expr, int *perrcode, a_e_h_t *math_hooks) lasttok = TOK_NUM; /* Plus and minus are binary (not unary) _only_ if the last - * token was as number, or a right paren (which pretends to be + * token was a number, or a right paren (which pretends to be * a number, since it evaluates to one). Think about it. * It makes sense. */ if (lasttok != TOK_NUM) { @@ -611,7 +615,7 @@ arith(const char *expr, int *perrcode, a_e_h_t *math_hooks) break; } } - /* We don't want a unary operator to cause recursive descent on the + /* We don't want an unary operator to cause recursive descent on the * stack, because there can be many in a row and it could cause an * operator to be evaluated before its argument is pushed onto the * integer stack. */ -- cgit v1.1