compiler: Accept numeric literals with leading zeroes.

When a numeric literal with leading zeroes was seen in the parser, it
    would only be accepted if it were a valid hex or octal literal.  Any
    invalid numeric literal would be split up into multiple tokens: the
    valid hex/octal literal followed by the rest of the characters.
    Instead, when scanning a numeric literal with leading zeroes, always
    accept the number and give an appropriate error if the accepted number
    does not fit in the expected base.
    
    Fixes golang/go#11532, golang/go#11533.
    
    Reviewed-on: https://go-review.googlesource.com/13791

From-SVN: r227193
This commit is contained in:
Ian Lance Taylor 2015-08-25 21:17:47 +00:00
parent 60c47c0073
commit 387b37c1ac
2 changed files with 9 additions and 3 deletions

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@ -1,4 +1,4 @@
14ca4b6130b9a7132d132f418e9ea283b3a52c08
f97d579fa8431af5cfde9b0a48604caabfd09377
The first line of this file holds the git revision number of the last
merge done from the gofrontend repository.

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@ -1047,7 +1047,7 @@ Lex::gather_number()
pnum = p;
while (p < pend)
{
if (*p < '0' || *p > '7')
if (*p < '0' || *p > '9')
break;
++p;
}
@ -1060,7 +1060,13 @@ Lex::gather_number()
std::string s(pnum, p - pnum);
mpz_t val;
int r = mpz_init_set_str(val, s.c_str(), base);
go_assert(r == 0);
if (r != 0)
{
if (base == 8)
error_at(this->location(), "invalid octal literal");
else
error_at(this->location(), "invalid hex literal");
}
if (neg)
mpz_neg(val, val);