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
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@ -1,4 +1,4 @@
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14ca4b6130b9a7132d132f418e9ea283b3a52c08
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f97d579fa8431af5cfde9b0a48604caabfd09377
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The first line of this file holds the git revision number of the last
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merge done from the gofrontend repository.
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@ -1047,7 +1047,7 @@ Lex::gather_number()
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pnum = p;
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while (p < pend)
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{
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if (*p < '0' || *p > '7')
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if (*p < '0' || *p > '9')
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break;
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++p;
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}
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@ -1060,7 +1060,13 @@ Lex::gather_number()
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std::string s(pnum, p - pnum);
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mpz_t val;
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int r = mpz_init_set_str(val, s.c_str(), base);
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go_assert(r == 0);
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if (r != 0)
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{
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if (base == 8)
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error_at(this->location(), "invalid octal literal");
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else
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error_at(this->location(), "invalid hex literal");
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}
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if (neg)
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mpz_neg(val, val);
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