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Don't use results of quoted command substitution in adjacent variable expansion
Given set var a echo "$var$(echo b)" the double-quoted string is expanded right-to-left, so we construct an intermediate "$varb". Since the variable "varb" is undefined, this wrongly expands to the empty string (should be "ab"). Fix this by isolating the expanded command substitution internally. We do the same when handling unquoted command substitutions. Fixes #8849
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3 changed files with 6 additions and 0 deletions
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@ -19,6 +19,7 @@ Deprecations and removed features
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Scripting improvements
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----------------------
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- Quoted command substitution that directly follow a variable expansion (like ``echo "$var$(echo x)"``) no longer affect the variable expansion (:issue:`8849`).
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- ``math`` can now handle underscores (``_``) as visual separators in numbers (:issue:`8611`, :issue:`8496`)::
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math 5 + 2_123_252
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@ -758,7 +758,9 @@ static expand_result_t expand_cmdsubst(wcstring input, const operation_context_t
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whole_item.reserve(paren_begin + 1 + sub_res_joined.size() + 1 +
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tail_item.completion.size());
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whole_item.append(input, 0, paren_begin - have_dollar);
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whole_item.push_back(INTERNAL_SEPARATOR);
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whole_item.append(sub_res_joined);
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whole_item.push_back(INTERNAL_SEPARATOR);
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whole_item.append(tail_item.completion.substr(const_strlen(L"\"")));
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if (!out->add(std::move(whole_item))) {
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return append_overflow_error(errors);
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@ -63,3 +63,6 @@ echo "($(echo A)B$(echo C))"
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echo "quoted1""quoted2"(echo unquoted3)"$(echo quoted4)_$(echo quoted5)"
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# CHECK: quoted1quoted2unquoted3quoted4_quoted5
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var=a echo "$var$(echo b)"
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# CHECK: ab
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