Qyriad
59bf6825ef
ExprVar::show() assumes it has a name. dynamic inherits do not
necessarily (ever?) have a name.
Change-Id: If10893188e307431da17f0c1bd0787adc74f7141
35 lines
770 B
C++
35 lines
770 B
C++
#include <sstream>
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#include <string_view>
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#include <gtest/gtest.h>
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#include "tests/libexpr.hh"
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#include "nixexpr.hh"
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#include "ref.hh"
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namespace nix
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{
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using namespace testing;
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struct ExprPrintingTests : LibExprTest
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{
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void test(Expr const & expr, std::string_view expected)
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{
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std::stringstream out;
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expr.show(state.symbols, out);
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ASSERT_EQ(out.str(), expected);
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}
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};
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TEST_F(ExprPrintingTests, ExprInheritFrom)
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{
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// ExprInheritFrom has its own show() impl.
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// If it uses its parent class's impl it will crash.
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auto inheritSource = make_ref<ExprVar>(state.symbols.create("stdenv"));
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ExprInheritFrom const eInheritFrom(noPos, 0, inheritSource);
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test(eInheritFrom, "(/* expanded inherit (expr) */ stdenv)");
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}
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}
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