Fix nameless struct handling.

BUG=401296 The addition of layout qualifier handling in es3 development inadvertently broke handling of nameless structs. Change-Id: I805bab7a981d1f7f6227ae043720296fc3454662 Reviewed-on: https://chromium-review.googlesource.com/211860Reviewed-by: Zhenyao Mo <zmo@chromium.org> Tested-by: Shannon Woods <shannonwoods@chromium.org> Conflicts: src/compiler/translator/UniformHLSL.cpp Change-Id: I2f9fca71cef383c1a48c9bcdbe1083ac3222fac0 Reviewed-on: https://chromium-review.googlesource.com/212153Reviewed-by: Zhenyao Mo <zmo@chromium.org> Tested-by: Shannon Woods <shannonwoods@chromium.org>

Fix nameless struct handling.
f4126d2b · Shannon Woods · 75551cf9 · f4126d2b
Commit f4126d2b authored Aug 11, 2014 by Shannon Woods
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OutputHLSL.cpp src/compiler/translator/OutputHLSL.cpp +5 -1

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--- a/src/compiler/translator/OutputHLSL.cpp
+++ b/src/compiler/translator/OutputHLSL.cpp
@@ -582,7 +582,11 @@ void OutputHLSL::header()
        else
        {
            const TStructure *structure = type.getStruct();
-            const TString &typeName = (structure ? structureTypeName(*structure, false, false) : typeString(type));
+            // If this is a nameless struct, we need to use its full definition, rather than its (empty) name.
+            // TypeString() will invoke defineNameless in this case, but layout qualifiers, if relevant, will not
+            // be taken into account.
+            const TString &typeName = ((structure && !structure->name().empty()) ?
+                                        structureTypeName(*structure, false, false) : typeString(type));
            const TString &registerString = TString("register(") + RegisterPrefix(type) + str(registerIndex) + ")";