Use same type in both cases of '?:' operator
Both possible result values need to be convertible to the same type. Some compilers fail to recognize that they can construct std::string from the empty string literal, so state it explicitly.
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@ -178,7 +178,7 @@ int cmCoreTryCompile::TryCompileCode(std::vector<std::string> const& argv)
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const char* rulesOverrideBase = "CMAKE_USER_MAKE_RULES_OVERRIDE";
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const char* rulesOverrideBase = "CMAKE_USER_MAKE_RULES_OVERRIDE";
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std::string rulesOverrideLang =
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std::string rulesOverrideLang =
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rulesOverrideBase + (lang ? std::string("_") + lang : "");
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rulesOverrideBase + (lang ? std::string("_") + lang : std::string(""));
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if(const char* rulesOverridePath =
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if(const char* rulesOverridePath =
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this->Makefile->GetDefinition(rulesOverrideLang.c_str()))
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this->Makefile->GetDefinition(rulesOverrideLang.c_str()))
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{
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{
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