std references shouldn't be fully qualified
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02a8ca8773
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@ -572,7 +572,7 @@ namespace foo {
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class Bar { // We want googletest to be able to print instances of this.
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...
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// Create a free inline friend function.
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friend ::std::ostream& operator<<(::std::ostream& os, const Bar& bar) {
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friend std::ostream& operator<<(std::ostream& os, const Bar& bar) {
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return os << bar.DebugString(); // whatever needed to print bar to os
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}
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};
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@ -580,7 +580,7 @@ class Bar { // We want googletest to be able to print instances of this.
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// If you can't declare the function in the class it's important that the
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// << operator is defined in the SAME namespace that defines Bar. C++'s look-up
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// rules rely on that.
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::std::ostream& operator<<(::std::ostream& os, const Bar& bar) {
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std::ostream& operator<<(std::ostream& os, const Bar& bar) {
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return os << bar.DebugString(); // whatever needed to print bar to os
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}
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@ -601,7 +601,7 @@ namespace foo {
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class Bar {
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...
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friend void PrintTo(const Bar& bar, ::std::ostream* os) {
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friend void PrintTo(const Bar& bar, std::ostream* os) {
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*os << bar.DebugString(); // whatever needed to print bar to os
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}
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};
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@ -609,7 +609,7 @@ class Bar {
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// If you can't declare the function in the class it's important that PrintTo()
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// is defined in the SAME namespace that defines Bar. C++'s look-up rules rely
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// on that.
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void PrintTo(const Bar& bar, ::std::ostream* os) {
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void PrintTo(const Bar& bar, std::ostream* os) {
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*os << bar.DebugString(); // whatever needed to print bar to os
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}
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