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std references shouldn't be fully qualified

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This commit is contained in:
Vadim Berezniker 2018-08-17 10:49:10 -07:00
parent 02a8ca8773
commit 0796415314
1 changed files with 4 additions and 4 deletions
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8
googletest/docs/advanced.md
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@ -572,7 +572,7 @@ namespace foo {
class Bar { // We want googletest to be able to print instances of this. class Bar { // We want googletest to be able to print instances of this.
... ...
// Create a free inline friend function. // Create a free inline friend function.
friend ::std::ostream& operator<<(::std::ostream& os, const Bar& bar) { friend std::ostream& operator<<(std::ostream& os, const Bar& bar) {
return os << bar.DebugString(); // whatever needed to print bar to os return os << bar.DebugString(); // whatever needed to print bar to os
} }
}; };
@ -580,7 +580,7 @@ class Bar { // We want googletest to be able to print instances of this.
// If you can't declare the function in the class it's important that the // If you can't declare the function in the class it's important that the
// << operator is defined in the SAME namespace that defines Bar. C++'s look-up // << operator is defined in the SAME namespace that defines Bar. C++'s look-up
// rules rely on that. // rules rely on that.
::std::ostream& operator<<(::std::ostream& os, const Bar& bar) { std::ostream& operator<<(std::ostream& os, const Bar& bar) {
return os << bar.DebugString(); // whatever needed to print bar to os return os << bar.DebugString(); // whatever needed to print bar to os
} }
@ -601,7 +601,7 @@ namespace foo {
class Bar { class Bar {
... ...
friend void PrintTo(const Bar& bar, ::std::ostream* os) { friend void PrintTo(const Bar& bar, std::ostream* os) {
*os << bar.DebugString(); // whatever needed to print bar to os *os << bar.DebugString(); // whatever needed to print bar to os
} }
}; };
@ -609,7 +609,7 @@ class Bar {
// If you can't declare the function in the class it's important that PrintTo() // If you can't declare the function in the class it's important that PrintTo()
// is defined in the SAME namespace that defines Bar. C++'s look-up rules rely // is defined in the SAME namespace that defines Bar. C++'s look-up rules rely
// on that. // on that.
void PrintTo(const Bar& bar, ::std::ostream* os) { void PrintTo(const Bar& bar, std::ostream* os) {
*os << bar.DebugString(); // whatever needed to print bar to os *os << bar.DebugString(); // whatever needed to print bar to os
} }