fmt: treat \r\n as \n in Scan

When scanning input and "white space" is permitted, a carriage return
followed immediately by a newline (\r\n) is treated exactly the same
as a plain newline (\n). I hope this makes it work better on Windows.

We do it everywhere, not just on Windows, since why not?

Fixes #5391.

R=golang-dev, dsymonds
CC=golang-dev
https://golang.org/cl/12142043

src/pkg/fmt/doc.go

@@ -215,6 +215,10 @@
 	stops if it does not, with the return value of the function
 	indicating the number of arguments scanned.
 
+	In all the scanning functions, a carriage return followed
+	immediately by a newline is treated as a plain newline
+	(\r\n means the same as \n).
+
 	In all the scanning functions, if an operand implements method
 	Scan (that is, it implements the Scanner interface) that
 	method will be used to scan the text for that operand.  Also,

src/pkg/fmt/scan.go

@@ -437,6 +437,9 @@ func (s *ss) skipSpace(stopAtNewline bool) {
 		if r == eof {
 			return
 		}
+		if r == '\r' && s.peek("\n") {
+			continue
+		}
 		if r == '\n' {
 			if stopAtNewline {
 				break

src/pkg/fmt/scan_test.go

@@ -192,6 +192,10 @@ var scanTests = []ScanTest{
 	{"-.45e1-1e2i\n", &complex128Val, complex128(-.45e1 - 100i)},
 	{"hello\n", &stringVal, "hello"},
 
+	// Carriage-return followed by newline. (We treat \r\n as \n always.)
+	{"hello\r\n", &stringVal, "hello"},
+	{"27\r\n", &uint8Val, uint8(27)},
+
 	// Renamed types
 	{"true\n", &renamedBoolVal, renamedBool(true)},
 	{"F\n", &renamedBoolVal, renamedBool(false)},