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Silence a MSVC warning in class Tie

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With warning level 4 MSVC complains that a default
assignment operator could not be generated due to
member 'file' is a reference (warning C4512).

Use a pointer instead of a reference and move
struct Tie outisde class Logger while there.

No functional change.

Signed-off-by: Marco Costalba <mcostalba@gmail.com>
This commit is contained in:
Marco Costalba 2012-07-08 10:22:42 +01:00
parent 6b5322ce00
commit 6becc81446
1 changed files with 25 additions and 25 deletions
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50
src/misc.cpp
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@ -94,33 +94,33 @@ void dbg_print() {
/// usual i/o functionality and without changing a single line of code!
/// Idea from http://groups.google.com/group/comp.lang.c++/msg/1d941c0f26ea0d81
struct Tie: public streambuf { // MSVC requires splitted streambuf for cin and cout
Tie(streambuf* b, ofstream* f) : buf(b), file(f) {}
int sync() { return file->rdbuf()->pubsync(), buf->pubsync(); }
int overflow(int c) { return log(buf->sputc((char)c), "<< "); }
int underflow() { return buf->sgetc(); }
int uflow() { return log(buf->sbumpc(), ">> "); }
streambuf* buf;
ofstream* file;
int log(int c, const char* prefix) {
static int last = '\n';
if (last == '\n')
file->rdbuf()->sputn(prefix, 3);
return last = file->rdbuf()->sputc((char)c);
}
};
class Logger {
Logger() : in(cin.rdbuf(), file), out(cout.rdbuf(), file) {}
~Logger() { start(false); }
struct Tie: public streambuf { // MSVC requires splitted streambuf for cin and cout
Tie(streambuf* b, ofstream& f) : buf(b), file(f) {}
int sync() { return file.rdbuf()->pubsync(), buf->pubsync(); }
int overflow(int c) { return log(buf->sputc((char)c), "<< "); }
int underflow() { return buf->sgetc(); }
int uflow() { return log(buf->sbumpc(), ">> "); }
int log(int c, const char* prefix) {
static int last = '\n';
if (last == '\n')
file.rdbuf()->sputn(prefix, 3);
return last = file.rdbuf()->sputc((char)c);
}
streambuf* buf;
ofstream& file;
};
Logger() : in(cin.rdbuf(), &file), out(cout.rdbuf(), &file) {}
~Logger() { start(false); }
ofstream file;
Tie in, out;