Well I just noticed that by changing the position -in microsoft visual studio- through "seekp" I implicitelly also change the read-position, when handling files.
I am wondering however if this is "portable" behaviour? Can I expect the position of reading & writing to be always the same? And consequently: will tellp & tellg always return the same value?
For file positions they are the same. In other words there is only one pointer maintained.
From 27.9.1.1p3:
A joint file position is maintained for both the input sequence and
the output sequence.
So, seekg and seekp are interchangeable for file streams. However, this is not true for other types of streams, as they may hold separate pointers for the put and get positions.
Update: So from all the comments and everything, it seems that for fstream, seekp and seekg use the same pointer. But for stringstream and probably other non-file based streams, they are separate.
Original Post:
Doesn't work for me on linux with g++ 4.7.2. They seem to be independent:
#include <sstream>
#include <iostream>
int main(int, char**) {
std::stringstream s("0123456789");
std::cout << "put pointer: " << s.tellp() << std::endl;
std::cout << "get pointer: " << s.tellg() << std::endl;
std::cout << std::endl;
s.seekp(2);
std::cout << "put pointer: " << s.tellp() << std::endl;
std::cout << "get pointer: " << s.tellg() << std::endl;
std::cout << std::endl;
s.seekg(4);
std::cout << "put pointer: " << s.tellp() << std::endl;
std::cout << "get pointer: " << s.tellg() << std::endl;
std::cout << std::endl;
}
Output:
put pointer: 0
get pointer: 0
put pointer: 2
get pointer: 0
put pointer: 2
get pointer: 4
Also the behaviour you describe sounds like it doesn't comply with the quotes here:
Sets the position of the get pointer. The get pointer determines the
next location to be read in the source associated to the stream.
and here:
Sets the position of the put pointer. The put pointer determines the location in the
output sequence where the next output operation is going to take
place.
Related
This is my first C++ related question and I'm new to character pointers and their usage. I think I've got it down but for an assignment the required output for this program is
So each first and last name is a character pointer of 20 characters (I could probably size it down but whatever) and when I output it now it looks like
cout << stu[i]->first << " " << stu[i]->last << " " << (float)stu[i]->mean << endl; and outputs the same thing as above but with a single space between each piece of data. How would I get it to print out the whitespace of the rest of the char pointer so it creates nice neat columns?
Thanks!
There is no magic whitespace in memory a char* points to. If you want to align your output you could use std::setw():
#include <iostream>
#include <iomanip>
int main()
{
char const *foo{ "Jamie" };
char const *bar{ "Reynolds" };
std::cout << std::setw(10) << foo << std::setw(10) << bar << '\n';
}
Many warning messages (via std::cout) might be printed out during the process. Is there a way to postpone the printing of the warning messaged in the end of the program? There are huge amount of the processing information will be printed. I'm planing to have all the warnings together in the end rather than scattered around.
More background:
code is already there.
there are about 50 warning messages within the code (in case if there is some sort of delay( ) function, I don't want to add 50 times, would be nice if there is an globally delaye/postpone function for stand output)
Thanks
One way to do it is to send everything to a stringstream, and then print at the end.
For example:
#include <iostream>
#include <sstream>
int main(){
int i = 5, j = 4;
std::stringstream ss;
std::cout << i * j << std::endl;
ss << "success" << std::endl;
std::cout << j + i * i + j << std::endl;
ss << "failure" << std::endl;
std::cout << ss.str() << std::endl;
return 0;
}
Output:
20
33
success
failure
If you're just trying to delay all printing of std::cout what you can do is redirect standard out to a string stream that acts as a buffer. It's pretty simple and avoids all of the dup, dup2, and piping stuff that one might be inclined to try.
#include <sstream>
// Make a buffer for all of your output
std::stringstream buffer;
// Copy std::cout since we're going to replace it temporarily
std::streambuf normal_cout = std::cout.rdbuf();
// Replace std::cout with your bufffer
std::cout.rdbuf(buffer.rdbuf());
// Now your program runs and does its thing writing to std::cout
std::cout << "Additional errors or details" << std::endl;
// Now restore std::cout
std::cout.rdbuf(normal_cout);
// Print the stuff you buffered
std::cout << buffer.str() << std::endl;
Also in the future, you should really use a buffer for errors from the start OR at a minimum write errors and logging to std::cerr so that your normal runtime print outs aren't cluttered with errors.
This question already has answers here:
Restore the state of std::cout after manipulating it
(9 answers)
Closed 4 years ago.
C++ steam objects have state. If one write a piece of code like
using namespace std;
cout << hex << setw(8) << setfill('0') << x << endl;
forgetting setting the stream state back. This will cause problems in some other unrelated codes. It's tedious to do "set" and "set back" pair matching. Besides from that, it seems to me it's also against convention behind RAII.
My question is: is it possible, with only a thin layer of wrapping, to make those state manipulations RAII-like. That is, right after the end of an expression by semicolon, stream state is automatically set back to default.
Update: Following the link provided by #0x499602D2, one workaround might be something like
#include <boost/io/ios_state.hpp>
#include <ios>
#include <iostream>
#include <ostream>
#define AUTO_COUT(x) {\
boost::io::ios_all_saver ias( cout );\
x;\
}while(0)
Then one can use the macro like
AUTO_COUT(cout << hex << setw(8) << setfill('0') << x << endl);
BTW, it might be a good idea to add a lock field to those saver class of boost::io::ios_state, in case funny things occur in a multi-threading program. Or they have already done so?
I'm going to suggest an alternative approach. The manipulators apply to the std::[i|o]stream instance, but they do nothing with regards to the std::[i|o]streambuf which is managed by that std::[i|o]stream.
Therefore, you can create your own std::[i|o]stream, which will have its own formatting state, but writing to the same buffer std::cout uses:
#include <iostream>
#include <iomanip>
int main()
{
std::cout << std::hex << 32 << "\n";
std::ostream os(std::cout.rdbuf());
os << 32 << "\n" << std::hex;
std::cout << std::dec;
os << 32 << "\n";
std::cout << 32 << "\n";
}
Output:
20
32
20
32
Live on Coliru
This uses only features from standard library, and since the original stream is not touched, applying manipulators is trivially thread safe (because each thread operates on a different stream). Now, the actual writes and reads' thread safety depends on the thread safety of the managed stream buffer.
I once wrote a utility class for my personal use. (I don't know whether it is as perfect as the boost code probably is but it worked for me – so, I dare to share.)
#include <iostream>
#include <iomanip>
/** provides a helper class to work with streams.
*
* It stores current format states of a stream in constructor and
* recovers these settings in destructor.
*
* Example:
* <pre>
* { // backup states of std::cout
* IOSFmtBackup stateCOut(std::cout);
* // do some formatted output
* std::cout
* << "dec: " << std::dec << 123 << std::endl
* << "hex: " << std::hex << std::setw(8) << std::setfill('0')
* << 0xdeadbeef << std::endl;
* } // destruction of stateCOut recovers former states of std::cout
* </pre>
*/
class IOSFmtBackup {
// variables:
private:
/// the concerning stream
std::ios &_stream;
/// the backup of formatter states
std::ios _fmt;
// methods:
public:
/// #name Construction & Destruction
//#{
/** constructor.
*
* #param stream the stream for backup
*/
explicit IOSFmtBackup(std::ios &stream):
_stream(stream), _fmt(0)
{
_fmt.copyfmt(_stream);
}
/// destructor.
~IOSFmtBackup() { _stream.copyfmt(_fmt); }
// disabled:
IOSFmtBackup(const IOSFmtBackup&) = delete;
IOSFmtBackup& operator=(const IOSFmtBackup&) = delete;
//#}
};
int main()
{
{ // backup states of std::cout
IOSFmtBackup stateCOut(std::cout);
// do some formatted output
std::cout
<< "dec: " << std::dec << 123 << std::endl
<< "hex: " << std::hex << std::setw(8) << std::setfill('0')
<< 0xdeadbeef << std::endl
<< "123 in current: " << 123 << std::endl;
} // destruction of stateCOut recovers former states of std::cout
// check whether formatting is recovered
std::cout << "123 after recovered: " << 123 << std::endl;
return 0;
}
Compiled and tested on ideone (life demo).
Output:
dec: 123
hex: deadbeef
123 in current: 7b
123 after recovered: 123
I'm using this code to output nodes of a huffman tree to a text file with a certain formatting. All the node outputs within the if block run as expected, but the first output in the else block is missing the '0' fill character after the "L:". It should output "L:076" but instead is outputting "L: 76". The cout looks correct but the text file isn't. All future loops through the else block output like they should, it's only the first loop that is missing the fill character. Here's a picture of my output
void preOrder(node* tree, std::ofstream& of) {
if (tree->label > 0) {
of << "I:" << tree->label << " ";
}
else {
std::cout.width(3);
std::cout << std::right;
std::cout.fill('0');
std::cout << int(tree->ch) << std::endl;
of << "L:";
of << of.fill('0');
of << std::right;
of << int(tree->ch);
of << " ";
return;
}
preOrder(tree->left, of);
preOrder(tree->right, of);
}
From cppreference.com:
The second form (2) sets fillch as the new fill character and returns the fill character used before the call.
"The second form" is the non-const version, that applies here. So my guess (I never used fill myself and I cannot compile your code as it is) would be that the call is correctly applied and then you put the old fill character (blank space presumably) to the stream, because you do:
of << of.fill('0');
Also, I noticed that you dont set the width of of.
Because you're hiding something naughty from us.
#include <iostream>
int main()
{
std::cout.width(3);
std::cout << std::right;
std::cout.fill('0');
std::cout << 3 << std::endl;
return 0;
}
Outputs 003 (live example).
Please provide an MCVE and I'll edit my answer to help you.
I am using Qt, and I have an unsigned char *bytePointer and want to print out a number-value of the current byte. Below is my code, which is meant to give the int-value and the hex-value of the continuous bytes that I receive from a machine attached to the computer:
int byteHex=0;
byteHex = (int)*bytePointer;
qDebug << "\n int: " //this is the main issue here.
<< *bytePointer;
std::cout << " (hex: "
<< std::hex
<< byteHex
<< ")\n";
}
This gives perfect results, and I get actual numbers, however this code is going into an API and I don't want to use Qt-only functions, such as qDebug. So when I try this:
int byteHex=0;
byteHex = (int)*bytePointer;
std::cout << "\n int: " //I changed qDebug to std::cout
<< *bytePointer;
std::cout << " (hex: "
<< std::hex
<< byteHex
<< ")\n";
}
The output does give the hex-values perfectly, however the int-values return symbols (like ☺, └, §, to list a few).
My question is: How do I get std::cout to give the same output as qDebug?
EDIT: for some reason the symbols only occur with a certain Qt setting. I have no idea why it happened but it's fixed now.
As others pointed out in comment, you change the outputting to hex, but you do not actually set it back here:
std::cout << " (hex: "
<< std::hex
<< byteHex
<< ")\n";
You will need to apply this afterwards:
std::cout << std::dec;
Standard output streams will output any character type as a character, not a numeric value. To output the numeric value, convert to a non-character integer type:
std::cout << int(*bytePointer);