I'm learning about the static cast command so I made this simple program. The two numbers being divided are Integers so I used a float cast to force it to do floating point division, however when I run the program the result is Integer division with the decimal being truncated. I am coding in visual c++ and using Visual Studio 2013.
Thanks for any help :)
void PrintAnwser(Fraction Fract)
{
using namespace std;
float Anwser = static_cast<float>(Fract.firstNumber / Fract.secondNumber);
cout << "The result of " << Fract.firstNumber << " " << "Divided by " <<Fract.secondNumber
<< " " << "is " << Anwser << endl;
}
The cast is happening after the (integer) division is completed. You need to cast one of the inputs to a float, so that the division will be floating point:
float Anwser = Fract.firstNumber / static_cast<float>(Fract.secondNumber);
Related
I've been trying to take a .txt document with three number entries, read those entries as strings and convert those entries in ints, then put them into an int array, but had no success in doing so and i have no clue as to why. Note that the entries as well as some variable names are pre determined by the assignment, additionally we have to use the std::stoi("string") command, which i am not familiar with nor has any syntax been provided to us (which is especially strange since we are usually not allowed to stray to far from the lecture material)
What I excpected to happen is that the numbers from the .txt file were converted into an array, however what actually happened is that an "unhandled exception" (my apologies if that term does not make sanes we have to programm in our native language) occured and the string library opened itself, marking the error on line 107. The problematic line in my code seems to be "auftraegearray[i++] = std::stoi(MengeanAuftraegen);"
int main()
{
std::fstream Auftraege;
Auftraege.open("Auftraege37.txt", std::ios::out);
Auftraege << "10" << std::endl;
Auftraege << "1" << std::endl;
Auftraege << "20" << std::endl;
Auftraege.close();
int i = 0;
int auftraegearray[4];
std::string MengeanAuftraegen;
Auftraege.open("Auftraege37.txt", std::ios::in);
while (!Auftraege.eof())
{
getline(Auftraege, MengeanAuftraegen);
std::cout << MengeanAuftraegen << std::endl;
auftraegearray[i++] = std::stoi(MengeanAuftraegen);
}
Auftraege.close();
I am working on a money program. Currently I am trying to make a part in the program that displays a list that shows every transaction that was made in the session. My problem is that when I convert the money amount to a string instead of displaying something like 100.35 in when converted to a string it instead displays something like 100.35000000. I was wondering if there was any way I could make the program drop the additional zeros? Here is a sample of how I convert the numbers to a string
int main(){
double samplemoney=100.35;
string sample="Today we made $";
string comsample;
comsample=sample+std::tostring(money)+".";
cout<<comsample<<endl;
return 0;
}
In my main program this part is handled with a class but as I said earlier it seems like no matter what the money value I put in is it will display a series of zero and I want my program to drop the unnecessary zeros.
Let's say you have this number:
double number = 190.0391000;
If the problem is displaying the value you may use
std::cout << std::setprecision(5) << number
where f is the number you want to show.
If your problem is having a string with a finite precision you can use sprintf() in the following way:
char arr[128];
sprintf(arr,"%3.2f", number);
std::string formattedstring(arr);
or in a more C++ oriented way something like this
std::stringstream strn;
strn.precision(2);
strn << std::fixed << number;
and then you get the string in the following way:
std::string formattedstring = strn.str();
I attach the full text of program here... tested on my machine:
#include <sstream>
#include <iostream>
int main() {
float number=100.24324232;
std::stringstream strn;
strn.precision(2);
strn << std::fixed << number;
std::cout << strn.str() << "\n";
return( 0 );
}
Short question:
How can setting the _EM_INVALID exception flag on the FPU result in different values?
Long question:
In our project we have turned off floating point exceptions in our Release build, but turned on ZERODIVIDE, INVALID and OVERFLOW using _controlfp_s() in our Debug build. This is in order to catch errors if they are there.
However, we would also like results of numerical calculations (involving optimisation algorithms, matrix inversion, Monte Carlo and all sorts of things) to be consistent between Debug and Release build to make debugging easier.
I would expect that the setting of the exception flags on the FPU should not affect the calculated values - only whether exceptions are thrown or not. But after working backwards through our calculations I can isolate the below code example that shows that there is a difference on the last bit when calling the log() function.
This propagates to a 0.5% difference in the resulting value.
The below code will give the shown program output when adding it to a new solution in Visual Studio 2005, Windows XP and compile in Debug configuration. (Release will give a different output, but that's because the optimiser reuses the result from the first call to log().)
I hope that someone can shed a bit of light on this. Thanks.
/*
Program output:
Xi, 3893f76f, 7.4555176582633598
K, c0a682c7, 7.44466687218
Untouched
x, da8caea1, 0.0014564635732296288
Invalid exception on
x, da8caea2, 0.001456463573229629
Invalid exception off
x, da8caea1, 0.0014564635732296288
*/
#include <float.h>
#include <math.h>
#include <limits>
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
unsigned uMaskOld = 0;
errno_t err;
cout << std::setprecision (numeric_limits<double>::digits10 + 2);
double Xi = 7.4555176582633598;
double K = 7.44466687218;
double x;
cout << "Xi, " << hex << setw(8) << setfill('0') << *(unsigned*)(&Xi) << ", " << dec << Xi << endl;
cout << "K, " << hex << setw(8) << setfill('0') << *(unsigned*)(&K) << ", " << dec << K << endl;
cout << endl;
cout << "Untouched" << endl;
x = log(Xi/K);
cout << "x, " << hex << setw(8) << setfill('0') << *(unsigned*)(&x) << ", " << dec << x << endl;
cout << endl;
cout << "Invalid exception on" << endl;
::_clearfp();
err = ::_controlfp_s(&uMaskOld, 0, _EM_INVALID);
x = log(Xi/K);
cout << "x, " << hex << setw(8) << setfill('0') << *(unsigned*)(&x) << ", " << dec << x << endl;
cout << endl;
cout << "Invalid exception off" << endl;
::_clearfp();
err = ::_controlfp_s(&uMaskOld, _EM_INVALID, _EM_INVALID);
x = log(Xi/K);
cout << "x, " << hex << setw(8) << setfill('0') << *(unsigned*)(&x) << ", " << dec << x << endl;
cout << endl;
return 0;
}
This is not a complete answer, but it is too long for a comment.
I suggest you isolate the code that does the questionable calculations and put it in a subroutine, preferably in a source module that is compiled separately. Something like:
void foo(void)
{
double Xi = 7.4555176582633598;
double K = 7.44466687218;
double x;
x = log(Xi/K);
…Insert output statements here…
}
Then you would call the routine with different settings:
cout << "Untouched:\n";
foo();
cout << "Invalid exception on:\n";
…Change FP state…
foo();
This guarantees that the same instructions are executed in each case, eliminating the possibility that the compiler has for some reason generated separate code for each sequence. The way you have compiled the code, I suspect it is possible the compiler may have used 80-bit arithmetic in one case and 64-bit arithmetic in another, or may have used 80-bit arithmetic generally but converted some result to 64-bit in one case but not another
Once that is done, you can partition and isolate the code further. E.g., try evaluating Xi/K once before any of the tests, storing that in a double, and passing it to foo as a parameter. The tests whether the log call differs depending on the floating-point state. I suspect that is the case, as it is unlikely the division operation would differ.
Another advantage of isolating the code this way is that you could step through it in the debugger to see exactly where behavior diverges. You could step through it, one instruction at a time, with different floating-point states simultaneously in two windows and examine the results at each step to see exactly where the divergence is. If there is no divergence by the time you reach the log call, you should step through that, too.
Incidental notes:
If you know Xi and K are close to each other, it is better to compute log(Xi/K) as log1p((Xi-K)/K). When Xi and K are close to each other, the subtraction Xi-K is exact (has no error), and the quotient has more useful bits (the 1 that we already knew about and some zero bits following it are gone).
The fact that slight changes in your floating-point environment cause a .5% change in your result implies your calculations are very sensitive to error. This suggests that, even if you make your results reproducible, the errors that necessarily exist in floating-point arithmetic cause your result to be inaccurate. That is, the final error will still exist, it just will not be called to your attention by the difference between two different ways of calculating.
It appears in your C++ implementation that unsigned is four bytes but double is eight bytes. So printing the encoding a double by aliasing it to an unsigned omits half of the bits. Instead, you should convert a pointer to the double to a pointer to const char and print sizeof(double) bytes.
I have two integers and trying to pass them to cout.
int a =1;
int b= 3;
cout<<a&b;
Compiler tells:
Error 2 error C2676: binary '&' : 'std::basic_ostream<_Elem,_Traits>' does not define this operator or a conversion to a type acceptable to the predefined operator
But a&b returns int that is understandable for '<<' operator.
Why this error rises?
Due to operators precedence, you need to use parenthesis:
cout << (a & b)
The << operator binds more tightly than &, so omitting the parenthesis makes the
compiler undertand it as (cout << a) & b, which explains the error report: The & operator can't be used with a stream (the returned object from cout << a) and an int.
This is a precedence problem, if I'm not mistaken. Try using cout << (a&b); and see if it doesn't work a bit better.
you can do it like this: :) or did i misunderstood? (a)
int a =1;
int b= 3;
cout<<a << "&" << b;
I have a function in my program that outputs a data structure that consists of three doubles in two formats, one text and one binary.
When I run the program in debug and release modes, I end up with different binary outputs but identical text outputs. What is going on?
Here is the binary output code:
void outputPoints(xyz* points, string description, int length, param parameters)
{
stringstream index;
index.str("");
index << setw( 3 ) << setfill( '0' ) << parameters.stage;
string outputName = parameters.baseFileName + " " + index.str() + " " + description + ".bin"; // create file name
ofstream output; // create output object
cout << "Output " << outputName.c_str() << "...";
output.open(outputName.c_str(), std::ios::binary | std::ios::out); // open or create file for output
output.write(reinterpret_cast<char*>(points), (sizeof(xyz) * length));
output.close(); // close output object
cout << "done" << endl;
}
The debug build usually initializes variables with some patterns. Usually data allocated has the content CDCD, deleted objects are overwritten with FEEE. The CDCD pattern is overwritten when you initialize your variables. The release build doesn't initiliaze with these patterns.
It's worth to check your program for uninitialized variables. You can define a Dump function that just prints the (fist few bytes of) the suspected variables.
I don't know whether you got a solution for your issue and I did not look at your code.
I had the same issue because I was adding unsigned char and unsigned short and saving into unsigned short. I changed all variables to unsigned short and the issue was solved.