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# double to int, a bit confused...

 P: n/a Hi all I am writing a small app that uses real numbers all over the place for calculations. But when it comes to displaying values it is better to only display an it, (especially when it comes to % values and so on). so I first thought that the best way would be... .... double x = 1.23456789012345; int y = (int)x; but that does not always work, in fact when the double has a lot of decimal places I get a rather nasty looking number. I am assuming that the casting could get it wrong if there are too many decimal places. so I thought I'd use the ceil(...); and floor(...); functions .... y = (x>=0)?(int((floor(x))) : (int(ceil(x)))); the logic been that the decimal places are removed by both functions and the casting then works just fine. but that raises a few questions. Why does the cast not work some times? is the floor/ceil method "guaranteed" to work? What would be the preferred/best method? Many thanks in advance. Simon Jul 22 '05 #1
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 P: n/a "Simon" wrote in message news:2p************@uni-berlin.de... Hi all I am writing a small app that uses real numbers all over the place for calculations. But when it comes to displaying values it is better to only display an it, (especially when it comes to % values and so on). so I first thought that the best way would be... ... double x = 1.23456789012345; int y = (int)x; but that does not always work, in fact when the double has a lot of decimal places I get a rather nasty looking number. I am assuming that the casting could get it wrong if there are too many decimal places. so I thought I'd use the ceil(...); and floor(...); functions ... y = (x>=0)?(int((floor(x))) : (int(ceil(x)))); the logic been that the decimal places are removed by both functions and the casting then works just fine. but that raises a few questions. Why does the cast not work some times? is the floor/ceil method "guaranteed" to work? What would be the preferred/best method? The C++ Standard, 4.9[1] says that a rvalue of a floating point type can be converted to a rvalue of an integral type by truncating (discarding the fractional part). However, as expected, if the truncated value cannot be represented in the integral type, the behavior is undefined. Are you getting abnormal results when using very large double values ? Also, why are you using C-style casts ? It is preferred to use the C++ casts: double x = 1.234; int y = static_cast(x); Have you tested the code with other compilers ? Vladimir Ciobanu Jul 22 '05 #2

 P: n/a "Simon" wrote in message news:2p************@uni-berlin.de... I am writing a small app that uses real numbers all over the place for calculations. But when it comes to displaying values it is better to only display an it, (especially when it comes to % values and so on). so I first thought that the best way would be... ... double x = 1.23456789012345; int y = (int)x; but that does not always work, in fact when the double has a lot of decimal places I get a rather nasty looking number. Could you provide actual examples of these "nasty looking" numbers ? One thing that might be happening is that the number is too large to be stored in an int... I am assuming that the casting could get it wrong if there are too many decimal places. so I thought I'd use the ceil(...); and floor(...); functions ... y = (x>=0)?(int((floor(x))) : (int(ceil(x)))); the logic been that the decimal places are removed by both functions and the casting then works just fine. but that raises a few questions. Why does the cast not work some times? is the floor/ceil method "guaranteed" to work? What would be the preferred/best method? floor/ceil will work, but if the number is beyond the range of int, the same problem will occur when the conversion is made. This said, if you are only performing these conversions for display purposes, it may be possible to specify the format for the output of the double value instead of converting to an int. Consider for example: using namespace std; //after include double d = 12.3456; cout << setprecision(0) << setiosflags(ios_base::fixed) << d << endl; printf("%.0f\n",d); // if you still prefer c-style output hth, Ivan -- http://ivan.vecerina.com/contact/?subject=NG_POST <- email contact form Brainbench MVP for C++ <> http://www.brainbench.com Jul 22 '05 #3

 P: n/a "Simon" wrote in message news:2p************@uni-berlin.de... Hi all I am writing a small app that uses real numbers all over the place for calculations. But when it comes to displaying values it is better to only display an it, (especially when it comes to % values and so on). so I first thought that the best way would be... ... double x = 1.23456789012345; int y = (int)x; but that does not always work, in fact when the double has a lot of decimal places I get a rather nasty looking number. 1. How do you know it isn't working? 2. It works fine with g++ 3. Post a complete, small program that shows the symptom (example below) and we'll look at it further. #include int main( ) { double x = 1.23456789012345; int y = (int)x; std::cout << x << ":" << y << std::endl; return 0; } -- Gary Jul 22 '05 #4

 P: n/a > 1. How do you know it isn't working? I could see it. :). 2. It works fine with g++ I am using VC6++, on windows XP. (I know everybody has strong feeling about MS products but that's not the point here I hope). 3. Post a complete, small program that shows the symptom (example below) and we'll look at it further. Well, it seems to work fine on my XP pro machine. Before I have to totally bury my head in shame, I will try and look at more windows machine. The problem was reported on a Win98SE machine. I guess I'll also need to look if I am not trying to work with doubles that are out of int range. Regards Simon Jul 22 '05 #5

 P: n/a > The C++ Standard, 4.9[1] says that a rvalue of a floating point type can be converted to a rvalue of an integral type by truncating (discarding the fractional part). However, as expected, if the truncated value cannot be represented in the integral type, the behavior is undefined. Sorry, can I be rude and ask to see that section of the standard? Simon Jul 22 '05 #6

 P: n/a "Simon" wrote in message news:2p************@uni-berlin.de... The C++ Standard, 4.9[1] says that a rvalue of a floating point type can be converted to a rvalue of an integral type by truncating (discarding the fractional part). However, as expected, if the truncated value cannot be represented in the integral type, the behavior is undefined. Sorry, can I be rude and ask to see that section of the standard? Yes. Here it is: 4.9[1]: "An rvalue of a floating point type can be converted to an rvalue of an integer type. The conversion truncates; that is, the fractional part is discarded. The behavior is undefined if the truncated value cannot be represented in the destination type." Vladimir Ciobanu Jul 22 '05 #7

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