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Yes, I was talking about what is described in your resources. You can do this: // define a floating-point literal in hex and print it in decimal float
by enriquto 3y ago
Yes, I was talking about what is described in your resources. You can do this:
// define a floating-point literal in hex and print it in decimal
float x = 0x1p-8; // x = 1.0/256
printf("x = %g\n", x); // prints 0.00390625
// define a floating point literal in decimal and print it in various ways
float y = 0.3; // non-representable, rounded to closest float
printf("y = %g\n", y); // 0.3 (the %g format does some heuristics)
printf("y = %.10f\n", y); // 0.3000000119
printf("y = %.20f\n", y); // 0.30000001192092895508
printf("y = %a\n", f); // 0x1.333334p-2
- codetrotter 3y agoSo for example if you make a variable that has the value parent commenter used 100000.000000000017 And then you print it. Does it preserve the exact value?
- enriquto 3y agoYour question is ambiguous for two different reasons. First, this value is not representable as a floating-point number, so there's no way that you can even store it in a float. Second, once you have a float variable, you can print it in many different ways. So, the answer to your question is, irremediably, "it depends what you mean by exact value". If you print your variable with the %a format, then YES, the exact value is preserved and there is no loss of information. The problem is that the literal that you wrote cannot be represented exactly. But this is hardly a fault of the floats. Ints have exactly the same problem: int x = 2.5; // x gets the value 2 int y = 7/3; // same thing
- codetrotter 3y agoSo in other words, is it fair to say that this situation is not much different from what you get with Python?