simple C++ question

I am a beginning C++ programmer and I would like to know one thing about variables that is not clear in the book I have.

When I declare a variable as double do I have to include a decimal?

for example in the book they use :

double speed;
speed = 139000.0;

if you enter speed = 139000 , will it not still remain a double value since it was declare as double?

This last example my better illustrate my question.
suppose I declare the wide char variable 'a' like so:

wchar_t = ch;
ch = L'a';

since I already declared the variable as wide, why do I have to still include the L in front of 'a'?

does this mean that my double variable is converted to an int value because it has no decimal?

Hi code_slinger and welcome to the forum and to c++ :wave:

Well, in the case of the double, it's just more clear and accurate to assign a double to a double. You could use:
speed = 139000;
which converts the 13900 to a double and the assigns it to the variable speed. So why not give a double from the beginning?

The point is that you can make errors here easily. For example:
speed = 99/10;
will result in "speed=10.0", because the compiler sees 2 integer values getting devided (99/10 = 9.9) and because they are integer, the compiler rounds them to 10. So the correct version would be:
speed = 99.0/10.0;

See the point?


It's equal with
wchar_t = ch;
ch = L'a';

'a' is a char (not a wchar_t), so you have to convert it. In this case I think the compiler isn't able to convert a wchar from a char, and
ch='a'
will fail.

Thank you, I believe I understand now. So the conversion to double happens after the variable value is "stored"? From your example if you assign speed = 99/10 the 99 and 10 are not converted before the division so the compiler rounds as it would with integers.

Thak you very much, I wish the book had used that. I guess its obvious?
anyway good example. (am I understanding the example correctly?)

Quote:

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Originally Posted by code_slinger

Thank you, I believe I understand now. So the conversion to double happens after the variable value is "stored"? From your example if you assign speed = 99/10 the 99 and 10 are not converted before the division so the compiler rounds as it would with integers.

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Exactly, if you wanted to calculate the exact division you should do speed = 99.0/10.0 or speed = 99/10.0. In the last example the compiler will auto-cast the 99 to double because you divide by a double.

Quote:

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Originally Posted by martho

speed = 99/10;
will result in "speed=10.0"

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It results in 9, not 10. Integer division truncates the fractional portion. Rounding of native types never occurs in C++ unless you call a function to implement it. The op should Google for "Usual Arithmetic Conversions" and "C++" for starters.

Quote:

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Originally Posted by Sef

It results in 9, not 10. Integer division truncates the fractional portion. Rounding of native types never occurs in C++ unless you call a function to implement it. The op should Google for "Usual Arithmetic Conversions" and "C++" for starters.

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Exactly.
As a side note, a common trick to perform rounding by exploiting the truncate feature is:

Code:

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int i = int(99/10+0.5);

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The above is for positive numbers. For negative numbers, you need to subtract 0.5.

[QUOTE=Marc G]Exactly.
As a side note, a common trick to perform rounding by exploiting the truncate feature is:

Code:

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int i = int(99/10+0.5);

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Won't work for this particular example (as is) but I know what you meant :)

Quote:

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Originally Posted by Sef

Won't work for this particular example (as is) but I know what you meant :)

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I know, as I said it was a side note on your "Rounding of native types never occurs in C++" :wave: