 Live2Win Korea (South). February 02 2008 13:29. Posts 6016 | Profile Blog | |
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 Chill Canada. February 02 2008 13:35. Posts 12186 | Profile Blog |
I think
X _= Y means X = X _ Y
So in your example: a = a*b |
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 Saracen United States. February 02 2008 13:37. Posts 2310 | Profile Blog |
yeah...
x+=5
is
x=x+5
so i guess what chill said  |
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| Saracen United States. February 02 2008 13:40. Posts 2310 | Profile Blog | |
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 Daveed United States. February 02 2008 13:42. Posts 226 | Profile Blog |
| If this is a Java question, why don't you open up eclipse or whatever you use, and try some of those expressions? |
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 FreeZEternal Korea (South). February 02 2008 13:45. Posts 3005 | Profile | |
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 fusionsdf Canada. February 02 2008 13:51. Posts 14042 | Profile Blog |
you ask about multiplication but not modulo? strange 
Last edit: 2008-02-02 13:53:01 |
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| Saracen United States. February 02 2008 13:52. Posts 2310 | Profile Blog |
On February 02 2008 13:51 fusionsdf wrote:you ask about multiplication but not modulo? strange
modulus doesn't really matter in this question |
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| fusionsdf Canada. February 02 2008 13:53. Posts 14042 | Profile Blog |
yeah but I always found +=, -=,/=, *= intuitive.
modulo confused me quite a bit when I first saw it :O |
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| FreeZEternal Korea (South). February 02 2008 14:03. Posts 3005 | Profile |
| The line c = a % 3 + c / b; is useless here. -.- |
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| Saracen United States. February 02 2008 14:19. Posts 2310 | Profile Blog |
On February 02 2008 14:03 FreeZEternal wrote:
The line c = a % 3 + c / b; is useless here. -.-
exactly
that's why b=2 and c=1 |
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| Live2Win Korea (South). February 02 2008 14:25. Posts 6016 | Profile Blog |
On February 02 2008 14:19 Saracen wrote:
Show nested quote +On February 02 2008 14:03 FreeZEternal wrote:
The line c = a % 3 + c / b; is useless here. -.-
exactly that's why b=2 and c=1
why? I don't get why it's useless.
a *= b;
a = a * b
a = 5 * 2
so,
a = 10
c = a % 3 + c / b
c = (10 % 3) + (6 / 2)
c = (1) + (3)
so,
c = 4
c = b--;
c = 2--
so,
c = 2
b = 1
Thus, a = 10, b = 1, c = 2
I'm not sure what "d" is. What's the default for that value? |
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| fusionsdf Canada. February 02 2008 14:33. Posts 14042 | Profile Blog |
On February 02 2008 14:25 Live2Win wrote:
Show nested quote +On February 02 2008 14:19 Saracen wrote:
On February 02 2008 14:03 FreeZEternal wrote:
The line c = a % 3 + c / b; is useless here. -.-
exactly that's why b=2 and c=1
why? I don't get why it's useless. a *= b; a = a * b a = 5 * 2 so, a = 10 c = a % 3 + c / b c = (10 % 3) + (6 / 2) c = (1) + (3) so, c = 4 c = b--; c = 2-- so, c = 2 b = 1 Thus, a = 10, b = 1, c = 2 I'm not sure what "d" is. What's the default for that value?
because you are setting c to some long equation
and then the very next step you set c to one less than b
In other words, you saved a value to c, didnt use it, and saved over it
as for d, it depends on the language. Java protects you, so it should be 0.0Last edit: 2008-02-02 14:34:01 |
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| fusionsdf Canada. February 02 2008 14:38. Posts 14042 | Profile Blog |
for:
public static void main(String args[]) {
char alpha;
alpha = “g”;
}
you have to use single quotes for chars (at least you do in C++ and I assume its the same here)
so it should be
public static void main(String args[]) {
char alpha;
alpha = 'g';
}
or you can combine the two steps to
public static void main(String args[]) {
char alpha = 'g';
}
which will initialize it with a value.
as a follow up to d, java protects you, but some languages don't. be careful relying on the value of a variable you didnt give a value to yet :O |
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 Macavenger United States. February 02 2008 14:49. Posts 911 | Profile Blog |
| For people saying the final variables are c=1 and b=2, look at it more carefully. b-- means evaluate b, then decrement b. At the start of that step b = 2, so c is set to 2, then b is decremented, giving c = 2 and b = 1. Most of you are doing c = b - 1; which is very different from c = b--;. For c to end up 1 using the -- operator, it would need to be c = --b; which would set both b and c to 1. |
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 BottleAbuser Korea (South). February 02 2008 16:29. Posts 1386 | Profile Blog |
+ Show Spoiler +
double a = 5, b = 2, c = 6, d;
a *= b; // equivalent to a = a * b, or a = 5 * 2, or a = 10
c = a % 3 + c / b; // equivalent to c = (10 mod 3) + (6 / 2), or c = (1) + (3), or c = 4
c = b– –; //equivalent to c = b; b = b - 1, or c = 2, b = 2 - 1, or c = 2; b = 1
End values:
a = 10
b = 1
c = 2
d is uninitialized.
Also, to set char alpha to lowercase g:
this will work:
char alpha = 'g';
this will also work:
char alpha; alpha = 'g';
this also will work:
char alpha; alpha = 71; // 71 is ASCII for 'g' Last edit: 2008-02-02 16:32:06 |
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| BottleAbuser Korea (South). February 02 2008 16:54. Posts 1386 | Profile Blog |
Um, what? You'll get a compile error if you try to access d, I think. d is uninitialized.
If d wasn't a primitive type, it would be equal to null. Since it is a primitive type, its value is uninitialized and you'd get a compile error. |
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 prOxi.swAMi Australia. February 02 2008 17:08. Posts 1941 | Profile Blog |
oh damn im way too late huh? T_T
anyway bottleabuser is right.
compiler would kick your face for trying to make use of d when there's no gaurantee that it won't be null. However, assuming d was initialized as 0, you'd get a = 10 b = 1 c = 2 d = 0
and the char one .. so ez.
char alpha = 'g';
is this for school?Last edit: 2008-02-02 17:24:27 |
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 Cambium Canada. February 02 2008 17:15. Posts 9999 | Profile Blog |
it's post decrement,
so c is 1 more than b.
and BottleAbuser is right. You can't compile your code unless you have d initialized. |
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