basic question on *=

[sanctus]

I'm now since a couple of weeks debugging a program and am now at the point where I don't know anymore what could be wrong (actually it compiles but does not do what I want)...
Anyway, if i want to write a=b*(c+d) and write the following code:

a=b;
a*=c+d;

does the program what I want or does it see a=b*c+d?
I really hope it is the latter possibility, because that risks to solve my problem... thanks in advance either way

[Savage]

I'm now since a couple of weeks debugging a program and am now at the point where I don't know anymore what could be wrong (actually it compiles but does not do what I want)...
Anyway, if i want to write a=b*(c+d) and write the following code:

a=b;
a*=c+d;

does the program what I want or does it see a=b*c+d?
I really hope it is the latter possibility, because that risks to solve my problem... thanks in advance either way

It's same as a=a*(c+d) and becaue of a=b it is same as a=b*(c+d);

Savage

[Silent1Mezzo]

I'm now since a couple of weeks debugging a program and am now at the point where I don't know anymore what could be wrong (actually it compiles but does not do what I want)...
Anyway, if i want to write a=b*(c+d) and write the following code:

a=b;
a*=c+d;

does the program what I want or does it see a=b*c+d?
I really hope it is the latter possibility, because that risks to solve my problem... thanks in advance either way

a=b;
a*=c+d;

Does what you want it a = b * (c + d)
If you want it to be more readable you could just assign c+d to another variable and then do a *= var; That would produce the same results but maybe be a little more easy to read.

PS Savage beat me to it.

[sanctus]

It's same as a=a*(c+d) and becaue of a=b it is same as a=b*(c+d);

Savage

So a*=c+d is equivalent to a*=(c+d) ?

EDIT: and you beat me to my answer to savage :)

Thanks very much both of you.

[weaknessforcats]

This code:

a=b;
a*=c+d;

is equivalent to:
[code=cpp]
a = b*(c+d);
[/code]

This is a precedence issue. The *= operator has a precedence of 16 whereas addition has a precedence of 6. That means the addition is done before the *=.

Personally, I like your original method:

[code=cpp]
a = b*(c+d);
[/code]

because it's easier to read.

[ssorower]

Just thought that this precedence list

http://www.cppreferenc e.com/operator_preced ence.html

might help to justify why a*=c+d is equivalent to a=a*(b+c)

--Sorower

[sanctus]

T

Personally, I like your original method:

[code=cpp]
a = b*(c+d);
[/code]

because it's easier to read.

I agree that it easier to read, but if the b,c,d terms are very long (about one line) and the program is huge (260 MB of code) and you haven't written the program just modified it, which means that you might not know all the variables' names defined elsewhere and inherited to where you are, makes it difficult and risky to define for every b,c,d term a variables. So it leaves you no choice.
I would though never do it if it would be easily avoidable...

Also thanks about the explication of why via the precedences.

[sanctus]

Just thought that this precedence list

http://www.cppreferenc e.com/operator_preced ence.html

might help to justify why a*=c+d is equivalent to a=a*(b+c)

--Sorower

Thanks, such a list clears up also some other things, bookm arked it right away.

[Savage]

a=b;
a*=c+d;

Does what you want it a = b * (c + d)
If you want it to be more readable you could just assign c+d to another variable and then do a *= var; That would produce the same results but maybe be a little more easy to read.

PS Savage beat me to it.

Muhahahahahah

Savage

[Savage]

So a*=c+d is equivalent to a*=(c+d) ?

EDIT: and you beat me to my answer to savage :)

Thanks very much both of you.

We are more than happy to help u.

Savage