Difference between int* a; and int *a;

[sharensla]

is their any body tell me the Difference between int* a; and int *a; ?

[Dani]

I don't believe there is a difference. I think it's just a matter of coding standards.

[programmersbook]

It's the same.

[William Hemsworth]

Yep, almost all spaces (except those between keywords) are ignorable, so in the end the compiler would just see it as

int*a

anyway.

[Narue]

>is their any body tell me the Difference between int* a; and int *a; ?
From a language standpoint there's no difference. However, placing the asterisk by the variable makes more sense when you have multiple variables in a declaration:

int *p, *q;

The reason this makes more sense is because when the asterisk is paired with the type, it's easy to forget that it only applies to the first variable:

int* p, q; // q should be a pointer, but it's not

And of course, when you fix that error, it just looks funny:

int* p, *q;

If you only have a single variable per declaration, it's not an issue and you can do whatever you think makes more sense. The problems only show up with multiple variables per declaration.

[sharensla]

thank you very much

[gdharani7686]

wat happens if there is more than one variable

[corby]

dont matter. same thing its like sayin a*b versus a * b....its the same thing except for where the space is. the compiler wont care for a space in that situation

[Fbody]

dont matter. same thing its like sayin a*b versus a * b....its the same thing except for where the space is. the compiler wont care for a space in that situation

This is relevant to the OP and the previous post's questions how???
This:

int* a, b;

Is not the same as this:

int *a, *b;

The first one produces an integer pointer called "a" and an integer called "b". It is a confusing and misleading formatting style that makes it look like the '*' extends to "b" when, in actuality, it does not.

The second produces 2 integer pointers called "a" and "b". The '*' operator associates with the variable name, not the datatype. It's area of effect does not extend past the first comma operator (',') to the right.