How to pass a pointer to a function that can change to addr of the pointer?

//in this way:

Code:

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void ChangeAddr( int **b ){
  *b = (int*) malloc( ... );
}

void main(){
  int *a ;
  ChangeAddr( a );
}

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// like this??
// THANK Q>

Well you shouldn't cast the return of malloc. Also you shouldn't use void main()
But yes that would do it.

Not quite.

Code:

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ChangeAddr( a ); -> ChangeAddr( &a );

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Heh I was looking at the ChangeAddr function I didn't even notice the call.

Quote:

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Originally Posted by Thantos

Well you shouldn't cast the return of malloc. Also you shouldn't use void main()
But yes that would do it.

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OK, well, now I've understood, FAQ:"The return type of main() must always be an int, this allows a return code to be passed to the invoker."

and ,i'm sorry, , What is the meaning of cast?

thx.

Quote:

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Originally Posted by L.O.K.

What is the meaning of cast?

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Preceding an expression by a parenthesized type name converts the value of the expression to the named type.

FAQ > Explanations of... > Casting malloc

Quote:

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Originally Posted by Dave_Sinkula

Preceding an expression by a parenthesized type name converts the value of the expression to the named type.

FAQ > Explanations of... > Casting malloc

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Oh, thx so much.