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This question already has an answer here:

When defining a function like this:

void  myFunction(arguments){
   // some instructions
}

What is the deference between using type name[] and type *name as the function's argument.

When using type name[] and calling the function does it make a copy to the name array or just point to the passed array.

Is the name array's subscript needs to be set. If not what's the difference.

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marked as duplicate by Kerrek SB, Marc B, Rahul Tripathi, Jens Gustedt, Shafik Yaghmour Nov 26 '13 at 18:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1  
@Kerrek_SB It's not the same question. –  Andrei B Nov 26 '13 at 13:54

3 Answers 3

up vote 1 down vote accepted

Either way only a pointer is passed.

void f(int array[]);

is synonymous to passing a const pointer

void f( int * const array);

They become non-synonymous when passing two dimensional arrays with bounds

void f(int array[][n]);

is synonymous to 

void f(int (* const array)[n]);

But different to 

void f(int **array);
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int array[][n] is an array of pointers to int where int **array is a pointer to pointer to int. Is that right? –  rullof Nov 26 '13 at 14:14
    
@rullof int array[][n] is an array of/a pointer to an undefined amount of arrays of n ints (single block of memory). While int ** array is an array of pointers to int (each of which can point to it's own separate block of memory holding an undefined amount of ints). –  Sergey L. Nov 26 '13 at 14:17
    
For the first one you'r right just it's an undefined amount of int not array of int. the second one (int **array) i don't think so, it's just a pointer to pointer to int. array is just the name otherwise you means something else –  rullof Nov 26 '13 at 14:23
3  
An int array[] parameter is not adjusted to int * const array. It is just int *array; there is no const. If the parameter were int array[const], then it would be adjusted to int * const array. Per C 2011 (N1570) 6.7.6.3 7: “A declaration of a parameter as ‘‘array of type’’ shall be adjusted to ‘‘qualified pointer to type’’, where the type qualifiers (if any) are those specified within the [ and ] of the array type derivation.” –  Eric Postpischil Nov 26 '13 at 14:59
1  
This is wrong. Eric Postpischil is right. array is int * –  newacct Nov 26 '13 at 22:44
up vote 1 down vote

There is no difference to my knowledge which is actually a problem in this situation:

int func(int a[20]) {
    sizeof(a); // equals sizeof(int*)!!
}

Therefore I always prefer the pointer form, as the array form can cause subtle confusion.

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up vote -2 down vote

In C++ it's the same. Different is in declaration;

char *s = "String"; // allocated and fill
char str[80];      // allocated memory
char *p1;          // not allocated memory
p1 = str;
share|improve this answer
    
The statements in this answer are about declarations within a block. The question is about function parameters, which behave differently. –  Eric Postpischil Nov 26 '13 at 15:01

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