Implicit conversion between pointers?

I am reading K&R book. In section 5.3 there is this code:

Code:

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    int a[10];
    int *pa;
    pa=a;

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I think that should be:

Code:

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    int a[10];
    int *pa;
    pa=(int *)a;

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Because pa is a pointer to an integer and a is a pointer to an array of integers. Of course the first code works maybe because it happens an implicit cast (conversion). Am I right?

thanks

A pointer to an array of integer is an int *, so there is no need for conversion, since pa and a are of the same type.

Thanks

Quote:

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Originally Posted by nonlinearly

Because pa is a pointer to an integer and a is a pointer to an array of integers.

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pa is a pointer to an int. a is an array, not a pointer to an array of integers. This declares a pointer to an array of 10 int:

Code:

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int (*p)[10];

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Quote:

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Originally Posted by nonlinearly

Of course the first code works maybe because it happens an implicit cast (conversion). Am I right?

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Yes, a is converted to a pointer to its first element, i.e., a pointer to an int. The cast is unnecessary.

Quote:

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Originally Posted by Tibo-88

A pointer to an array of integer is an int *, so there is no need for conversion, since pa and a are of the same type.

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No, that is not true. pa and a are not of the same type. It is easy to demonstrate by running this program:

Code:

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#include <stdio.h>

int main(void)
{
    int a[10];
    int *pa;
    printf("%u %u\n", sizeof(a), sizeof(pa));
    return 0;
}

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if a and pa were of the same type, then sizeof(a) == sizeof(pa) must be true.

Sorry for my mistake, I guess you're right! But what is the difference of type then? They're both int* and if you do pa=a, you can access pa[x] as well as if you did a[x], so pa sort of becomes an array too. Could you explain me that please?

Quote:

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Originally Posted by Tibo-88

But what is the difference of type then? They're both int* and if you do pa=a, you can access pa[x] as well as if you did a[x], so pa sort of becomes an array too. Could you explain me that please?

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As I mentioned, a is converted to a pointer to its first element. Concerning the notation, pa[x] is equivalent to *(pa + x).

I came back...
I think that the original question should be with 2 dimensional array like:

Code:

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int array[2][3];
int (*p2a)[3];
p2a=(int (*)[3])array;

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In this situation the cast is necessary...!!!

No, the cast is still unnecessary.

Sorry my fault.. I will correct it

Huh? What are you trying to do now? Your code snippet is wrong because a pointer to an array of 3 ints is not convertible to a pointer to an int.

SORRY... I meant

Code:

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int array[2][3];
int *ptr;
ptr=(int *)array;

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Ah, then yes, there is no implicit conversion.

The difference is that with this approach we can not refer to an element using subscript notation (for example ptr[1][2] is not valid).
thanks for your responses... you have helped me to understand very well...

Quote:

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Originally Posted by nonlinearly

The difference is that with this approach we can not refer to an element using subscript notation (for example ptr[1][2] is not valid).

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It isn't really a difference in approach: it is the difference between using a pointer to an array of int and using a pointer to an int.

Quote:

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Originally Posted by nonlinearly

thanks for your responses... you have helped me to understand very well...

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You're welcome :)