Re: [algogeeks] pointer and array
array is passed a pointer in the function, hence sizeof(arr)==sizeof(*arr)
On Fri, Jul 23, 2010 at 9:10 PM, tarak mehta tarakmeht...@gmail.com wrote:
int arr[]={1,2,3,4};
k=sizeof(arr)/sizeof(*arr);
value of k=4;
however
void hell(int arr[]);
main()
{
int arr[]={1,2,3,4};
hell(arr);
}
void hell(int arr[])
{
printf(%d,sizeof(arr)/sizeof(*arr));
}
output of hell() is 1. why???
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Re: [algogeeks] pointer and array
@tarak mehta: if u wanna understand, try passing a char array to a function n do de same... On Sun, Jul 25, 2010 at 9:59 AM, Manjunath Manohar manjunath.n...@gmail.com wrote: @Apporve... yeah u r right :)sizeof ptr is always 2 in 16 bit compilers, i.e, the sizeof an address is 2.and the sizeof(int)=2..i.e sizeof(*arr)=2..hope u got it now.. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.comalgogeeks%2bunsubscr...@googlegroups.com . For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] pointer and array
sizeof(arr) is 4.. o.e the number of elements in the array
size of *arr is the size of any normal pointer i.e 4(in case of 32-bit
compilers)
so the answer is 1
On Sat, Jul 24, 2010 at 9:52 AM, ravi gupta ravikant0...@gmail.com wrote:
On Sat, Jul 24, 2010 at 9:40 AM, tarak mehta tarakmeht...@gmail.comwrote:
int arr[]={1,2,3,4};
k=sizeof(arr)/sizeof(*arr);
value of k=4;
however
void hell(int arr[]);
main()
{
int arr[]={1,2,3,4};
hell(arr);
}
void hell(int arr[])
{
printf(%d,sizeof(arr)/sizeof(*arr));
}
output of hell() is 1. why???
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When array is passed as an argument, only a pointer to the first element
of the array is passed. Therefore the parameter int arr[] in void hell(int
arr[]) is just a pointer, hence the result .
I hope it answers your query.
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Re: [algogeeks] pointer and array
when arrays are passed as arguments to a function,the starting address of the array is passed like a pointer, thus sizeof(arr)=2..thus 2/2=1..this is the precise reason for always specifying the column length in the definition of function when functions have arrays as one of the arguments.. Hope i made any sense.. :) -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] pointer and array
void hell(int arr[]);
main()
{
int arr[]={1,2,3,4,5};
hell(arr);
}
void hell(int arr[])
{
printf(%d,sizeof(arr)/sizeof(*arr));
}
even this gives 1 !!
@manjunath ur idea seems correct..but could u plz elaborate a bit
On Sat, Jul 24, 2010 at 10:51 PM, Manjunath Manohar
manjunath.n...@gmail.com wrote:
when arrays are passed as arguments to a function,the starting address of
the array is passed like a pointer,
thus sizeof(arr)=2..thus 2/2=1..this is the precise reason for always
specifying the column length in the definition of function when functions
have arrays as one of the arguments..
Hope i made any sense.. :)
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Re: [algogeeks] pointer and array
@tarak:
You can see it like this. When we create an array then 'a' points to the
whole array not just the first element so it returns the size of the whole
array.
when you pass the array though by default in c it is pass by value but as
you are passing the address of the array so it acts like pass by reference.
So the formal parameter acts as a pointer when you pass the address of the
array to it.
And you know that the size of a pointer is always equal to the size of int.
On Sun, Jul 25, 2010 at 12:31 AM, tarak mehta tarakmeht...@gmail.comwrote:
void hell(int arr[]);
main()
{
int arr[]={1,2,3,4,5};
hell(arr);
}
void hell(int arr[])
{
printf(%d,sizeof(arr)/sizeof(*arr));
}
even this gives 1 !!
@manjunath ur idea seems correct..but could u plz elaborate a bit
On Sat, Jul 24, 2010 at 10:51 PM, Manjunath Manohar
manjunath.n...@gmail.com wrote:
when arrays are passed as arguments to a function,the starting address of
the array is passed like a pointer,
thus sizeof(arr)=2..thus 2/2=1..this is the precise reason for always
specifying the column length in the definition of function when functions
have arrays as one of the arguments..
Hope i made any sense.. :)
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regards
Apoorve Mohan
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Re: [algogeeks] pointer and array
@Apporve... yeah u r right :)sizeof ptr is always 2 in 16 bit compilers, i.e, the sizeof an address is 2.and the sizeof(int)=2..i.e sizeof(*arr)=2..hope u got it now.. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] pointer and array
On Sat, Jul 24, 2010 at 9:40 AM, tarak mehta tarakmeht...@gmail.com wrote:
int arr[]={1,2,3,4};
k=sizeof(arr)/sizeof(*arr);
value of k=4;
however
void hell(int arr[]);
main()
{
int arr[]={1,2,3,4};
hell(arr);
}
void hell(int arr[])
{
printf(%d,sizeof(arr)/sizeof(*arr));
}
output of hell() is 1. why???
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When array is passed as an argument, only a pointer to the first element of
the array is passed. Therefore the parameter int arr[] in void hell(int
arr[]) is just a pointer, hence the result .
I hope it answers your query.
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