is that valid to use (int*)calloc(size-1,2)?
is that valid to use (int*)calloc(size-1,2)?
I am making a program where user can edit array.
I am new in programming.code is not finished yet.
Can I use calloc(size-1,2)? Is that valid ?
Does it creates array size-1?
calloc(size-1,2)
size-1
#include <stdlib.h>
#include <stdio.h>
int main()
int *pointer1,*pointer2,size,i,choice,j=0,index;
printf("Enter Elements of Arrays " );
scanf("%d",&size);
pointer1=(int*)calloc(size,2);
for(i=0;i!=size;i++)
scanf("%d",pointer1+i);
printf("Enter your choice n" );
scanf("%d",&choice);
switch(choice)
case 0:
printf("Enter Index for Deletation ");
scanf("%d",&index);
/* I know that code is not finish. but calloc(size-1,2) is that valid or not ?*/
pointer2=(int*)calloc(size-1,2);
for(i=0;i!=size-1;i++)
if(i!=index)
*pointer2+i=*pointer1+i+j;
else
j++;
for(i=0;i<=size;i++)
printf("n%d",*pointer2+i);
break;
default:
printf("Error!");
return 0;
Your
calloc call seems assume that sizeof(int) is 2, which might not be correct and certainly is not portable. If sizeof(int) is in fact 2, your calloc call allocates space for size-1 elements, but your loop accesses elements at index size-1 and size which will be beyond the end of the memory allocated by calloc.– Ian Abbott
Sep 5 '18 at 10:19
calloc
sizeof(int)
sizeof(int)
calloc
size-1
size-1
size
calloc
Your loop is
for(i=0;i<=size;i++) so you certainly don't want size-1 elements in the array, you need size+1 elements. Moreover pointer1[i+j] is likely to index beyond the bounds of the memory allocation although we don't see the allocation for pointer1.– Weather Vane
Sep 5 '18 at 10:20
for(i=0;i<=size;i++)
size-1
size+1
pointer1[i+j]
pointer1
That first loop will never finish as you have
i-- and i++ cancelling each other out, so it'll get stuck when i==index– Chris Turner
Sep 5 '18 at 10:27
i--
i++
i==index
2 Answers
2
Since you do not tell us what is pointer2, it's difficult to answer.
pointer2
But, if you write:
int *pointer2 = NULL;
size_t size = 10;
if (size-1 > 0)
pointer2 = calloc(size-1, sizeof *pointer);
The pointer2 will points to an portion of memory able to store size-1 integer (int) values.
pointer2
size-1
int
[m|c|re]alloc
2
calloc(..., 2);
pointer2
Function signature for calloc is as below.
void * calloc( size_t num, size_t size );
Make sure that num > 0 and size > 0
num > 0
size > 0
thanks for ans! but can i do calloc(size-1,2);. user will enter size.
– Owais Qureshi
Sep 5 '18 at 10:08
@OwaisQureshi I suggest
pointer2 = calloc(size + 1, sizeof *pointer2);– Weather Vane
Sep 5 '18 at 10:26
pointer2 = calloc(size + 1, sizeof *pointer2);
@WeatherVane but i want to delete element! that's why calloc(size-1,sizeof(pointer2));. thank you.
– Owais Qureshi
Sep 5 '18 at 10:47
@OwaisQureshi your code is
for(i=0;i<=size;i++){ ... pointer2[i]= ... etc. In the final iteration you are indexing pointer2[size] and therefore there must be size+1 elements.– Weather Vane
Sep 5 '18 at 10:50
for(i=0;i<=size;i++){ ... pointer2[i]= ...
pointer2[size]
size+1
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It might be but it is not the best way. Please read stackoverflow.com/questions/7097678/… for better ideas
– Antti Haapala
Sep 5 '18 at 10:03