Array size without sizeof operator

Question

I am trying to understand program below, but it's not clear to me.

    #include<stdio.h>     int main()     {         int a[]={1,2,3,4,5,6,9};         printf("sizeof array is %d\n",sizeof(a));         printf("size of array using logic is %d\n",((&a)[1]-a));         printf("value of (&a)[1] is %p \n",(&a)[1]);         printf("value of a is %p \n",a);         printf("address of a[0] is %p\n",&a[0]);         printf("address of a[1] is %p\n",&a[1]);         printf("address of a[2] is %p\n",&a[2]);         printf("address of a[3] is %p\n",&a[3]);         printf("address of a[4] is %p\n",&a[4]);         printf("address of a[5] is %p\n",&a[5]);         printf("address of a[6] is %p\n",&a[6]);     } 

Above code output is :

    sizeof array is 28     size of array using logic is 7     value of (&a)[1] is 0x7ffc4888e78c      value of a is 0x7ffc4888e770      address of a[0] is 0x7ffc4888e770     address of a[1] is 0x7ffc4888e774     address of a[2] is 0x7ffc4888e778     address of a[3] is 0x7ffc4888e77c     address of a[4] is 0x7ffc4888e780     address of a[5] is 0x7ffc4888e784     address of a[6] is 0x7ffc4888e788 

It's not clear to me why ((&a)[1]-a)) on second print statement is returning 7; it should be 0x7ffc4888e78c - 0x7ffc4888e770 which is 0x1c i.e 28 total size of array.

For reference I also tried printing (&a)[1] and a values which you can see in code. I tried also debugging.

Answer 1

If you cast (&a)[1] and a to long before the calculation, then you will get your expected result. As haccks commented, you're currently calculating the pointer difference.

// These two sizes will be the same printf("sizeof array is %ld\n",sizeof(a)); printf("size of array using logic is %ld\n",((long)(&a)[1]-(long)a)); 

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Explaining the Math

What's happening in this case, is &a is considered to be type int(*)[7].

Then, you reference (&a)[1], which translates to *((&a)+1). In English, this means "give me the point in memory 1 after the beginning of a." As &a happens to be type int(*)[7], that point is at the end of the array.

When you subtract a, the pointer to the beginning of the array, you are performing pointer arithmetic and take a base the size of an int (because a is an int array). So the expression ((&a)[1]-a) is counting how many int are between (&a)[1] and a.

An overview on pointer arithmetic can be found here.