Hello Sir,
I have a small question :
I want to find the size of structure, “node” with out sizeof operator.
In the code snippet mentioned below, the address of the 2 elements of array is printed and the offset is also correct, according to the size of the struct node.
However, when I subtract and print the offset value, the value is 1 instead of the actual size.
But it is correct in the second approach.
Can you please tell me what the issue in first case??
Code Snippet :
struct node {
int data;
struct node *link;
};
int main (){
struct node NodeAry[2], *ptrNode = 0;
printf (“Address of nodes is := %p %p\n”, &NodeAry[1],&NodeAry[0] ); //Prints correct difference
int size_of = (&NodeAry[1] – &NodeAry[0]); //<- This offset can be printed directly.
printf (“sizeof node is := %d\n”, size_of); //Wrong Size
*ptrNode++;
printf (“Sizeof Struct using ptr++ := %d\n”, ptrNode); //Correct Size
return 0;
}
Sample Output:
Address of nodes is := 0028ff34 0028ff2c
sizeof node is := 1 // <- This is wrong output.
Sizeof Struct using ptr++ := 8
I executed this in windows, using codeblocks and using gcc compiler.
Thanks in advance for the time took to go through this.
Thanks,
Jay
Dear S,
When you do ( &NodeAry[1] – &NodeAry[0] ), you are subtracting a pointer from another of the same type. The C compiler automatically divides the difference by the size of the item being pointed to. Hence you will always get value of 1 for any type of array for the expression above. This is simple to understand if you look at the following code snippet.
int x;
SomeType * p, * q;
q = p + x; // Note that the compiler adds ( x * sizeof (*p) ) to pointer p value to get q.
Hence, when you use the expression ( q – p ), you can see that the result must be x by simple transposition. This can be obtained only by divided the address difference by the sizeof (*p).
M.L.S. Shastry
Hi Sir,
Thanks for the explanatory answer.
In fact I was looking at the same matter as well.
I changed the line to : int size_of = ((int)&NodeAry[1] – (int)&NodeAry[0]);
And this gave me the expected result.
Thanks,
Jay
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