Possible Duplicate:
Sizeof an array in the C programming language?
I have been working on this for long, but can't seem to understand why this is happening.
void fun(char []);
void fun(char a[])
{
printf("Size is %d\n",sizeof(a));
}
int main ()
{
char arr[10] = { 0 };
printf("Size is %d\n",sizeof(arr));
fun(arr);
return 0;
}
Why is the Output:
Size is 10
Size is 4
The first sizeof reports the size of a pointer, the second the size of an array.
A parameter declared with array syntax (void fun(char [])) is actually a pointer parameter. The [] is just syntactic sugar.
It's because arrays when passed in functions get decayed into pointers in functions.
sizeof in fun() returns the size of the pointer a. Read about Arrays and Pointers.
a in fun() is actually a pointer copied by value.
You should pass the size of the array with as a separate argument.
You should try it this way:
#include <stdio.h>
void fun(char []);
void fun(char a[], int n)
{
// printf("Size is %d\n",sizeof(arr));
printf("Size is %d\n",n);
}
int main ()
{
char arr[10] = { 0 };
const int truesize = sizeof(arr)/sizeof(arr[0]); // try to understand this statement what it does - sizeof returns size in bytes
printf("Size is %d\n",truesize);
fun(arr,truesize); // This is how you should call fun
return 0;
}
sizeof is an operator, it computes it value at compile time. sizeof(char[]) is the size of the pointer.
