c++ sizeof command in function not working

Question

I am a newbie to c++ and got the following code:

#include <stdio.h>

void func(int n, double *x);

int main() 
{
      int n;
      double x[] = { 0.0, 0.0, 0.0 };

      int xsize = sizeof(x)/sizeof(double);
      printf("\n xsize in main = %i \n",xsize);

      func( n, &*x );

      return 0;      
}


void func(int n, double *x)
{
      int xsize = sizeof(x)/sizeof(double);
      printf("\n xsize in func = %i \n",xsize);
}

The output is this:

xsize in main = 3

xsize in func = 1

My question is:

Why is xsize not 3 in func, but 1 ???

Thank you very much!

score 0 · Accepted Answer · answered Mar 24 '20 at 10:56

0

Why is xsize not 3 in func, but 1 ???

Because the function argument is a pointer. And the size of a pointer divided by the size of a double is 1 on your system and not 3.

answered Mar 24 '20 at 10:56

eerorika

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Thank you, that helps! Is there any way to modify my func code, so it gives me the size of x ? – redsquirrel Mar 24 '20 at 10:58
@redsquirrel You can pass the size as an argument. – eerorika Mar 24 '20 at 11:17

c++ sizeof command in function not working

1 Answers1