Thread: sizeof function and pointers

I have this simple program:

Code:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>


static unsigned char cmd[]={
                0x01,0x80,0x00,0x00,0x00
             };


void printCmd(unsigned char cmd[])
{
	int i;
	
	printf("size: %zu", sizeof(cmd));
	for(i=0;i<5;i++)
	{
		printf("0x%x\n", cmd[i]);
	}
}


int main(int argc, char *argv[])
{
	int i,n;
	if ( argc != 2 )
	{
		printf("specify n sensors\n");
		return 1;
	}
	n = atoi(*++argv);
	
	for(i=1; i<n+1;i++)
	{	
		//reset cmd to default
		cmd[1] = 0x80; 
		
		cmd[1]|=i;
		printCmd(cmd);
		sleep(3);
	}


	return 0;
}

It gives this output:

size: 80x1
0x81
0x0
0x0
0x0
size: 80x1
0x82
0x0
0x0
0x0
size: 80x1
0x83
0x0
0x0
0x0

All is right except the size. Why does it give 80x1 as size instead of the digit 5?

You can't use sizeof on an array in a different scope than where you actually defined the array. Remember when you pass the array to your function the array decays to a pointer to the first element.

Also why are you using the global variable?

Jim