Bytewise reading of memory: “signed char *” vs “unsigned char *”

Question

One often needs to read from memory one byte at a time, like in this naive memcpy() implementation:

void *memcpy(void *dest, const void *src, si         


        

Answer 1

#include
#include

int main()
{

unsigned char a[4]={254,254,254,'\0'};
unsigned char b[4];
char c[4];

memset(b,0,4);
memset(c,0,4);

memcpy(b,a,4);
memcpy(c,a,4);
int i;
for(i=0;i<4;i++)
{
    printf("\noriginal is %d",a[i]);
    printf("\nchar %d is %d",i,c[i]);
    printf("\nunsigned char %d is %d \n\n",i,b[i]);
}

}

output is

original is 254
char 0 is -2           
unsigned char 0 is 254 


original is 254
char 1 is -2
unsigned char 1 is 254 


original is 254
char 2 is -2
unsigned char 2 is 254 


original is 0
char 3 is 0
unsigned char 3 is 0 

so here char and unsign both have the same value so it doesnt matter in this case

Edit

if you read anything as signed char still in that case most highre bit will also going to copy so it doesnt matter