memcpy copies array object to character pointer in C but not vice versa

Question

I am attempting to copy a character from an array to a character pointer by means of a function. Following is my code:

#include<stdio.h>
void pass(void* arg){
    char arr[1];
    arr[0]='X';
    memcpy((char *)arg,arr,1);
}

void main(){
    char *buf=malloc(sizeof(char));
    pass(buf);
    printf("%c",buf);
    free(buf);
}

Here, the pointer buf prints a garbage value. I tried the reverse of the above program as follows:

#include<stdio.h>
void pass(void* arg){
    char arr[1];
    memcpy(arr,(char *)arg,1);
    printf("%c",arr[0]);
}

void main(){
    char *buf=malloc(sizeof(char));
    buf[0]='X';
    pass(buf);
    free(buf);
}

This was successful and arr[0] prints the value as X. Could someone clarify why does the first case fail?

Turn your warnings knob up a notch.

n. m. could be an AI
– n. m. could be an AI

2015-12-13 09:54:51 +00:00
Commented Dec 13, 2015 at 9:54 — n. m. could be an AI
– n. m. could be an AI, Commented Dec 13, 2015 at 9:54

MikeCAT · Accepted Answer · 2015-12-13 05:24:01Z

3

You have to print *buf or buf[0] instead of buf to print the first element of the array pointed by buf.

Also note that passing data having type char* to be printed via %c in printf(), which expect data having type int, is undefined behavior due to type mismatch.

Try this:

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
void pass(void* arg){
  char arr[1];
  arr[0]='X';
  memcpy((char *)arg,arr,1);
}
int main(void){
  char *buf=malloc(sizeof(char));
  pass(buf);
  printf("%c",buf[0]);
  free(buf);
  return 0;
}

answered Dec 13, 2015 at 5:24

MikeCAT

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Comments

paxdiablo · Accepted Answer · 2015-12-13 05:30:43Z

3

Your problem lies here:

char *buf=malloc(sizeof(char));
:
printf("%c",buf);

because buf is a pointer to a character, not a character. The character is *buf, or buf[0] so you can fix it with:

printf ("%c", *buf);

As an aside, it's totally unnecessary to create arr and call memcpy in the first example when you can simply do:

void pass (void* arg) {
    ((char*)arg)[0] = 'X';
}

Dynamic memory allocation for a character in this case is also unnecessary when you can just replace the malloc line with a simple:

char buf[1];

(and get rid of the free line, of course).

edited Dec 13, 2015 at 5:30

answered Dec 13, 2015 at 5:25

paxdiablo

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