Thread: check my c code.

I am trying to format a code to print hello fred, so far it only prints hello. the code cannot be replaced with a simple printf statement. Can anybody help me figure out how to print fred too. the code is

Code:

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

//
// A short program to print the string "Hello, Fred"
//

int main (int argc, const char *argv[])
{
	char name[5];
	char ch = name[5];

	char *last_char = &name[5];
	ch = *last_char;

	char *src = "Fred";
	char *dst;
	dst = name;

	// copy the name
	while (*src)
	{	
                src != 0;
                ++src;
		*dst = *src;
	}


	char *msg = (char *) malloc (5);
	strcpy (msg, "Hello, ");
	strcat (msg, name);
	puts(msg);

	//free (last_char);
	free (msg);
	*last_char = 'A';

       //free (msg);
	msg = (char *) malloc (5);
	msg = src;

	puts (msg);
	return 0;
}

Code:

char name[5];
char ch = name[5];

char *last_char = &name[5];
ch = *last_char;

...

*last_char = 'A';

name[5] is an area of memory that you shouldn't access. The valid indexes for the name array are 0 through 4. The last character of the array is name[4].

Code:

char *msg = (char *) malloc (5);

You should avoid casting the return value of malloc in C.

Code:

char *msg = (char *) malloc (5);
strcpy (msg, "Hello, ");
strcat (msg, name);

The string "Hello, " requires a buffer of 8 characters (7 + 1 for the terminating null) to store properly. Concatenating "Fred" requires an additional 4 spaces so your buffer must be at least 12 characters but you've only allocated 5.

Code:

msg = (char *) malloc (5);
msg = src;

This is a memory leak, you've allocated space but then repoint the pointer elsewhere immediately after the allocation. You've now lost the pointer to that memory location and have no way to free it.