I created a program which at some point turned to be a buggy one and I can not find a good way to handle situations like in the following program:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int get_user_name( char *const dest, char *const src );
char *foo( char *variable ); /// foo() returns NULL
int main( void )
{
char arr[256] = { 0 };
char buffer[ 256 ] = { 0 };
char *const ptr = foo( arr ); /// foo() returned NULL here
if ( get_user_name( buffer, ptr ) == 0 ) /// get_user_name, should here return 0
{
printf("NULL returned from get_user_name()\n" );
exit( EXIT_FAILURE );
}else
{
printf( "Everithing is OK" );
}
}
int get_user_name( char *const dest, char *const src )
{
char *ret = strcpy( dest, src ); /// Here Segfault happens
/// There is no return here because of above Segfault
if ( ret == NULL )
{
return 0;
}else
{
return 1;
}
}
char *foo( char *variable )
{
if ( strlen( variable) < 1 )
{
return NULL; /// Here will return NULL because there is no Length
}
/// Some code here ...
return variable;
}
Here is use a demo function which returns NULL
to explain my problem.
I was thinking to check src
before past it to strcpy
but I really cannot understand why strcpy
returns dest
without check if src
is NULL
or not.
Why does this happens because I see that strcpy
manual say only about the return of dest
and nothing if fails:
RETURN VALUE
The strcpy() and strncpy() functions return a pointer to the destination string dest.
You handle the situation by first checking the parameters are not null. For example:
int get_user_name( char *dest, char *const src )
{
char *ret;
if (dest==0 || src==0) return 0;
strcpy( dest, src );
return 1;
}
Note also that dest
cannot be const
because it is written to.