Reputation: 111
I am trying to understand following piece of code, but I am confused between "\0"
and '\0'
.I know its silly but kindly help me out
#define MAX_HISTORY 20
char *pStr = "\0";
for(x=0;x<MAX_HISTORY;x++){
str_temp = (char *)malloc((strlen(pStr)+1)*sizeof(char));
if (str_temp=='\0'){
return 1;
}
memset(str_temp, '\0', strlen(pStr) );
strcpy(str_temp, pStr);
Upvotes: 9
Views: 4416
Reputation: 399753
Double quotes create string literals. So "\0"
is a string literal holding the single character '\0'
, plus a second one as the terminator. It's a silly way to write an empty string (""
is the idiomatic way).
Single quotes are for character literals, so '\0'
is an int
-sized value representing the character with the encoding value of 0.
Nits in the code:
malloc()
in C.sizeof (char)
, that's always 1 so it adds no value.NULL
typically.Upvotes: 10
Reputation: 213513
\0
is the null terminator character.
"\0"
is the same as {'\0', '\0'}
. It is a string written by a confused programmer who doesn't understand that string literals are always null terminated automatically. Correctly written code would have been ""
.
The line if (str_temp=='\0')
is nonsense, it should have been if (str_temp==NULL)
. Now as it happens, \0
is equivalent to 0, which is a null pointer constant, so the code works, by luck.
Taking strlen
of a string where \0
is the first character isn't very meaningful. You will get string length zero.
Upvotes: 6
Reputation: 121357
They are different.
"\0"
is a string literal which has two consecutive 0's and is roughly equivalent to:
const char a[2] = { '\0', '\0' };
'\0'
is an int
with value 0. You can always 0 wherever you need to use '\0'
.
Upvotes: 12