difference between “\\0” and '\\0'

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);

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:

Don't cast the return value of malloc() in C.
Don't scale allocations by sizeof (char), that's always 1 so it adds no value.
Pointers are not integers, you should compare against NULL typically.
The entire structure of the code makes no sense, there's an allocation in a loop but the pointer is thrown away, leaking lots of memory.

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'.

\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.

来源：https://stackoverflow.com/questions/40129319/difference-between-0-and-0

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null-character