String initialization with and without explicit trailing terminator

Question

What is the difference between

char str1[32] = \"\\0\";

and

char str2[32] = \"\";

Answer 1

Well, assuming the two cases are as follows (to avoid compiler errors):

char str1[32] = "\0";
char str2[32] = "";

As people have stated, str1 is initialized with two null characters:

char str1[32] = {'\0','\0'};
char str2[32] = {'\0'};

However, according to both the C and C++ standard, if part of an array is initialized, then remaining elements of the array are default initialized. For a character array, the remaining characters are all zero initialized (i.e. null characters), so the arrays are really initialized as:

char str1[32] = {'\0','\0','\0','\0','\0','\0','\0','\0',
                 '\0','\0','\0','\0','\0','\0','\0','\0',
                 '\0','\0','\0','\0','\0','\0','\0','\0',
                 '\0','\0','\0','\0','\0','\0','\0','\0'};
char str2[32] = {'\0','\0','\0','\0','\0','\0','\0','\0',
                 '\0','\0','\0','\0','\0','\0','\0','\0',
                 '\0','\0','\0','\0','\0','\0','\0','\0',
                 '\0','\0','\0','\0','\0','\0','\0','\0'};

So, in the end, there really is no difference between the two.