The ASCII value of '\0' is same as ASCII value of 0?

Question

I learned the ASCII value of '\0' is 0, and the ASCII value of 0 is 0x30, but when I try to print their ASCII value using printf, I get the same output:

printf("\'\\0\' : %d\n", '\0');
printf("\'\\0\' in hex : %x\n", '\0');
printf("0 : %d\n", 0);
printf("0 in hex: %x\n", 0);

output:

'\0' : 0
'\0' in hex : 0
0 : 0
0 in hex: 0

why?

Answer 1

The ASCII character '0' is different than the number 0. You are printing the integer 0 in your second pair of printfs instead of '0'.

Try this:

printf("'\\0' : %d\n", '\0');
printf("'\\0' in hex : %x\n", '\0');
printf("'0' : %d\n", '0');
printf("'0' in hex: %x\n", '0');

Also, you don't need to escape ' inside strings. That is, "'" is fine and you don't need to write "\'"

Answer 2

You confuse 0, '\0', and '0'.

The first two of these are the same thing; they just represent an int with value 0.

'0', however, is different, and represents an int with the value of the '0' character, which is 48.

Answer 3

printf("0 : %d\n", '0');

Print like this, you will get 48.

Answer 4

Yes, the character literal'\0' has value 0 and this is independent of the character set.

(K&R2, 2.3) "The character constant '\0' represents the character with value zero, the null character. '\0' is often written instead of 0 to emphasize the character nature of some expression, but the numeric value is just 0."

Answer 5

All of these answers seem to miss the point of \0 ...

Try this:

printf("This line\0 might not completely appear");

That's the reason for \0

So you can define a NULL terminated string ...

#define TEST_STRING "Test"

is quite different from

#define TEST_STRING "Test\0"

the latter being useful in certain cases.

Answer 6

The ASCII value for '\0' is indeed 0. But '\0' is different from '0' (notice the backslash, which is the escape symbol).

Answer 7

Your last line should be

printf("0 in hex: %x\n", '0');