C output 问题。谁能解释一下这个 output 吗？

Question

#include<stdio.h>
void main()
{
    printf("%d",'AA');
}

I was expecting an error there but the program ran and output was 16705. Can anyone please explain this?

Answer 1

Can anyone please explain this?

The 'AA' is a multi-character character constant. It has the type int . It's value is implementation-defined.

"Implementation" here is the compiler and your compiler has rules to which int value 'AA' is mapped to. The mapping seems to be easy. Because I don't know your compiler, I am guessing it. Consult your compiler documentation to be sure.

'AA' maps to a value 'A' << 8 | 'A' 'A' << 8 | 'A' . Bit shifted 'A' by a byte with another 'A' . You system most probably uses ASCII to represent characters. The 'A' maps in ASCII to a value 65 in decimal ( 0x41 in hex). Calculating 0x41 << 8 | 0x41 0x41 << 8 | 0x41 gives the value of 16705 in decimal. Because this is an int value, you can use %d to print the result. So your code is equivalent to printf("%d\n", 16705) .

Answer 2

'AA' is an exotic beast. It's a character literal, but ASCII does not have a single character 'AA' . This explains why you get a non-ASCII value instead.

Answer 3

this code run because the char data type is a number and you have request ho print the real number of 'AA'

https://en.wikipedia.org/wiki/C_data_types