C语言：从“ int”到“ char”的隐式转换将值从12592更改为48

Question

I have a char array in C

char value_numbers [] = {'2', '3', '4', '5', '6', '7', '8', '9', '10'};

but I get the following error messages in XCode

Implicit conversion from 'int' to 'char' changes value from 12592 to 48
Multi-character character constant

Does anyone know what this means?

Answer 1

12592 is 0x3130. That suggests that your C compiler represents characters with ASCII and sets the values of multiple-character character constants in a straightforward way, as if each character were a digit in a base-256 numeral.

To initialize an element of value_numbers with this value, the compiler must convert 12592 to a char . If char is unsigned, this is effectively done by taking just the low eight bits, which are 0x30 or 48, the code for '0' . (Mathematically, the remainder modulo 256 is taken.) If char is signed, the C standard requires the C implementation to define the result of converting the value (which may include signaling an exception instead of producing a value and continuing). Wrapping modulo 256 to a representable value is common.

Since your source code '10' represents the value 12592, but the compiler was forced to store a different value in the array, it warns you.

Note that the actual character encoding is implementation dependent (0 is 48 in ASCII but not in EBCDIC).