What causes a char to be signed or unsigned when using gcc?

The default depends on the platform and native codeset. For example, machines that use EBCDIC (mainframes usually) must use unsigned char (or have CHAR_BIT > 8) because the C standard requires characters in the basic codeset to be positive, and EBCDIC uses codes like 240 for digit 0. (C11 standard, §6.2.5 Types ¶2 says: An object declared as type char is large enough to store any member of the basic execution character set. If a member of the basic execution character set is stored in a char object, its value is guaranteed to be nonnegative.)

You can control which sign GCC uses with -fsigned-char or -funsigned-char options. Whether that’s a good idea is a separate discussion.

Character type char to be signed or unsigned, depending on the platform and compiler.

According to this reference link :

The C and C++ standards allows the character type char to be signed or unsigned, depending on the platform and compiler.

Most systems, including x86 GNU/Linux and Microsoft Windows, use signed char,

but those based on PowerPC and ARM processors typically use unsigned char.(29)

This can lead to unexpected results when porting programs between platforms which have different defaults for the type of char.

GCC provides the options -fsigned-char and -funsigned-char to set the default type of char.

On x86-64 Linux at least, it's defined by the x86-64 System V psABI

Other platforms will have similar ABI standards documents that specify the rules that let different C compilers agree with each other on calling conventions, struct layouts, and stuff like that. (See the x86 tag wiki for links to other x86 ABI docs, or other places for other architectures. Most non-x86 architectures have only one or two standard ABIs.)

From the x86-64 SysV ABI: Figure 3.1: Scalar Types

   C            sizeof      Alignment       AMD64
                            (bytes)         Architecture

_Bool*          1             1              boolean
-----------------------------------------------------------
char            1             1              signed byte
signed char
---------------------------------------------------------
unsigned char   1             1              unsigned byte
----------------------------------------------------------
...
-----------------------------------------------------------
int             4             4              signed fourbyte
signed int
enum***
-----------------------------------------------------------
unsigned int    4             4              unsigned fourbyte
--------------------------------------------------------------
...

* This type is called bool in C++.

*** C++ and some implementations of C permit enums larger than an int. The underlying type is bumped to an unsigned int, long int or unsigned long int, in that order.

Whether char is signed or not does actually directly affect the calling convention in this case, because of a currently-undocumented requirement which clang relies on: narrow types are sign or zero-extended to 32 bit when passed as function args, according to the callee prototype.

So for int foo(char c) { return c; }, clang will rely on the caller to have sign-extended the arg. (code + asm for this and a caller on Godbolt).

gcc:
    movsx   eax, dil       # sign-extend low byte of first arg reg into eax
    ret

clang:
    mov     eax, edi       # copy whole 32-bit reg
    ret

Even apart from the calling convention, C compilers have to agree so they compile inline functions in a .h the same way.

If (int)(char)x behaved differently in different compilers for the same platform, they wouldn't really be compatible.

gcc has two compile time options that control the behaviour of char:

-funsigned-char
-fsigned-char

It is not recommended to use any of these options unless you know exactly what you are doing.

The default is platform-dependent and is fixed when gcc itself is built. It is chosen for best compatibility with other tools that exist on that platform.

Source.

One important practical note is that the type of a UTF-8 string literal, such as u8"...", is an array of char, and it must be stored in UTF-8 format. Characters in the basic set are guaranteed to be equivalent to positive integers. However,

If any other character is stored in a char object, the resulting value is implementation-defined but shall be within the range of values that can be represented in that type.

(In C++, the type of the UTF-8 string constant is const char [] and it is not specified whether characters outside the basic set have numeric representations at all.)

Therefore, if your program needs to twiddle the bits of a UTF-8 string, you would need to use unsigned char. Otherwise, any code that checks whether the bytes of a UTF-8 string are in a certain range will not be portable.

It’s better to explicitly cast to unsigned char* than to write char and expect the programmer to compile with the right settings to configure that as unsigned char. However, you might use a static_assert() to test whether the range of char includes all the numbers from 0 to 255.