Universal character name from token concatenation

You create a universal character name by joining tokens with ## operator

Description

This defect occurs when two preprocessing tokens joined with a ## operator create a universal character name. A universal character name begins with \u or \U followed by hexadecimal digits. It represents a character not found in the basic character set.

For instance, you form the character \u0401 by joining two tokens:

#define assign(uc1, uc2, val) uc1##uc2 = val
...
assign(\u04, 01, 4);

Risk

The C11 Standard (Sec. 5.1.1.2) states that if a universal character name is formed by token concatenation, the behavior is undefined.

Fix

Use the universal character name directly instead of producing it through token concatenation.

Examples

Universal Character Name from Token Concatenation

#define assign(uc1, uc2, val) uc1##uc2 = val

int func(void) {
    int \u0401 = 0;
    assign(\u04, 01, 4); 
    return \u0401;
}

In this example, the assign macro, when expanded, joins the two tokens \u04 and 01 to form the universal character name \u0401.

Correction — Use Universal Character Name Directly

One possible correction is to use the universal character name \u0401 directly. The correction redefines the assign macro so that it does not join tokens.

#define assign(ucn, val) ucn = val

int func(void) {
    int \u0401 = 0;
    assign(\u0401, 4); 
    return \u0401;
}

Result Information

Group: Programming

Language: C | C++

Default: On for handwritten code, off for generated code

Command-Line Syntax: PRE_UCNAME_JOIN_TOKENS

Impact: Low