QA·C Help - 0674

	0674	[C] Initializer for pointer is of incompatible type.

Constraint violations		REFERENCE - ISO:C90-6.5.7 Semantics, ISO:C90-6.3.16.1 Simple Assignment - Constraints

This is a constraint error

A pointer has been initialized with an expression of incompatible type. The initializer must be an expression which points to a compatible (possibly less qualified) type.

For example:


/*PRQA S 750,753,759,2213,2216,3122,3132,3210,3205,3408,3447,3623,3625 ++ */
extern void foo(int p);

struct ST1 { int si; char sbuf[10];};
union  UN1 { int ui; int uj;};

extern struct ST1 st1a;
extern union  UN1 un1a;

extern int           *pi;
extern unsigned char *puc;
extern int            gi;
extern char           cbuf[10];

extern void test(void)
{
    char * pc1 = pi;             /* 0674                                          */
    char * pc2 = gi;             /* 0674                                          */
    char * pc3 = st1a;           /* 0674                                          */
    char * pc4 = un1a;           /* 0674                                          */
    char * pc5 = cbuf;           /* OK, array lvalue decays to pointer            */
    char * pc6 = "abc";          /* OK, string literal is of type 'array of char' */
    char * pc7 = puc;            /* 0674                                          */
}

Note that according to ISO:C Standard 'char' and 'unsigned char' (or 'signed char') are distinct types, even though 'char' may be implemented as 'unsigned char' (or 'signed char'). For this reason message 0674 is generated for the last definition in the example above. If your compiler incorrectly allows such a conversion, you can use QA C option -foldplainchar (compiler personality) to mimic this behaviour.

MISRA C:2012 Rules applicable to message 0674:

Rule-1.1 (Required)

The program shall contain no violations of the standard C syntax and constraints, and shall not exceed the implementation's translation limits