c - Returning an array of char pointers -

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I am trying to return an array of char * in a function. The code is simplified in the case of a test, in which a four array is cloned, in which instead of changing the letters indicates those characters.

/ * * code.c * / #include & lt; Stdio h & gt; Four * mapcountercopy (four CEN []); Int main () {char cTest [] = {'c', 't', 's', 't'}; Four * CPT [] = MakePaperCopy (CTT); Printf ("% p% c", CPTist, * CPTST); Fflush (stdout); Return 0; } Char * makePointerCopy (char cIn []) {char * cOut [sizeof (cin) / sizeof (Cen [0])]; Int iCntr; (ICntr + 0; iCntr & lt; for sizeof (cin) / sizeof (Cen [0]); iCntr ++) [iCntr] = cIn + iCntr; Returning cottage; }

On the one hand, what the compiler has to say about this code snippet:

Invalid initial (but char * CPT [] = MakePaperCopy (CTT); )

Why does this happen?

Because makes a copy of a copypoint a four * Is not a four * [] .

You should be able to change that line to: 

  char * cPTest = makePointerCopy (cTest);

More specifically, instead of a few things about the reasons, you receive this error message, it is that those who started the array need to stabilize in compile-time .

Even if the declaration file is not in the scope, it will be illegal in both C90 and C99. The C90 automatically creates the initial demand for compilation-time for ARA and registration. And for both C90 and C99, a letter array is required, in which a) a string literally, or b) a brace-attached initial list.

However, this type of mismatch is a real problem.


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