c - ポインターの配列の警告「互換性のないポインター型からの初期化」

Question

ここで私が求めていることは、非常に簡単に解決できます。コードは正常に動作しますが、この警告は私を悩ませています!

//initailize array elements
    char ZeroA[6]   = {0xC0,0x07,0x40,0x04,0xC0,0x07,};
    char OneA[6]    = {0x80, 0x04, 0xC0, 0x07, 0x00, 0x04,};
    char TwoA[6]    = {0x40, 0x07, 0x40, 0x05, 0xC0, 0x05,};
    char ThreeA[6]  = {0x40, 0x05, 0x40, 0x05, 0xC0, 0x07,};
    char FourA[6]   = {0x80, 0x03, 0x00, 0x02, 0x80, 0x07,};
    char FiveA[6]   = {0xC0, 0x05, 0x40, 0x05, 0x40, 0x07,};
    char SixA[6]    = {0xC0,0x05,0x40,0x05,0x40,0x07,}; 
    char SevenA[6]  = {0x40,0x04,0x40,0x03,0xC0,0x00,};
    char EightA[6]  = {0xC0,0x07,0x40,0x05,0xC0,0x07,};
    char NineA[6]   = {0xC0,0x05,0x40,0x05,0xC0,0x07,};
    char TenA[6]    = {0x00,0x01,0x80,0x03,0x00,0x01,};

int *mCount;     //address holder
char var = 4;    //Just random number for illustration

int *XYZ[11]={&ZeroA,&OneA,&TwoA,&ThreeA,&FourA,&FiveA,&SixA,&SevenA,&EightA,&NineA,&TenA};

mCount = XYZ[Var];