Question

我在这里问的问题很容易解决。代码工作正常，但这个警告让我烦恼！

//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];

Answer 1

随着amine.ahd的延续，mCount需要成为一个char指针。

x = [1,2,3]
y = [0,*x,4,5]

插入几行代码进行测试。虽然它印有白色字符，但似乎有效。更重要的是，没有编译警告。

Answer 2

ZeroA，因此有一个指向char元素数组的指针。 &ZeroA保存指向数组ZeroA的指针的地址，因此要保留它，您需要char **

在您的示例中执行此操作的正确方法如下： char *XYZ[11]={ZeroA,OneA,TwoA,ThreeA,FourA,FiveA,SixA,SevenA,EightA,NineA,TenA};

指针数组中的警告“从不兼容的指针类型初始化”

2 个答案: