C语言 从不兼容的指针类型初始化 [默认启用] - 有什么问题？
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Initialization from incompatible pointer type [enabled by default] - whats wrong?
提问by user3139356
I have this warning in my code:
我的代码中有这个警告：
initialization from incompatible pointer type [enabled by default]
从不兼容的指针类型初始化[默认启用]
code is as follows:
代码如下：
/*
* Method return some number :-)
* It is TEST METHOD
*/
int check(char array[]) {
int num = 0; // my number
char **elem_p = array; // test
while (*elem_p) { // it is while
num++;
elem_p++;
}
return num; // my return
}
What is wrong? How can I fix this? Thank you. Test method is not relevant, is a sample.
怎么了？我怎样才能解决这个问题？谢谢你。测试方法不相关，是一个样本。
回答by haccks
What is wrong?
怎么了？
arrayis of type char *(pointer to char) but you are using it to initialize char **(pointer to pointer to char) type variable elem_p.
array是类型char *（指向 的指针char），但您使用它来初始化char **（指向指向 的指针的指针char）类型变量elem_p。
How can I fix this?
我怎样才能解决这个问题？
Make pointer and pointee (object to be pointed) compatible to each other. Declare elem_pas char *;
使指针和指针对象（要指向的对象）相互兼容。声明elem_p为char *;
char *elem_p = array;
回答by haccks
char array[]will be rewritten as char* array. You either need char* array[]or char** array, but there is no difference in a function declaration. You probably intended to only have one asterisk, as haccks points out.
char array[]将被重写为char* array. 您需要char* array[]或char** array，但函数声明没有区别。正如 hackks 指出的那样，您可能打算只使用一个星号。
char* array[]) {
int num = 0; // my number
char **elem_p = array; // test
回答by Thiago Hirai
I think you meant char *elem_p = array;. For more information, see 'Pointers and Arrays' at http://www.cplusplus.com/doc/tutorial/pointers/.
我想你的意思是char *elem_p = array;。有关更多信息，请参阅http://www.cplusplus.com/doc/tutorial/pointers/ 上的“指针和数组” 。
