"c" is a character literal having the array type char[2]. It is represented in memory like array { 'c', '\0' }. Used as an initializer expression it is implicitly converted to a pointer to its first element of the type char *. So in this declaration
char z = "c";
that is equivalent to the declaration
char z = &"c"[0];
you are trying to initialize the object z having the type char with a pointer of the type char *. Instead of the string literal you need to use an integer character constant like
char z = 'c';
In calls of printf like this
printf("%d\n %lf\n %c\n", m, fx, z);
the length modifier l in this format %lf is redundant and has no effect. You may write
printf("%d\n %f\n %c\n", m, fx, z);
Also you should cast pointers to the type void * in calls of printf like this
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“ C”是具有数组类型
char [2]的字符字面字符。它在内存中表示，例如数组{'c'，'\ 0'}。用作初始化器表达式，它被隐式转换为指针，转换为其类型char *的第一个元素。因此，在此声明中，您
要试图初始化对象
z具有类型char的指针的指针chode> char *。您需要使用字符串字符串，您需要使用一个整数字符常数，例如printf这样的呼叫中的length Modifier
l以这种格式％lf是多余的，没有效果。您也可以写作，您应该在
printf的调用中将指针投入到类型void *了
printf的呼叫，似乎代替 意思是这个电话
"c" is a character literal having the array type
char[2]. It is represented in memory like array{ 'c', '\0' }. Used as an initializer expression it is implicitly converted to a pointer to its first element of the typechar *. So in this declarationthat is equivalent to the declaration
you are trying to initialize the object
zhaving the typecharwith a pointer of the typechar *. Instead of the string literal you need to use an integer character constant likeIn calls of
printflike thisthe length modifier
lin this format%lfis redundant and has no effect. You may writeAlso you should cast pointers to the type
void *in calls ofprintflike thisAnd it seems instead of this call of
printfyou mean this call