预计“uint32_t的”,但参数的类型为“* uint32_t的” [英] expected ‘uint32_t’ but argument is of type ‘uint32_t *’
问题描述
我用C新,试图调用功能,但它给我的错误,我无法理解为什么
INT set_price(安培;是亮度> type.name);
块引用>它返回我
预计uint32_t的,但参数的类型为* uint32_t的。警告:传递参数'诠释set_price'将指针整数,未作投
当指针
house_list *颜色= NULL;
块引用>和
名称在结构定义为
uint32_t的名称;
块引用>原来的函数接受
INT set_price(uint32_t的名字){搜索结果/的做的东西在这里的/结果}
块引用>
块引用>我该怎么办错了吗?如果struct成员,名称定义为uint32_t的,我定义一个指针的颜色,比我相信,我需要使用&安培;之前是亮度>类型和使用点之前的名字是不是?
感谢您
解决方案set_price(安培;是亮度> type.name);
删除&放大器;你会没事的。
set_price(是亮度> type.name);
set_price
要求整数作为参数,而不是一个指针为整数。我建议你应该阅读一个好的C书。
I am new in C, trying to call a function, but it gives me error that I can not understand why
int set_price(&colour->type.name);
it returns me
expected ‘uint32_t’ but argument is of type ‘uint32_t *’. warning: passing argument ‘int set_price’ makes integer from pointer without a cast
where the pointer is
house_list *colour = NULL;
and name is defined in struct as
uint32_t name;
the original function accept
int set_price(uint32_t name)
{
/do something here/
}
what do I do wrong? If in the struct member, name is defined as uint32_t, and I defined a pointer colour, than I believe that I need to use & before colour->type and use dot before name isn't it?
Thank you
解决方案set_price(&colour->type.name);
remove the & and you'll be fine
set_price(colour->type.name);
set_price
expects an integer as an argument, not a pointer to integer.I suggest that you should read a good C book.
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