xcode 被调用的对象“char*”类型不是函数或函数指针
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called object 'char* 'type is not a function or function pointer
提问by Itzik984
I have this function which is supposed to set a certain time format to the given char*:
我有这个函数，它应该为给定的 char* 设置特定的时间格式：
static void timeStamp(char* time)
{
time(&strTime);<---ERROR
curTime = std::localtime(&strTime);
strftime(time, 8, "%H:%M::", curTime);
}
strTime and curTime were declared like this:
strTime 和 curTime 声明如下：
tm* curTime;
time_t strTime;
but for some reason i get:
但出于某种原因，我得到：
called object type 'char*' is not a function or function pointer
on the marked place.
在标记的地方。
any idea why?
知道为什么吗？
im using xCode by the way.
顺便说一下，我正在使用 xCode。
回答by Dancrumb
Your function parameter timeis a pointer to a char.
您的函数参数time是一个指向字符的指针。
However, in your function body, you're trying to treat it if it were a function that you can call. That's what the error is saying...
但是，在您的函数体中，如果它是一个可以调用的函数，您就会尝试处理它。这就是错误所说的......
the object of type
char *is not a function or a function pointer [therefore, I can't call it!]
类型的对象
char *不是函数或函数指针[因此，我不能调用它！]
Essentially, you've hidden the timefunction by having a local variable of the same name. I'd recommend changing your function parameter's name.
本质上，您time通过使用同名的局部变量隐藏了该函数。我建议更改函数参数的名称。
回答by Daniel Fischer
The function parameter
函数参数
static void timeStamp(char* time)
{
time(&strTime);<---ERROR
// ...
}
shadows the time()function. Rename the parameter.
阴影time()功能。重命名参数。
回答by Eight
static void timeStamp(char* time)
here, this char* timeparameter is hidding the function time().you will need to rename it.
在这里，这个char* time参数隐藏了函数。time()你需要重命名它。
回答by Gart
The function timeis also the name of your parameter to timeStampfunction. The compiler tries to call the char* parameter as if it were a function.
该函数 time也是要timeStamp运行的参数的名称。编译器尝试调用 char* 参数，就好像它是一个函数一样。
回答by Richard J. Ross III
You have redefined the symbol time, to fix, try the following:
您已经重新定义了符号time，要修复，请尝试以下操作：
static void timeStamp(char *time)
{
time_t strTime;
{ // brackets here are important!
extern time_t time(time_t *);
time(&strTime);
}
// ...
}
We add brackets here to add additional scope to the function, and then we tell the compiler that hey, in this block, time is not a variable, its a function, and that allows the compiler to work fine with your current variable name, as well.
我们在此处添加括号以向函数添加额外的作用域，然后我们告诉编译器hey, in this block, time is not a variable, its a function，这样编译器也可以使用您当前的变量名正常工作。
Optionally, you can also use the global namespace specifier, like this:
或者，您还可以使用全局命名空间说明符，如下所示：
static void timeStamp(char *time)
{
::time(&strTime);
// ...
}
回答by glglgl
Because timeis a char*which you cannot call - look at your parameter list.
因为time是char*你不能调用的 - 查看你的参数列表。
Just rename that and your problem will be gone.
只需重命名，您的问题就会消失。
