'setfill' is not a member of 'std'
时间: 2023-12-25 08:03:19 浏览: 30
This error message usually occurs when the code is trying to use the setfill() function from the iomanip header file but it is not properly including the appropriate standard library namespace.
To fix this error, you can add the following line at the beginning of your code:
```
#include <iomanip>
using namespace std;
```
This will ensure that the code has access to the setfill() function from the iomanip header file and also properly includes the standard library namespace.
相关问题
std::setfill
std::setfill is a manipulator function in C++ that sets the fill character for output streams. It is used in conjunction with other output formatting functions like std::setw to specify the character that should be used to fill the empty spaces when values are printed with fixed width.
Here is an example of how std::setfill can be used:
```cpp
#include <iostream>
#include <iomanip>
int main() {
int number = 123;
std::cout << "Default fill: " << std::setw(10) << number << std::endl;
std::cout << "Set fill to '-': " << std::setfill('-') << std::setw(10) << number << std::endl;
std::cout << "Set fill to '*': " << std::setfill('*') << std::setw(10) << number << std::endl;
return 0;
}
```
Output:
```
Default fill: 123
Set fill to '-': ----123
Set fill to '*': *******123
```
In the example, std::setfill('-') and std::setfill('*') are used to set the fill character to '-' and '*', respectively. The std::setw(10) sets the width of the output to 10 characters.
javafx The method setFill(Paint) in the type Shape is not applicable for the arguments (Color)
这个错误是因为在使用`Shape`类的`setFill`方法时,传入的参数类型不正确。`setFill`方法的参数应该是一个`Paint`类型的对象,而你传入的是一个`Color`类型的对象。
解决这个错误的方法是,将`Color`类型的对象转换成`Paint`类型的对象。可以使用`Color`类的`valueOf`方法来将`Color`对象转换成`Paint`对象,具体代码如下:
```
shape.setFill(Color.valueOf("#FF0000"));
```
其中,`#FF0000`是一个十六进制的颜色值,你可以根据你需要设置的颜色来修改这个值。