C++ 没有上下文类型信息的重载函数| 无法解析基于转换为“int”类型的重载函数“swap”
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overloaded function with no contextual type information | cannot resolve overloaded function 'swap' based on conversion to type 'int'
提问by Jonathan Dewein
I am trying to write my own bubble sort algorithm as an exercise. I do not understand the two error messages. Can anyone point out the problem with my code?
我正在尝试编写自己的冒泡排序算法作为练习。我不明白这两个错误消息。谁能指出我的代码的问题？
// Bubble sort algorithm
#include <iostream>
#include <iomanip>
using namespace std;
void bubbleSort(int array[], int arraySize); // bubbleSort prototype
int main(void)
{
const int arraySize = 10;
int array[arraySize] = {2,3,6,5,7,8,9,3,7,4};
cout << "Unsorted: ";
for(int i = 0; i < arraySize; ++i)
cout << setw(5) << array[i];
cout << "Sorted: " << bubbleSort(array, arraySize);
}
void bubbleSort(int array[], int arraySize)
{
const int max = arraySize;
int swap = 0;
for(int i = 0; i < max; ++i)
{
if(array[i] > array[i + 1])
{
swap = array[i + 1];
array[i + 1] = array[i];
array[i] = swap;
}
else
break;
}
}
回答by Alexander Kondratskiy
I see that you are using
我看到你正在使用
using namespace std;
So when you type
所以当你输入
array[i] = swap;
The compiler cannot disambiguate whether you are referring to the std::swapfunction or your int swapvariable. In fact it looks like it assumed you were referring to the function and tried to somehow convert it to type int. Try renaming your variable to something else.
编译器无法区分您指的是std::swap函数还是int swap变量。事实上，它似乎假设您指的是该函数并试图以某种方式将其转换为 type int。尝试将您的变量重命名为其他名称。
In general, try to stay away from usingdirectives, to avoid name collisions like this.
一般来说，尽量远离using指令，以避免像这样的名称冲突。
回答by Nawaz
array[i] = swap;
This line is causing problem. It is better to change the name of swaplocal variable, as there exists already a function with same name, in stdnamespace which is brought into scope by the line using namespace std;which is to be avoided, anyway.
这条线引起了问题。最好更改swap局部变量的名称，因为在std命名空间中已经存在一个具有相同名称的函数using namespace std;，无论如何，该函数被要避免的行带入范围。
I would also suggest you to declare the variable, inside the if-block where it is actuallyused:
我还建议您在实际使用它的 if 块内声明变量：
if(array[i] > array[i + 1])
{
//declare temp here where it is actually used!
int temp = array[i + 1];
array[i + 1] = array[i];
array[i] = temp;
}
Best practice: reduce the scope local variables by delaying their declarations, which means declare them where they are actually used. Do not declare them in the beginning of the function.
最佳实践：通过延迟声明来减少局部变量的范围，这意味着在实际使用它们的地方声明它们。不要在函数的开头声明它们。
Another way to fix the problem in your code is to give the compiler a contextwhich you can by doing this (though I wouldn't suggest this solution; it is just for you to know):
解决代码中问题的另一种方法是为编译器提供一个上下文，您可以这样做（尽管我不建议使用此解决方案；它仅供您了解）：
array[i] = (int)swap; //giving compiler contextual type information
When you cast swapto int, the compiler can know that swaprefers to the local variable, not the function which is defined in stdnamespace.
当您强制转换swap为 时int，编译器可以知道它swap指的是局部变量，而不是在std命名空间中定义的函数。
回答by Mahesh
cout << "Sorted: " << bubbleSort(array, arraySize);
The return type of the function is void. There is nothing to print for. If you need to print the sorted array, you need to iterate over the array elements after the function call.
函数的返回类型是void。没有什么可打印的。如果需要打印排序后的数组，则需要在函数调用后遍历数组元素。
