The correct syntax to call the function passing a and its size would be as shown below. There is no need to have the square brackets [] when passing a to the function.
Additionally, for passing the size of the array as the second argument, you can either use std::size which is available with C++17 or use the expression sizeof a / sizeof a[0] also shown below.
//----------v---------------->no need for the square brackets []
bubbleSort(a, std::size(a)); //use std::size(a) with C++17
Or
bubbleSort(a, sizeof a / sizeof a[0]); //works with all C++ versions
When you pass an array as an argument into a function, you simply pass it as if it were a variable. In this case, simply just a would do. This is because arrays "decay" into pointers so a is a "pointer" to your array.
发布评论
评论（3）
正确调用函数传递
a及其大小的正确语法将如下所示。当将a传递给该功能时，无需让Square Brackets[]。此外，要将数组的大小作为第二个参数，您可以使用
std :: size，它可以使用c ++ 17使用，也可以使用Expressiona / sizeof a [0] < / code>的尺寸也如下所示。或
工作demo
The correct syntax to call the function passing
aand its size would be as shown below. There is no need to have the square brackets[]when passingato the function.Additionally, for passing the size of the array as the second argument, you can either use
std::sizewhich is available withC++17or use the expressionsizeof a / sizeof a[0]also shown below.Or
Working demo
当您将数组作为参数传递到函数中时，您只需将其传递，就好像它是变量一样。在这种情况下，只有
a就可以做到。这是因为 arrays”> arrays“ decay'to pointers so代码>是您数组的“指针”。此外，我建议除
sizeof（a [0]）以 sizeof函数返回字节中的大小When you pass an array as an argument into a function, you simply pass it as if it were a variable. In this case, simply just
awould do. This is because arrays "decay" into pointers soais a "pointer" to your array.In addition, I recommend dividing by the
sizeof(a[0])to get the full length as the sizeof function returns the size in bytes您可以传递对数组的引用，而不是通过指针和长度。
您也不需要
[]即可为a;Rather than pass a pointer and a length, you could pass a reference to the array.
You also don't need
[]to namea;