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C++ passing arrays to a method problem

开发者 https://www.devze.com 2022-12-17 08:55 出处:网络
Hey there I have this function: void vdCleanTheArray(int (*Gen_Clean)[25]) { } I wanted to know what kind of array does it accepts and clear it.

Hey there I have this function:

void vdCleanTheArray(int (*Gen_Clean)[25])
{

}

I wanted to know what kind of array does it accepts and clear it.

Sorry, I 开发者_StackOverflowhave a little knowledge in c++, someone just asked me.

Thanks!


It accepts an array declared as

int array[25];

which should be passed "by pointer", i.e. by applying the & operator to the array

vdCleanTheArray(&array);

As for how to clean it... As I understand, it is the above function that should clean it and you are supposed to write it, right? Well, inside the function you access the array by using the dereference operator * as in

(*Gen_Clean)[i]

and you just "clear" it as you would clear any other array. Depends on what you mean by "clear". Fill with zeros? If so, then just write a cycle from 0 to 24 (or to sizeof *Gen_Clean / sizeof **Gen_Clean) that would set each element to zero.


void vdCleanTheArray(int (*Gen_Clean)[25]) can be described as 'a function taking a pointer to an array of 25 integers and returning nothing.'

Assuming by 'clear' you mean set all values to 0, you could do it with something like this:

for (int i = 0; i < 25; i++) {
  (*Gen_Clean)[i] = 0;
}

Note that in C++, C-style arrays cannot be resized. It's better to use std::vector if you need a resizable array, or std::array if you need a fixed-size array.

If you're new to C++, you probably shouldn't be around a function like this. The way the first parameter is typed isn't something you'll regularly (or should ever) see in C++.

To those wondering about the 'int (*Gen_Clean)[25]' syntax, I believe it's typed like this to enforce that the pointer being passed actually points to an array of 25 ints, as normally C++ would degrade a parameter 'int *Gen_Clean[25]' into a pointer-to-a-pointer, rather than a pointer-to-an-array.

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