Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
276 views
in Technique[技术] by (71.8m points)

Confusion about pointers and references in C++

I have a bunch of code like this:

#include <iostream>
using namespace std;

void swap(int *a, int *b) {
    int temp = *a;
    *a = *b;
    *b = temp;
}

int main() {
    int a;
    int b;
    a = 7;
    b = 5;
    swap(a, b);
    cout << a << b;

    return 0;
}

This code does the swapping process as what I exactly wanted to swap 2 numbers, but when I want two numbers from the user as follows;

int a;
int b;
cin >> a;
cin >> b;
swap(a, b);
cout << a << b;

the compiler gives me an error about int to int* error which is as expected. Why does the first code do the right swapping although I didn't use the method with & operator?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

In the first example, std::swap is called, because of your using namespace std. The second example is exactly the same as the first one, so you might have no using.

Anyway, if you rename your function to my_swap or something like that (and change every occurence), then the first code shouldn't work, as expected. Or, remove the using namespace std and call std::cin and std::cout explicitly. I would recommend the second option.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...