将指针传递给函数? [英] Passing Pointers to Functions?

问题描述

#include 无效交换(int *a，int *b){int *k = a;a = b;b = k;}int main() {整数阿尔法= 5;整数β = 34；交换(阿尔法，贝塔)；std::cout <<阿尔法<

第一个问题:我如何将值传递给具有指针作为传递参数的函数.

第二个问题:如何交换.它交换指针.但是当它出来时，alpha 和 beta 的值不应该仍然保持不变.

第三个问题:为什么当我传递指针时该函数不起作用，而当我传递常规变量时该函数起作用.

第四个问题:在函数void swap(int *a, int* b)中是int *a和int *b引用?

解决方案

这将是最好的例子，说明为什么我们不应该在全局级别包含命名空间 std...

swap (alpha,beta);

会调用 std::swap 而不是你自己的 swap.

<块引用>

第三个问题:为什么当我传递指针时该函数不起作用，而当我传递常规变量时该函数起作用

因为您的 swap 没有正确编写.应该是:

void swap (int *a, int *b) {//最好你还应该检查指针是否不为 NULL.内部温度；温度 = *a;*a = *b;*b = 温度；}

<块引用>

第四个问题:在函数void swap(int *a, int* b)中是int *a和int *b引用?

引用的语法是:

void swap ( int& a, int& b );

所以不，他们不是.它们是指针.

#include <iostream>

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

int main() {
    int alpha = 5;
    int beta = 34;

    swap (alpha,beta);

    std::cout << alpha << std::endl;
    std::cout << beta << std::endl;

    int *ab = new int();
    *ab = 34;
    int *cd = new int();
    *cd =64;

    swap(ab,cd);

    std::cout << *ab << std::endl;
    std::cout << *cd << std::endl;
}

First Question: How am I able to pass values to a function which has pointers as passing parameters.

Second Question: How is it able to swap. It swaps the pointers. But when it comes out shouldn't the values of alpha and beta still remain same.

Third Question: Why doesn't the function work when I pass pointers but works when I pass regular variables.

Fourth Question: In the function void swap(int *a, int* b) are int *a and int *b references?

解决方案

And that would serve the best examples for why we should not include namespace std at global level…

swap (alpha,beta);

would call std::swap rather than your own swap.

Third Question: Why doesn't the function work when I pass pointers but works when I pass regular variables

Because your swap isn't properly written. It should be:

void swap (int *a, int *b) {
    // Preferably you should also check if pointers are not NULL.
    int temp;
    temp = *a;
    *a = *b;
    *b = temp;
}

Fourth Question: In the function void swap(int *a, int* b) are int *a and int *b references?

The syntax of reference is:

void swap ( int& a, int& b );

So no, they aren't. They are pointers.

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