如何将指向动态数组的指针传递给void函数？

Question

更新了问题以反映答案和我的研究

感谢卡萨布兰卡和沃利克的回答。成功！！我最终必须编辑我的strcpy和strcat以及更改我的Displayname（char c_fullname）函数并研究更多关于我得到的其他错误（包括编译器和运行时），但现在我有工作代码！感谢大家的帮助和耐心。我最后的问题是我应该把我的函数声明和实现（但不是我的函数调用）写成“Displayname（char * c_fullname）”而不是“Displayname（char * c_fullname）”。简单的错误，但绝对是一个昂贵的错误，因为我的程序没有完全传递字符串。再次感谢everyne的帮助。我知道如果没有它，我会走得这么远。

我正在为可能遇到同样问题的其他人发布我已完成的工作代码：

#include <iostream>
#include <cstring>
using namespace std;

void DisplayName(char* c_fullname);

int main()
{
//Declaration of variables
const int n_SIZE = 25;          //Size declaration for first, middle, and last name arrays
char c_firstname[n_SIZE];       //Firstname array
char c_middlename[n_SIZE];      //middlename array
char c_lastname[n_SIZE];        //lastname array
int n_namesize = 0;             //size declaration for dynamic array
char *c_fullname;               //pointer for fullname. Later will be allocated an array size
bool b_test = true;             //while loop test value
char c_exittest;                //test value for user prompt

//Explanation of program to user
cout <<"This program inputs a user's first, middle, and last name.\nNext, the names are dynamically allocated to a single array\nand printed to the screen" <<endl <<endl;

while (b_test == true)
{
    cout <<"Please enter a first name: ";   //User inputs firstname
        cin >>c_firstname;
    cout <<"Please enter a middle name: ";  //User inputs middlename
        cin >>c_middlename;
    cout <<"Please enter a last name: ";    //User inputs lastname
        cin >>c_lastname;
    cout <<endl;                            //endl for aesetics

    //Testing of stored variables
    /*
    cout <<c_firstname;
    cout <<c_middlename;
    cout <<c_lastname;
    cout <<endl <<endl;

    cout <<strlen( c_firstname ) <<endl; 
    cout <<strlen( c_middlename ) <<endl; 
    cout <<strlen( c_lastname ) <<endl;
    */


    //setting size of n_namesize for length of array name[]
    n_namesize = (strlen(c_firstname) + strlen(c_middlename ) + strlen( c_lastname ) + 3);   //strlen used to determine length of each element + 3 for spaces
    c_fullname = new char[n_namesize];          //allocates memory to dynamic array 

    //Testing of n_namesize
    //cout <<n_namesize <<endl <<endl;

    //Concatenating strings to array fullname[n_namesize]
    strcpy_s(c_fullname, n_namesize, c_firstname );     //copies firstname to fullname
    strcat_s(c_fullname, n_namesize, " ");              //concatenates space to fullname
    strcat_s(c_fullname, n_namesize, c_middlename);     //concatenates middlename to fullname
    strcat_s(c_fullname, n_namesize, " ");              //concatenates space to fullname
    strcat_s(c_fullname, n_namesize, c_lastname);       //concatenates lastname to fullname

    //Testing of fullname[]
    //cout <<"Name: "<<c_fullname <<endl <<endl;

    //Passing pointer of array to function
    DisplayName(c_fullname);

    //Prompt users to enter a new name
    cout <<"Do you wish to enter a different name?" <<endl;
    cout <<"Enter 'y' for yes, or 'n' for no" <<endl;
    cin >>c_exittest;

    //User wants to enter a new name
    if (c_exittest == 'y')
    {
        delete[] c_fullname;    //deallocates memory assigned to array
    }

    //User does not want to enter a new name
    if (c_exittest == 'n')
    {
        b_test = false; //exits while loop
    }
}
}
void DisplayName(char* c_fullname)
{
cout <<"The full name entered is: "<<c_fullname <<endl;
}

嘿大家，

感谢您以前的帮助。我有另一项任务（过期）给我带来了问题。我认为赋值是将动态数组的指针传递给函数，然后函数将显示字符串。我的指示是“将构造的数组传递给一个名为void DisplayName（char * Name）的函数”但是，我不断收到错误。我的代码是：

#include <iostream>
#include <cstring>
using namespace std;

void DisplayName(char *c_fullname);

int main()
{
//Declaration of variables
const int n_SIZE = 25;      //Size declaration for first, middle, and last name arrays
char c_firstname[n_SIZE];       //Firstname array
char c_middlename[n_SIZE];      //middlename array
char c_lastname[n_SIZE];        //lastname array
int n_namesize = 0;          //size declaration for dynamic array
char c_fullname[100];           
     //fullname array. size initialy set to 100
char *c_pointer;                //pointer for dynamic array
bool b_test;                     //while loop test value
char c_exittest;                //test value for user prompt

//Explanation of program to user
cout <<"This program inputs a user's first, middle, and last name.\nNext, the names are dynamically allocated to a single array\nand printed to the screen" <<endl <<endl;

while (b_test = true)
{
    cout <<"Please enter a first name: ";   //User inputs firstname
        cin >>c_firstname;
    cout <<"Please enter a middle name: ";  //User inputs middlename
        cin >>c_middlename;
    cout <<"Please enter a last name: ";    //User inputs lastname
        cin >>c_lastname;
    cout <<endl;                            //endl for aesetics

    //Testing of stored variables
    /*
    cout <<c_firstname;
    cout <<c_middlename;
    cout <<c_lastname;
    cout <<endl <<endl;

    cout <<strlen( c_firstname ) <<endl; 
    cout <<strlen( c_middlename ) <<endl; 
    cout <<strlen( c_lastname ) <<endl;
    */

    //setting size of n_namesize for length of array name[]
    n_namesize = (strlen(c_firstname) + strlen(c_middlename ) + strlen( c_lastname ) + 3);   //strlen used to determine length of each element + 3 for spaces
    c_pointer = new char[n_namesize];                                                       //allocates memory to dynamic array
    c_pointer = c_fullname;                                                                 //sets pointer to dynamic array     

    //Testing of n_namesize
    //cout <<n_namesize <<endl;

    //Concatenating strings to array fullname[n_namesize]
    strcpy_s(c_fullname, c_firstname);      //copies firstname to fullname[]
    strcat_s(c_fullname, " ");              //concatenates space to fullname[]
    strcat_s(c_fullname, c_middlename);     //concatenates middlename to fullname[]
    strcat_s(c_fullname, " ");              //concatenates space to fullname[]
    strcat_s(c_fullname, c_lastname);       //concatenates lastname to fullname[]

    //Testing of fullname[]
    cout <<"Name: "<<c_fullname <<endl <<endl;

    //Passing pointer of array to function
    DisplayName(*c_fullname);

    //Prompt users to enter a new name
    cout <<"Do you wish to enter a different name?" <<endl;
    cout <<"Enter 'y' for yes, or 'n' for no" <<endl;
    cin >>c_exittest;

    //User wants to enter a new name
    if (c_exittest == 'y')
    {
        delete[] &c_fullname;   //deallocates memory assigned to array
    }

    //User does not want to enter a new name
    if (c_exittest == 'n')
    {
        b_test = false; //exits while loop
    }
}
}

void DisplayName(char *c_fullname)
{
cout <<"The full name entered is: "<<&c_fullname <<endl;
}

我的错误是： 错误C2664：'DisplayName'：无法将参数1从'char'转换为'char *'

我试图更改几个部分以使其工作，所有这些都给我带来了其他错误。在这一点上，我真的陷入困境，所以寻找一些建议。 在此先感谢您的帮助，

Kula Dhad

Answer 1

您已使用*和&混合使用。您不需要任何运算符来传递指针，您只需直接传递它。这样：

DisplayName(*c_fullname);

应该是：

DisplayName(c_fullname);

同样，char *已经是一个字符串，因此您可以直接打印它。 &获取的地址不是您想要的地址。这样：

cout <<"The full name entered is: "<<&c_fullname <<endl;

应该是：

cout <<"The full name entered is: "<<c_fullname <<endl;

最后，您应该 NOT delete内存未使用new分配。在您的情况下，c_fullname已在堆栈上分配，因此您应删除此行：

delete[] &c_fullname;

更新：也许我之前不清楚。该函数仍然定义为char *：

void DisplayName(char *c_fullname);

但你只需将其称为：

DisplayName(c_fullname);

  

将动态数组传递给DisplayName（char *）函数以显示和释放动态分配的数组

为此，您只需删除现有的静态数组（char c_fullname[n_SIZE]）并使用new创建数组。在这种情况下，您还需要稍后使用delete：

char *c_fullname;
...
c_fullname = new char[n_SIZE];
...
DisplayName(c_fullname);
...
delete [] c_fullname;

Answer 2

要修复，请更换

DisplayName(*c_fullname);

与

DisplayName(c_fullname);

c_fullname已经是指针类型。将*放在它前面取消引用它提供元素的内容，类型为char。

顺便说一句，删除＆amp; c_fullname-或与c_fullname相关的任何其他内容都会导致令人讨厌的运行时问题。无需删除自动数组。