* 類型的 C++ 參數與類型 ** 的參數不兼容

Question

您好，我是 C++ 新手，無法弄清楚為什么我的代碼會這樣做。 我已經在互聯網上搜索並找不到我需要的解決方案。 我感謝所有的幫助。 給我帶來問題的那一行是當我調用該函數時。 根據visual studios，它指出“'char*' 類型的參數與'char**' 類型的參數不兼容”。 它指的是 newArr。

#include <iostream>
#include <string>
#include <stdio.h>
#include <ctype.h>
#include "stdafx.h"

using namespace std;

bool isPalindrome(char *newArr[], int);

//int i = 0;
//char phrase; 
//char c;
bool palindrome;
bool tOf;
int numb;
char c;

const int length = 80; //const so that it can't be changed
char inarr[length]; //array set to a const length of 80
char newArr[length]; //array that will have no spaces

string str;

int main()
{

    cout << "This program tests if a word/phrase is palindrome.\n\n";
    cout << "Please enter your phrase (just letters and blanks, 
please):\n\n";

    cin.getline(inarr, length);
    //cout << array; //spits out the array
    str = inarr; //turn into string
    numb = str.length();
    //cout << numb << "\n"; //how many characters in array

    for (int i = 0; i < (numb / 2) + 1; i++)
    {

        for (int j = 0; j < (numb / 2) + 1; j++)
        {
            newArr[j] = inarr[i];   //from old array to new array   

            c = newArr[j];
            newArr[j] = toupper(c); //change to all upper case

                                //cout << newArr[j];
            i += 2; //goes to every other index to skip space in string
        }

    }

    tOf = isPalindrome(newArr, numb); //calling of function

    if (tOf == true) //the response to true or false
    {
        cout << "\nYes, the phrase is a palindrome!";
    }
    else
    {
        cout << "\nNo, the phrase is not a palindrome!";
    }

    return 0;

}

bool isPalindrome(char *newArr[], int numb) //function to determine true or 
false
{
    for (int i = 0; i < (numb / 2) + 1; i++) //within the array...
    {
        if (newArr[i] != newArr[(numb / 2) - i]) //if first index != last 
and etc (iterates)
        {
            palindrome = false;
        }
        else
        {
            palindrome = true;
        }
    }
    return palindrome;
}

Answer 1

您正在嘗試將newArr （一個char * ）傳遞給isPalindrome() （它需要一個char ** ）。 這就是“'char*' 類型的參數與'char**' 類型的參數不兼容”的意思。

要解決這個問題，只需傳入一個char ** ； 你可以通過傳入newArr的地址而不是newArr本身來做到這一點：

tOf = isPalindrome(&newArr, numb); //calling of function

Answer 2

Brief將函數的函數簽名（包括定義和聲明）更改為

bool isPalindrome(char* newArr, int numb);

稱之為tOf = isPalindrome(newArr, numb);

詳情

如果您調用isPalindrome(newArr, numb) 。 您正在傳遞第一個元素的地址 &newArr[0] 。 所以你在函數定義中應該可以選擇元素的地址。 因此 *newArr

此外，您的函數將使用數組算術驗證詳細信息。 沒關系。

輸出

$ ./a.out
This program tests if a word/phrase is palindrome.

Please enter your phrase (just letters and blanks, please):

Palindrome

No, the phrase is not a palindrome!
$ ./a.out
This program tests if a word/phrase is palindrome.

Please enter your phrase (just letters and blanks, please):

YeseY

Yes, the phrase is a palindrome!

$

Answer 3

您好，我是C ++的新手，無法弄清楚為什么我的代碼這樣做了。 我在互聯網上進行搜索，找不到所需的解決方案。 我感謝所有的幫助。 給我問題的那一行是當我調用該函數時。 根據視覺工作室，它指出“類型為'char *'的參數與類型為'char **'的參數不兼容”。 它指的是newArr。

#include <iostream>
#include <string>
#include <stdio.h>
#include <ctype.h>
#include "stdafx.h"

using namespace std;

bool isPalindrome(char *newArr[], int);

//int i = 0;
//char phrase; 
//char c;
bool palindrome;
bool tOf;
int numb;
char c;

const int length = 80; //const so that it can't be changed
char inarr[length]; //array set to a const length of 80
char newArr[length]; //array that will have no spaces

string str;

int main()
{

    cout << "This program tests if a word/phrase is palindrome.\n\n";
    cout << "Please enter your phrase (just letters and blanks, 
please):\n\n";

    cin.getline(inarr, length);
    //cout << array; //spits out the array
    str = inarr; //turn into string
    numb = str.length();
    //cout << numb << "\n"; //how many characters in array

    for (int i = 0; i < (numb / 2) + 1; i++)
    {

        for (int j = 0; j < (numb / 2) + 1; j++)
        {
            newArr[j] = inarr[i];   //from old array to new array   

            c = newArr[j];
            newArr[j] = toupper(c); //change to all upper case

                                //cout << newArr[j];
            i += 2; //goes to every other index to skip space in string
        }

    }

    tOf = isPalindrome(newArr, numb); //calling of function

    if (tOf == true) //the response to true or false
    {
        cout << "\nYes, the phrase is a palindrome!";
    }
    else
    {
        cout << "\nNo, the phrase is not a palindrome!";
    }

    return 0;

}

bool isPalindrome(char *newArr[], int numb) //function to determine true or 
false
{
    for (int i = 0; i < (numb / 2) + 1; i++) //within the array...
    {
        if (newArr[i] != newArr[(numb / 2) - i]) //if first index != last 
and etc (iterates)
        {
            palindrome = false;
        }
        else
        {
            palindrome = true;
        }
    }
    return palindrome;
}