Question

分配字符错误：

structplusclass.cpp: In constructor ‘Soldado::Soldado(char, unsigned int, char, char, char)’:
structplusclass.cpp:25:27: error: invalid conversion from ‘char’ to ‘const char*’ [-fpermissive]
  strcpy(mTodos.modelo, mol);
                           ^
In file included from /usr/include/c++/5/cstring:42:0,
                 from structplusclass.cpp:2:
/usr/include/string.h:125:14: note:   initializing argument 2 of ‘char* strcpy(char*, const char*)’
 extern char *strcpy (char *__restrict __dest, const char *__restrict __src)
              ^
structplusclass.cpp:27:29: error: invalid conversion from ‘char’ to ‘const char*’ [-fpermissive]
  strcpy(mTodos.material, mat);
                             ^
In file included from /usr/include/c++/5/cstring:42:0,
                 from structplusclass.cpp:2:
/usr/include/string.h:125:14: note:   initializing argument 2 of ‘char* strcpy(char*, const char*)’
 extern char *strcpy (char *__restrict __dest, const char *__restrict __src)
              ^
structplusclass.cpp:28:18: error: invalid conversion from ‘char’ to ‘const char*’ [-fpermissive]
  strcpy(mNome, ns);

代码：

#include <iostream>
#include <cstring>
#define SIZE 20

using namespace std;

struct faca {
    char modelo[SIZE];
    unsigned int peso;
    char material[15];
};

class Soldado {
    public:
        Soldado (char ns, unsigned int p, char mat, char arm, char mol);
        ~Soldado ();
        void Imprime();
    protected:
        faca mTodos;
        char mNome[SIZE];
        char mArmapri[SIZE];
};

Soldado::Soldado (char ns, unsigned int p, char mat, char arm, char mol) {
    strcpy(mTodos.modelo, mol);
    mTodos.peso = p;
    strcpy(mTodos.material, mat);
    strcpy(mNome, ns);
    strcpy(mArmapri, arm);
}

void Soldado::Imprime () {
    cout << "Soldado : " << mNome << ", " << "firearm : "
     << mArmapri << endl;
    cout << endl << " Faca : " << mTodos.modelo << ", " 
    << "lenght : " << mTodos.peso << ", " << "Material : "
    << mTodos.material << endl;
}

Soldado::~Soldado () {
    mTodos.peso = 0;
}

int main() {
    char names1[SIZE], fire[SIZE];
    char  names2[SIZE], mat[15];
    unsigned int eight;
    cout << "nome Soldado, nome arma de fogo " << endl;
    cin >> names1 >> fire;
    cout << "modelo faca e material " << endl;
    cin >> mat >> names2;
    cout << "peso" << endl;
    cin >> eight;
    Soldado brazil(names1, eight, mat, fire, names2);
    brazil.Imprime();



    return 0;
}

Answer 1

Soldado :: Soldado（）使用了错误的参数。

char names1[SIZE], fire[SIZE];

cout << "nome Soldado, nome arma de fogo " << endl;
cin >> names1 >> fire;

Soldado brazil(names1, ..., ..., fire, ...);

此处names1和fire都声明为char[SIZE]类型。这意味着有SIZE个类型char的项目在内存中排成一行（具体来说，在堆栈中，但这对于了解这一点并不重要）。但是然后你将这些传递给你的构造函数，你的构造函数期望这样：

Soldado::Soldado (char ns, ..., ..., char arm, ...) { ... }

此处，ns和arm被声明为char类型，这是char类型的单个项目，而不是像上面那样的列表。

解决此问题的最佳方法是将ns和arm（以及期待单词的其他参数）从char更改为char[SIZE] 。这将纠正您的编译器错误，并使您的代码正常工作。

然而，还有一点点。你的构造函数有几行代码：

strcpy(mTodos.modelo, mol);

只要您的用户永远不会输入超过19个字母的名称（留下一个char来标记结尾），这些将完美无缺。但是如果他们输入更长的名字，你的代码就会愉快地写出你为它保留的内存的末尾。你应该用`strncpy（）函数来保护你的记忆：

strncpy(mTodos.modelo, mol, SIZE);
mTodos.modelo[SIZE - 1] = '\0';

不幸的是，你需要第二行来确保字符串以null字符结尾，因为strncpy()不会一直为你做。

C ++错误字符赋值

1 个答案: